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Copyright © 2009,2010,2011 Joseph Mack
v20110113, released under GPLv3.
Abstract
Class lessons for a group of 910th graders who have taken my introductory programming course in python and who are working on the electricity course. This material was produced to help the students understand forces and energy.
Material/images from this webpage may be used, as long as credit is given to the author, and the url of this webpage is included as a reference. The diagrams are produced by the GPL'ed package dia.
Table of Contents
Since you don't get this in school nowadays, here's some basic mechanics to bring you up to speed.
Here are Newton's 3 laws of motion (http://en.wikipedia.org/wiki/Newton's_laws_of_motion).
Newton's 2nd law says that to accelerate a body, you must apply a force.
F = ma ; force = mass * acceleration 
Let's say we have a 1kg object. To accelerate this mass by 1m/sec^{2}, we need a force of 1N (newton).
Instead of supplying a force ourselves, we can let gravity do it for us. The acceleration due to gravity is 9.8m/sec^{2}. The symbol for acceleration due to gravity is g. The value of g should be committed to heart by everyone who lives on the surface of the earth. If we let this 1kg mass drop, then from it's acceleration, we know that it's being acted on by a force of 9.8N. Here's the formula for the force due to gravity acting on an object of mass m.
F = mg 
If we allowed a 2kg mass to drop, what force would it be accelerated by ^{[1]} ? The acceleration by gravity is the same no matter what the mass. This was proven by Galileo (http://en.wikipedia.org/wiki/Galileo_Galilei) who dropped unequal masses from the Tower of Pisa, both masses hitting the ground at the same time. (Apparently this story is apochrephal). Why don't a feather and a 1kg weight drop to the floor at the same speed ^{[2]} ?
Note  

You can get the same effect in air by using a massive object to push the air out of the way of the feather (instead of a feather, you can use a small piece of paper). If you orient a calendar or newspaper horizontally and drop it, a piece of paper sitting on top will drop as fast as the calendar. 
As a kid, I remember seeing a movie of a person in a vacuum chamber (and in a pressure suit) dropping different weights (and probably a feather) to show the same thing. Here is the experiment done with a Feather and Ball Bearing Dropped in Vacuum/Air (http://www.youtube.com/watch?v=_XJcZKoL9o). The experiment was repeated on the moon at by Apollo 15 Commander David Scott, who dropped a geologic hammer and a feather on the moon (http://www.youtube.com/watch?v=5C5_dOEyAfk). (Notice how much slower objects drop on the moon compared to earth.)
Work (energy) is force*distance.
W = F*s ; s is the symbol for distance 
Note  

energy is from the Greek  "work within" (see heat work energy http://www.engineeringtoolbox.com/heatworkenergyd_292.html). In the cgs (centimeter, gram, second) system, the unit of work is the erg. 1 erg is the work done moving a force of 1 dyne (accelerating 1 g by 1cm/sec^{2}) through 1cm. An erg is not much; 10^{7}ergs = 1 Joule. 
Work must be done to lift massive objects against gravity. If we lift the 1kg mass, 1m against gravity, we will do 9.8*1=9.8 Joules of work. If we drop the 1kg mass, 1m and let it land on our foot, it will release 9.8J of work into our foot (this energy will be used to rearrange our foot).
Power is work/time.
P = W/t 
If we lift the 1kg mass, 1m against gravity in 1 sec, we work at a rate of 9.8J/sec=9.8w.
The Jin Mao Tower (Golden Prosperity Building) in Shanghai has elevators(lifts) that travel at 9.1m/sec, taking 45secs to reach the top (88th floor). (If the elevators are all glass, this would be a great ride.) Assume the elevators weigh 1 Tonne (1 Mg) (I have no idea, I just made it up). What power is needed to run the elevator ^{[3]} ?
Note  

1HP (horsepower) = 746w. The elevator needs a 120HP motor. 
Standard voltages for motors are 110V, 240V, 415V and 1.5KV. If you were limited to 100A cables (any more current and the cables would be too thick), what voltage would you choose for your elevator motors ^{[4]} ?
Example: a hiker, all up weight (i.e. with pack) of 70kg ascends 1000m in 3hrs. What power level is the hiker operating at ^{[5]} ? The hiker is putting out energy at a rate that would light a small household light bulb. Assuming that the hiker returns that day and does the same amount of work descending (in terms of conservation of energy, the energy that the hiker expended gaining altitude is released on descending. However most hikers think that it's just as much work going down as going up), what is the total amount of energy expended by the hiker during the day ^{[6]} ? How much extra food (calories) should the hiker pack to fuel the day's hike ^{[7]} ? Alternately you could use the conversion between calories and joules: calorie (http://en.wikipedia.org/wiki/Calorie) which I used to calculate the 100W number in the first place. 1 calorie = 4.2kj.
total energy = 2 *686kJ = 2 * 686/4.2 calories = 326 calories 
Going to the pantry, you find that this amount of energy is in a cup (100g) of raisins (raisins are mostly water, so say this is 3050g of sugar). The hiker should pack about 15% more food than they use just hanging around the house.
Note  

This would be a reasonable day's exercise for most hikers. However there's not much energy used in ascending, and the hiker doesn't need to take any more food than they would need just to stay at home for the day. Most of our food goes to maintain body temperature. Admittedly on a hike more energy is used for other things than ascending. You have to balance your body (and pack) and move it up and down a few cm just to walk horizontally. 
Power * time = energy. A unit of energy is the kWhr  the amount of energy supplied to a device that uses 1kW for an hour.
1 kwhr = 1kj/sec * 3600secs = 3600kJ 
Last time I looked, 1kwhr of electricity cost 6c. What would it have cost the hiker to do the hike using electricity, rather than food and muscle, for the day's work ^{[8]} ? In How much coal is required to run a 100watt light bulb 24 hours a day for a year? (http://www.howstuffworks.com/question481.htm) we find that a burning a ton of coal produces 2,460 kWh of electricity. Assume an imperial ton = metric tonne = 1Mg. What weight of coal would been burned to generate the above amount of electricity ^{[9]} ? This is comparable to the weight of raisins eaten. With the heat of combustion of anthracite =50kj/g, sugar 5calories/g = 21kj/g, I would have expected the weight ratio coal/sugar=0.5, i.e. 15g coal. However, by the wonders of modern economics, getting your energy from coal is cheaper than getting from the sun via food. The amount a work a human can do pales by comparison to what can be done by burning tons of coal, that took millions of years to produce.
Look at this derivation
P = W/t = F*s/t = F*v ; power = force * velocity 
Let's try this on the Jin Mao elevators. What power are they putting out?
P = F * v = m * g * v = 10^3kg * 9.8m/sec^2 * 9.1m/sec = 90kW 
... the same answer we got earlier
Let's do some phyisical work. Requirements: set of scales, weights, watch with stopwatch function.
How powerful is a programmer? Find a set of stairs about 5m in height. Strap on a pack with weights (1015kg should do it  to make you do some work and reduce any advantage due to agility); weigh yourself and the pack. Time yourself running up the stairs. Calculate your power output.
Note  

I did this in 8th grade Physics. You can't use any advantage from agility, so you can't start back from the stairs to give you a running start, and you can't do the stairs 2 at a time. Presumably you could try variations on this experiment to see if these restrictions make any difference. 
Here's the results from our class. Height was 18 stairs of 18.75cm = 3.375m.
person all up weight,kg best time,sec work,J power,W power,HP Z 83.75 5.44 2770 509 0.68 J 77.75 5.94 2571 433 0.58 
The SI (System Internationale) is a scheme for describing physical quantities. The units are
Name symbol Quantity Symbol metre m length l kilogram kg mass m second s time t ampere A electric current I kelvin K thermodynamic temperature T candela cd luminous intensity Iv mole mol amount of substance n 
The scheme is dimensionally consistent (see dimensional analysis http://en.wikipedia.org/wiki/Dimensional_analysis). "dimensionally consistent" will be explained below by examples, but for the moment every physical quantity is some combination of these dimensions
If we have an equation, not only do the quantities and units have to agree on both side of the "=" sign, but Newton realised that the dimensions must agree as well (i.e. you can't add a kilogram to a metre). Fourier realised that physical laws, like F = ma, should be independent of the units employed to measure the physical variables.
We're all familiar with dimensions in an informal sense. We know that if our car uses fuel at x miles/gallon and we've done y miles, we expect the y/x will give us a result in gallons.
Let's look at the formula for velocity
v = d / t dimensions of v = l.t^1 
Velocity has the dimensions l.t^{1} (l/t).
What are the dimensions of force? We find
F = ma 
The dimensions of F then are m.l/t^{2} or l.m.t^{2}.
Let's look at the formula for power.
P = F * d / t 
From this we know that the definition of power must have the units of (newton*metre/sec). The dimensions of metre are l, the dimensions of sec are t, Using the dimensions of force that we just found,
P = m.l^2.t^3 
the dimensions of P are m.l.t^{2}.l/t = m.l^{2}.t^{3}.
A little earlier we did this calculation. Let's check it for dimensional correctness.
weight of coal burned = 0.38kwhr * 1 tonne / 2460 kwhr = 154g LHS required dimension = m dimensions of RHS = power.t.m/power.t On the RHS, it doesn't matter what the dimension of power is, as both the numerator and denominator have it and these two terms will cancel. As well the numerator and denominator both have t (which cancel). dimension of RHS = m The equation is dimensionally correct. 
Let's use dimensional analsys to determine the quantities that affect the period of a pendulum. From observation we know that the period of a pendulum is not dependant on the amplitude of the swing (as long as it is "small"), it's not affected by the temperature (as long as the length of the pendulum doesn't vary) or the length of time you observe it (i.e. we can experimentally eliminate the angle/amplitude, temperature and time as variables).
Galileo (http://galileo.rice.edu/sci/instruments/pendulum.html) determined that the period of a pendulum was independant of the amplitude of the swing. According to the story I heard when I was a kid, young Galileo was watching a chandelier in the cathedral in Pisa (of leaning tower fame, and where Galileo was born and lived all his life). Presumably he was in the cathedral to pray, but was distracted and instead timed the period of the chandelier with his pulse (there were no accurate time pieces then, portable or otherwise). He then poked the chandelier (how I don't remember, but he would have had to stop pretending he was praying and stand on something), to find that with its new, larger amplitude, the period of the pendulum was unchanged. (the story I heard gets some confirmation here http://physicshistory.suite101.com/article.cfm/galileos_pendulum and here, except that the wind changed the amplitude http://www.ehow.com/about_5292790_galileospendulum.html )
Galileo, later in life (by then he was blind), went on to develope the pendulum clock, a device accurate enough for astronomical timekeeping (and eventually solving the longitude problem).
Let's use dimensions to find the dependance of the period of a pendulum on
The formula for the period of a pendulum will thus be
t = f(weight of bob, length of pendulum, accel due to gravity) ie t = f(m,l,g) 
where f() is shorthand for a function to be determined. If f() is the sum of a bunch of terms, dimensional analysis says that each term must have the dimension t. Let's lump all these terms together; summing them will result in a dimensionless constant in front of a single function, whose dimension is t.
Note  

We're using the same symbols for mass, length, time and for their dimension(m,l,t). I'll try to differentiate the representation. 
After combining the sums, the only possibilty left is to multiply terms (actually powers of terms) in m,l,g. Mutliplying (powers of) the three terms above gives the period of a pendulum as
mutliplying the three terms above period = m^x.l^y.a^z 
where x,y,z are to be determined. (If x=1, then you'll be dividing, rather than multiplying by m). Dimensional analysis says that the dimensions on both sides of the equation must be the same. Checking dimensions for the period of the pendulum
period = m^x.l^y.a^z the dimension of "a" are l.t^2 thus writing out the dimensions m^0.l^0.t^1 = m^x.l^y.(l^z.t^2z) = m^x.l^(y+z).t^2z separating dimensions (they're independant remember) LHS RHS m^0 m^x l^0 l^(y+z) t^1 t^2z this gives us 3 simultaneous equations x = 0 (1) y + z = 0 (2) 2z = 1 (3) from (3) z = 0.5 (4) substituting (4) in (2) y = 0.5 from (1) x = 0 
the dimensional check shows that the period of a pendulum
Numbers are dimensionless, so dimensional analysis doesn't allow us to determine any constants in the formula for the period of a pendulum. The formula for the period of a pendulum then is (with k a constant whose value must be determined by other means)
t = k * sqrt(l/g) 
As it turns out k=2π and thus t = 2π√(l/g)
Let's check: A pendulum with a period of 1sec: what's the period of the same pendulum
Note  

The word "Jupiter" comes from "Ju" the word for sky (which has become "dieu", "diety") and "pater" (meaning father). Jupiter is the "sky father". (I can't remember where I read this.) 
In studying waves and periodic functions (like alternating current, or electromagnetic radiation), a useful unit is the Hz which is the number of oscillations/sec (also called the frequency). See oscillations (http://en.wikipedia.org/wiki/Oscillation). Here's the frequency of some common oscillations
type of oscillation frequency ringing of the earth following earthquake 1/54mins heart rate 40240beats/min audible sound 2020kHz middle C (physicist's scale) 256Hz A' (musician's scale) 440Hz electromagnetic radiation 1Hz>10^23Hz (1) BC radio (AM) 5501630kHz FM band 88108MHz TV 50800MHz (approx) cell phones 8001900MHz (only slices thereof) wifi, household remote phones, microwave ovens 2.5GHz 
There isn't a symbol for frequency in electrical engineering, but in wave mechanics the symbol is ν ("nu", pronounced "new") What is the dimension of frequency (the number of oscillations is a number; a number is dimensionless) ^{[12]} ?
Another useful parameter to characterise waves is the wavelength (symbol λ); the distance between successive peaks in the oscillations. You can see the distance between crests in ocean waves or ripples in a pond. However most wave motions are not detected as such by human senses. Here's some wavelengths of common waves
wave wavelength ocean waves 100mmany km middle C (physicist's scale) 4' (1.3m) BC radio 200600m SW radio 10010m visible light 390700nm Xrays 0.11nm gamma rays 1pm 
What's the dimension of wavelength ^{[13]} ?
The conversion between frequency and wavelength involves another simple physical quantity with which you will be familiar. Using dimensions, find (or confirm) this physical quantity ^{[14]} .
Dimensional analysis doesn't say anything about numbers (these are dimensionless), so the actual formula relating v,f,λ could have a constant in it (i.e. the formula could be v=5.4*f.λ). A simple thought experiment will show you that the constant is 1. The formula relating these three quantities is v = ν.λ.
For convenience, for simple illustrations, physicists use the velocity of sound as 1024'/sec and the frequency of middle C as 256Hz. What is the wavelength of middle C ^{[15]} ? The velocity of sound in water is 4.3 times higher and in iron is 5 times higher (speed of sound http://en.wikipedia.org/wiki/Speed_of_sound). What is the wavelength of middle C in water and in iron ^{[16]} ?
Transverse (i.e. moving sideways) waves in stretched strings (i.e. strings under tension) are well known; e.g. guitar, piano, violin. The frequency of the note produced is affected by the tension (force) in, the weight and the length of the string.
Note  

The tension is the force you'd have to pull on a string.
Here's one way of setting the tension if you want to test its effect.
One end of the string is fixed, the other runs over a smooth surface
(wheel, or polished curved metal). The end of the string holds a weight

The tension in a piano frame (plate) can exceed 20 tons (see piano http://en.wikipedia.org/wiki/Piano). Tuning a piano is a more difficult than a violin or guitar (where the instrument is stiff and you can tune each string independantly), in a piano, a stiff plate would be too heavy. In a piano, if you let off the tension in a string, the plate expands, raising the pitch of the other strings (but only a little bit). In a piano you can only adjust the strings a small amount at a time.
Using dimensional analysis, determine the effect on the frequency of a string, of the tension, the mass and the length of the string ^{[17]} . Does this equation agree with your prediction on the effect of mass, length and tension on the pitch of a string?
While musicians are interested in the pitch of a musical note, physicists deal with the propagation of waves in media of infinite extent (infinite at least in principle), so they're more interested in the velocity of the wave. You could rerun this derivation using velocity on the left hand side. Instead use your previous result v= ν.λ to determine the formula for the velocity of a transverse wave in a string (you should get the same result).
Note  

What's the wavelength of the wave in a string? Assume that you are only concerned with the fundamental mode of oscillation i.e. the length of the string is half the wavelength. (see the λ/2 wave at standing wave http://en.wikipedia.org/wiki/Standing_wave). The wavelength then is 2*l. 
Since we're only concerned with dimensions, the constant "2" here is irrelevant. Here's the result ^{[18]} . The constant "k" here is 1. So v=1/sqrt(T.M). Check the formula to show that changing the tension and mass/unit length of string have the expected effect on the velocity of the wave.
Note  

Did we use the value of the wavelength (2l)? No; the dimension of l is the same no matter how long it is. 
If you release a ball on top of one of the sides of a bathtub, it speeds up as it looses height. After crossing the bottom of the bathtub, it looses speed as it goes back up the other side, until (at least in a frictionless environment) it returns to the same height on the other side. The same happens to a pendulum or swing.
To lift the mass from the bottom to the top of the side of the bathtub, work had to be done against gravity. By the law of conservation of energy, the mass at height can now do the same amount of work descending to the bottom of the bathtub. The mass at height can potentially (i.e. if you let it go) do work. According to the law of conservation of energy, the mass has potential energy (P.E.).
What is a "law"?
When an observation (e.g. "energy is conserved") is repeatable and useful, but has no explanation and you aren't particularly bothered by the lack of explanation, then to get on with the other things you're more interested in, and get the unexplained observation out of your way, you call the observation a "law". Well known laws are "things fall down; opposites attract". That things fall has been known for so long and observed so consistently, that noone questions its truth.
A "law" is not like an axiom in mathematics, some unfathomable fundamental property of the physical universe, for which there is no hope of further understanding. A law is just something we don't understand, but can describe well (possibly very accurately). As for assigning fundamental properties to the universe; we don't know enough about the universe to be doing this just yet. We used to think matter was earth, air, fire and water; but later came atoms; and then subatomic particles. You don't think we're at the end yet do you?
Note Do not confuse being able to describe something accurately (e.g. the effect of gravity; we can build bridges that stay up; we can land spacecraft on remote objects in the solar system), with understanding it. Noone has a clue why massive objects attract each other. Similarly noone understands why mass has inertia (resistance to accelaration) or why the amount of inertia is equal to the amount of mass (in the formula F=ma, m is actually inertia). The two properties of mass, of mutual attraction and inertia, are well known and well described, but not understood. Laws are completely arbitary, but useful. When they prove to no longer be useful, they are discarded. There's no implication of truth in a law, only usefulness. The very existance of a law declares our complete ignorance of how things work there. At least this is how it is in the physical world. In the legal realm of the US, somehow truths become enshrined in law, indicating that the final resting place for truth is law.
Because the "law of conservation of energy" is a law, I can't justify it and I can't prove it. I'll just be using it.
The ball, after release from the side of the bathtub and rolling across the bottom of the bathtub, can by virtue of its motion, do work. The moving ball can collide with another ball of the same mass, sending the 2nd ball up to the release height on the opposite side of the bathtub. The moving ball at the bottom of the bathtub then has energy, called kinetic energy (K.E.) (http://en.wikipedia.org/wiki/Kinetic_energy).
Note  

The adjective kinetic is from the Greek kinesis meaning motion. The same root produces the word cinema (referring to motion pictures). 
By the law of conservation of energy, the P.E. at the top of the side of the tub is converted into an equal amount of K.E. at the bottom of the tub. The K.E. is then turned into to original amount of P.E. when the ball goes up the other side of the tub. At intermediate points in the ball's trip, the P.E. descreases as it descends (and increases as it ascends), while the opposite is happening to the K.E. Since energy is conserved, the total (K.E+P.E) is constant.
We know how to calculate P.E, it's the energy put into a mass by raising it through height h
P.E. = mgh 
If we see a ball whizzing past us, how do we calculate its K.E.? Let's start using dimensional analysis. All we know about the ball is its mass and velocity. Derive a formula for K.E. using these two measurements. Here's my answer ^{[19]} . We need to find the dimensionless constant k. To do this we need to understand the physics of the situation. We know that after falling h that the mass has gained K.E.=(lost P.E.)=mgh. Since K.E. is determined by velocity, we need to know the velocity of a mass after it falls a distance h. We don't know that yet.
Note  

From dimensional analysis we see there is no dependance of K.E. on direction i.e. the K.E. at any instant is the same whether the mass is moving up, down or sideways. 
By dropping weights from different heights into a block of clay, Gravesande (http://en.wikipedia.org/wiki/Willem_%27s_Gravesande) (early 1700's) determined that their penetration depth was proportional to the square of their impact speed (see kinetic energy http://en.wikipedia.org/wiki/Kinetic_energy).
Note  

Joe: I don't know how Gravesande timed the weight's fall (its impact velocity). Presumably he used an inclined plane because of the requirement to be able to time the weight's fall. However the velocity on impact would have been small. 
In the next sections, we're going to find first the velocity as a function of time, then the distance as a function of time. From that we'll figure out velocity as a function of distance. This will allow us to determine the K.E. gained as a function of height dropped, giving us the dimensionless constant for K.E.
Let's calculate the distance s, travelled by an object moving at constant velocity u, for time t. By the definition of velocity
u = initial velocity s = distance travelled t = time then u = s/t thus s = ut 
Let's see if there's any other way we can derive this relation. Here's the graph of the velocity of the object as a function of time.
 ^    v .......................u   _______________________t time> 
Here's the position (displacement) s of the object as a function of time, with the starting position s=0 at t=0.
 o  o ^  o   o s  o  o  o o_______________ t, secs > 
This graph is a straight line (the relationship between x and t, under constant v is linear). If the object has an initial displacement (wasn't at the origin at t=0, but instead was at x=x_{0} at t=0) then the relationship is still linear (i.e. the line is of the form y=mx + b).
 o  o ^  o   o x  o  o  o x0 !o    _______________ t, secs > 
What's the area under the line on the vt graph?
 ^    v .......................u /////////////////////// /////////////////////// /////////////////////// t time> 
The area is vt. What is the dimension of the area?
dimension of vt = (l/t).t = l 
Dimensional analysis tells us that the area under the line (the line is a straight line) is the type of quantity (distance) that we're after. The two xt graphs immediately above the vt area graph, only differ in the starting position of the mass, and are both described by the one vt graph. The area under the vt graph describes the change in position (i.e. the distance moved) during the time t. There is a dimensionless constant to determine. From the definition of velocity, the dimensionless constant is 1. Thus for constant velocity, the area under the vt curve is the distance. (It's true for varying velocity too, but we haven't proven it.)
Let's find v=f(t) for an accelerating body.
The definition of acceleration is the change in velocity per unit time. A stationary object accelerated at 1m/sec^{2} will have a velocity of 1m/sec after 1sec, 2m/sec after 2sec... Here's the graph of its velocity
2  o  o ^  o   o v 1  o m/s  o  o 0 o_______________ 0 1 2 t, secs > 
We know that a falling body is accelerating at a rate g, which allows us to calculate the velocity at time t.
v = at a = g v = gt 
Let's see what we can find from the area under curves (here straight lines). Here's a graph of the acceleration of an object falling in a gravitational field g for t secs.
 ^    a .......................g   _______________________t time> 
In this graph, the acceleration (=g) is constant.
From the definition of acceleration, the velocity for the accelerated object (assumed to have v=0 at t=0) is
 o  o ^  o   o v  o  o  o o_______________ t, secs > 
This graph is a straight line (the relationship between v and t under constant a is linear). If the object has an initial velocity then the relationship is still linear (i.e. is of the form y = mx + b).
 o  o ^  o   o v  o  o  o o    _______________ t, secs > 
What's the area under the line on the at graph?
 ^    a .......................g /////////////////////// /////////////////////// /////////////////////// t time> 
The area is gt. What's the dimensions of the area ^{[20]} ? The area under the curve tells us something about velocity, but what? The two vt lines immediately above the at area graph are described by the one at graph. The area under the at graph describes the change in velocity during the time t. This is reasonable, considering the definition of accelaration (the change in velocity/sec). From the definition of accelaration, the change in velocity is a*t, which is the area under the at graph. This makes the dimensionless constant 1.
Newton's Laws of Motion (http://en.wikipedia.org/wiki/Newton's_laws_of_motion) are (here's the word "law" again)
From this we see that acceleration is independant of the initial velocity. Accelaration only causes a change in velocity. The equation above is better written
v: final velocity u: initial velocity a: accel t: time v  u = change in velocity v  u = at v = u + at ; the usual form of the equation 
Example: you throw a ball upwards with an initial velocity of 9.8m/sec. How long does it take for the ball to stop moving?
Inputs u = 9.8m/sec v = 0 m/sec a = g t? v = u + at t = (v  u)/a t = (09.8)/9.8 sec = 1 sec 
Hmm, the ball stopped 1 sec before we threw it. What went wrong here? Velocity is a vector quantity  it has direction (speed is a scalar quantity; it has magnitude but no direction).
Note  

the words vector, scalar are adjectives. They are used routinely as nouns e.g. this is a vector, this is a scalar. 
Accelaration is a vector too. To use vectors, we need a frame of reference to establish a direction. The ball is moving (going up and down) in a 1dimensional frame of reference, the vertical frame. We can arbitarily assign up as being the +ve direction, or down as being the +ve direction.
Note  

No matter which frame of reference we use (up is +ve, or up is ve), we will get the same answer. The initial velocity is in the opposite direction to the acceleration, so velocity and acceleration will have the opposite sign, no matter which direction is chosen as the +ve direction. 
Let's assign up as the +ve direction. Here's our problem now
Inputs u = 9.8m/sec v = 0 m/sec a = g t? v = u + at t = (v  u)/a t = (09.8)/9.8 sec = 1 sec 
The ball stops moving (i.e. its velocity is 0) 1 sec after you release it. Using the eqn v = u + at, plot the velocity of the ball for the first 2 secs of its flight.
9.8 x  x ^  x   x v 0 x m/s  1 x 2  x 9.8  x t, secs > 
The ball starts (at t=0) with a +ve velocity (it's going up). The velocity decreases till at t=1 sec the ball is not moving (i.e. v=0). After that, the ball will start falling back to earth (v<0). The velocity of the ball is decreasing linearly through the whole flight.
Note  

The speed initially decreases till the ball stops and then the speed increases again. 
If we'd chosen the down direction as the +ve frame of reference, then what would have been happening to the velocity, the speed ^{[21]} ?
At what time would the ball have a velocity 19.6m/sec?
Inputs u = 9.8m/sec v = 19.6m/sec a = g t? v = u + at 19.6 = 9.8 + 9.8t t = (19.69.8)/9.8 sec = 3 sec 
Here's the velocity of the ball at t=0,1,2,3secs
t,secs v,m/sec 0 9.8 1 0 2 9.8 3 19.6 
At what time would the ball have a velocity 19.6m/sec?
Inputs u = 9.8m/sec v = 19.6m/sec a = g t? v = u + at 19.6 = 9.8 + 9.8t t = (19.69.8)/9.8 sec = 1 sec 
What does the ve time mean? It means that 1 sec before the ball came past your hand, it was moving upward at 19.6m/sec
Note  

The equation is a continuous function of t and covers t for ∞<t<+∞. Into the equation, you plugged; the velocity at t=0 and the acceleration. The equation handles the rest. You didn't specify anything special about the behaviour of the ball for t<0 and therefore the equation specifies v for all t. Equations with discontinuous properties (e.g. for t<0; v=0) are tedious to write and use for calculations. Unless you arrange it otherwise, equations are continuous in the independant variable (here t). 
This equation doesn't tell us where the ball was, when it was moving upward at 19.6m/sec. We'll need an equation, to be derived below, to figure that out. Ordinary experience says that the ball would have been coming up from below.
Example: In the US, in 2009, the average new car accelerates from 0 to 60mph (http://en.wikipedia.org/wiki/0_to_60_mph) (97kmph) in 8.4sec. Assuming that the acceleration is constant (and the car is in a vacuum), what is the acceleration during this period relative to g ^{[22]} ?
The weight of a US car is about 1.5tonnes. What power is being used by the car when it starts to move ^{[23]} ? What power is being used when the car is doing 97kmph (and still accelerating at the same rate) (give answers in kW and HP; 1HP=746w) ^{[24]} ?
Note  

The same car at highway speed uses about 40kW. Cars use less fuel on freeways than driving around town, where power used accelerating in traffic is lost as heat in braking. The rest of the highway fuel economy comes from the engine being designed to be most efficient at highway speeds. 
In the python course, we did some calculations involving a Boeing 747400ER taking off. This time we have competent programmers loading the plane. Here's some numbers
Note  

Jet fuel has a relative density of 0.9. The mass of fuel then is 241,140l*0.9kg/l=217,000kg. Over half the takeoff weight of the plane is fuel. FIXME: stats for train 
Calculate the acceleration in m/sec^2, and relative to g, during the takeoff run and the power output at the moment of takeoff (assume the plane accelerates at a constant rate). Here's my answer ^{[25]} . A diesel train engine outputs about 12Mw.
Note  

takeoff is the most difficult point of the flight as far as the power budget is concerned. The plane is at maximum weight (it will land almost empty of fuel at half the weight); the plane is flying at about 1/3 of its cruising speed (the wings are supplying less lift, and have the flaps down to increase lift at the expense of increased drag). Once the plane reaches any speed at all, no runway is long enough for it to stop by braking and there is nothing (e.g. nets) that can stop it. If there is an emergency at takeoff, the plane is too heavy to land (the wings with all the fuel will tear off, the plane erupting in a fireball). 
An example in two dimensions: You shoot a projectile (canon, arrow, kick a football, hit a golf ball, throw a raw egg, tomatoe) in a vacuum, with initial velocity u, at an angle ϑ (theta) to the horizontal.
Here's the initial path of the ball.
 /   /  / / v  u / e  / r  / t  / /theta  horizontal 
Note  

The path of the ball is a parabola (ascii art parabola shown),
but we won't be proving this till a bit later.

We'll be looking at the path of the ball, both as a function both of time t and of horizontal position x. For the rest of the discussion, we will be use s (the horizontal distance) to describe the position on the xaxis. Here's the initial path of the ball as function of t.
 /  / ^  /   / h  /  /  / /theta  t, secs > 
Here's the initial path of the ball as a function of s (the horizontal distance)
 /  / ^  /   / h  /  /  / /theta  s, m > 
Note that both graphs of the initial path (with s or t on the abscissa) are similar; only the scale on the xaxis changes. You can do this because the horizontal velocity is constant (there is no horizontal force/acceleration on the ball). The scales on the two graphs are related by s=u_{horizontal}.t, i.e. s=const.t.
  ^    v,hor u m/sec     t, secs > s, m 
The ball is acted on by the force of gravity, accelerating downwards. Here's the velocity of the ball as a function of both time and distance (two graphs are combined into one since s and t are related by a constant). The ball has an initial +ve velocity (it's going up), it slows down till it stops, and then the velocity becomes ve. The velocity decreases linearly through the flight (becoming more ve).
Note  

Thinking about the speed only confuses the matter, but here it is anyhow: the speed decreases till the top of the arc (when the ball has zero velocity), then increases again as the ball falls back to earth. While speed is useful for driving a car, or walking; if you want numbers and something is being acted on by a force, you need to think in terms of velocity, not speed. Nature thinks in terms of velocity; if you want to converse with nature, you should use its language. Speed is fine for social situations, but nowhere else. 
u x  x ^  x   x v,vert 0 x m/sec  x  x  x u  x t, secs > s, m 
Here's the path of the ball again.
 /  / ^  /   / o o h  / o o  /o o  /o o /theta o  t, secs > 
At what angle ϑ, do you have to launch your projectile, to make it go the furthest horizontal distance, before hitting the ground?
First we need to find a frame of reference.
multidimensional space, degrees of freedom
We live in 3D space.
The previous problems have been 1D; we threw a ball up in the vertical direction, or we accelerated a car or a cargo plane in the horizontal direction. These are 1D problems; the position of the object is described by its height (or distance along the road/runway). The object is said to have 1 degree of freedom, i.e. only 1 parameter is needed to describe it's position (or velocity).
In the current problem, the ball is launched at an angle, and is moving in 2D space. Its motion in any direction is independant of its motion in any other direction (you can change the ball's x independantly of its y).
A counter example, where x,y of an object in 2D space, can't be changed independantly, is a ball fixed to the circumference of a rotating disk, or to the end of of a pendulum. For a small change in x there will be only one possible new value for y, i.e.x,y cannot be changed independantly. For the case of the rotating disk or the pendulum, x,y are completely determined by the angle of rotation of the disk (or pendulum rod). This system is said to have only 1 degree of freedom, i.e. only 1 parameter (here the angle of rotation) is needed to determine the position of the ball. While the ball at the circumference of the disk moves in 2D space, its movement only has 1 degree of freedom i.e. it's a 1 parameter problem.The motion of the arrow/golfball is in 2D and because movement in the x,y directions is independant, the ball has 2 degrees of freedom. We need two parameters to describe the position (or velocity) of the ball.
Note We need two parameters because the problem has 2 degrees of freedom, not because it's a 2D problem. We have 2 degrees of freedom because the ball is moving in 2D space AND because the ball can move independantly in each of the two dimensions.
How about a 1D problem with 2 degrees of freedom? Imagine two train carriages, and instead of being joined by a standard coupler, are joined by a spring. The two carriages are sent down the rail line by pushing the rear carriage. The two carriages will oscillate about their mean position, by virtue of the spring, while their center of mass travels along the rail at constant velocity. You can choose two frames of reference
The first frame of reference is the simplest; the problem resolves into two separate and well studied problems
 parameter_1: the center of mass of the two carriages; parameter_2: the distance between the two carriages.
 parameter_1: the position of carriage_1; parameter_2: the position of carriage_2.
 the center of mass travelling at constant velocity
 a pair of oscillating masses connected by a spring
In a 1D problem there is only 1 possible frame of reference (well, you can offset the origin). In 2D there are an infinite number of them. We can choose any two directions (the frame of reference). to describe the position of the arrow. Life is simpler if we choose orthogonal directions (directions at right angles), and simpler yet if we choose one of the directions parallel to the acceleration of gravity. However we could choose one direction along (or at right angles to) the direction of the ball's initial movement and the other along the ground. We can choose any frame of reference we like and so we choose the one that gets the answer the fastest (or simplest).
Note Some frames of reference make life easier. The Copernican heliocentric frame of reference is simpler than the Ptolemaic geocentric system. Using the heliocentric frame of reference, Kepler was able to determine that the orbits of the planets around the sun were ellipses, a problem that for 2000yrs had confounded people using the geocentric system. (It turns out that objects attracted under the inverse square field of gravity move along conic sections, i.e. circle, elipse, parabola, hyperbola. Kepler found one of them, the elipse. (In principle, objects gravitationally attracted to the sun can have circular, parabolic or hyperbolic orbits.) While the heliocentric frame of reference is the simplest system to describe objects orbiting the sun, for objects that are earth bound (or near earth), the geocentric system is the most useful. Amateur astronomers need to know where and when to look up at night; if you want to launch a satellite, go to the moon or calculate sunrise, sunset or the tides, the assumption that the earth is the center of the universe is simplest.
When you're learning, you'll be shown the frame of reference that will solve the problem quickest. You'll soon get the idea of what to look for. You may have to translate the axes, rotate the problem, use nonorthogonal axes before you find a frame of reference in which the problem becomes tractable.
From past experience, the two directions to choose for our frame of reference are parallel to the acceleration of gravity (the y axis) and at right angles to it (the x axis, where there is no acceleration and the ball has constant velocity). For the positive directions, we'll use the Cartesian convention; for the positive y direction, we'll choose up; for the positive x direction, we'll throw the ball to the right. Since movement in two orthogonal directions is independant, we've resolved the problem into two components
 the vertical movement: the ball moves under the acceleration of gravity, just like it did in our previous 1D problems.
 the horizontal movement: the ball moves with constant velocity.
Note I said that movement in the two orthogonal directions is independant. What is actually happening is that movement in the direction parallel to the field of gravity (where a force acts on the ball) is independant of movement at right angles to the field of gravity (where no force acts on the ball). These two directions are orthogonal. Movement in the two orthogonal directions is independant, because movement parallel and at right angles to gravity is independant, and because the two directions are orthogonal. The problem to be solved then is our previous 1D problem, where we threw a ball up in the air, but now as well, the ball is moving sideways. We calculate the position/velocity of the ball for each component direction and then we combine the two results to see where the ball is and how it's moving. We need to find how long it takes for the ball to return to earth in the vertical direction, and then figure out how far the ball has travelled sideways in that time.
The initial velocity u is at an angle ϑ to the horizontal. We've separated the problem into two simpler problems; the behaviour in the x direction and in the y direction. We need the initial conditions for the two separate problems. and these are the component of the initial velocity in the two directions.
Note  

To be precise, we say "the component of the velocity in the x direction", rather than "the velocity in the x direction". This is to make clear that the object has velocity in other directions and that for the moment we are only looking at a component of the velocity, not the velocity. In casual conversation you can say "the velocity in the x direction" as long as it's clear that you're talking about the components. 
Note  

This next section, to find the components of initial velocity,
may seem a little laboured.
The result may be obvious already.
If not, after you understand the principle,
the result will be obvious and you can use it in a problem without justification
(the result is obvious to everyone else, at least after it has been explained).
However you need to know how to prove the result, so that you know it's true.
The results are

First we're going to find the component of the position in each direction. Then to find the components of velocity, we'll divide the change in position, by the change in time. Imagine that the ball/arrow is at a distance l from the origin or that the ball is on the end of a stick of length l, whose base is at the origin, and which is leaning at an angle ϑ to the horizontal.
  o  / ^  /   l / y  /  /  / /theta  x > 
To find the component of the position of the end of the stick in the y direction, you move off to the right and see how high the end of the stick appears, giving you h. (the ascii art to the right is an eyeball looking at the top of the stick)
 \ o O  / / ^  /    l /  y  /  h  /   /  /theta   x > 
You'll need a bit of trig here.
h = l*sin(theta) o / /  l /  /  h /  /  /theta   
Since the ball is moving, at time=dt later , the ball has moved a distance dl and has changed height by dh. The hypoteneuse of the small triangle, showing the movement of the ball, is also at an angle ϑ to the horizontal, so
dh = dl * sin(theta)  \ o O  dl / dh /  o  / ^  /    l /  y  /  h  /   /  /theta   x > 
To find the component of velocity in the y direction, we divide both sides by dt.
dh = dl * sin(theta) divide both sides by dt (dh/dt) = (dl/dt) * sin(theta) therefore vvert = u.sin(theta) 
The component of initial velocity in the y direction is u.sin(ϑ).
 o  /  _ / ^ ^  / /    u / u.sin(theta) h  /   /   /  /theta   t, secs > 
To find the horizontal velocity, we need to know the component of the position in the x direction.
h = l*cos(theta) o / /  l /  /  /  /  /theta   w 
Following the same process as above, we find that the component of initial velocity in the x direction is u.cos(ϑ). (Make sure you can do the derivation yourself.)
 >/ u.cos(theta)  _ / ^  / /   u / h  /  /  / /theta  t, secs > 
Now we have the components of velocity in both directions and we can use the equation v=u+at to find the velocity at some other time. First find the time of the flight (hint ^{[26]} ), then find the horizontal distance moved in this time. This trig identity will help.
sin(A)*cos(A) = sin(2A)/2 
Then maximize the distance as a function of ϑ. (sin(ϑ) has values from 1 to +1. The maximum value occurs with x=90°.)
Do the problem in 3 parts.
First analyse the problem in the direction parallel to the acceleration of gravity. Find the time it takes for the ball to return to earth ^{[27]} .
Next analyse the problem in the direction orthogonal to the acceleration of gravity (i.e. in the horizontal plane). Find the distance the ball moves horizontally, in the time you calculated for the ball to fall back to earth ^{[28]} .
The term 2.u^{2}/g is constant for this problem. The only variable available to maximize the distance is theta. Using the trig identity sin(A)*cos(A) = sin(2A)/2, find the distance as a function of sin(2ϑ) ^{[29]} . Next find the angle which maximizes the distance travelled ^{[30]} .
Now plug this value for ϑ into the distance formula, to get the horizontal distance travelled ^{[31]} .
Let's try some numerical examples. The world record speed for hitting a golfball is 328kph (golf ball http://en.wikipedia.org/wiki/Golf_ball). How far would this ball travel if hit at 45° in a vacuum ^{[32]} ? The longest golf drive in air is 377m ( distance of the longest driven golf ball http://hypertextbook.com/facts/2004/CrystalCuevas.shtml). It would seem then that air resistance halves the distance of a long drive.
A good golfer can hit a 200m drive (long drive http://en.wikipedia.org/wiki/Long_drive). On the moon, the acceleration due to gravity is about 1/6th that of earth Acceleration of gravity due to the Moon How far should a normal golfer be able to drive a golf ball on the moon http://hypertextbook.com/facts/2004/MichaelRobbins.shtml. ^{[33]} ?
Astronaut Alan Shepherd encumbered by a bulky space suit, did a one handed sandtrap shot with a 6iron on the moon, shooting the ball a distance quoted as 800m (Moon Trivia http://home.hiwaay.net/~krcool/Astro/moon/moonglossary/moontrivia.htm). Assume the distance was on the full, rather than the distance the ball then bounced (if it did bounce), do you believe the distance ^{[34]} ?
Note  

This example, of lobbing a canon ball, comes from an often ignored part of the history of physics. Once physicists understood acceleration, they were in great demand to calculate artillery trajectories. Lagrange was the professor of mathematics at The Royal Artillery School in Turin (Lagrange http://en.wikipedia.org/wiki/Joseph_Louis_Lagrange). I believe Galileo received some of his support ($) for doing this. Physicists (and less so other scientists) have always been happy to receive money to find better ways to destroy things. 
How do we know that the result we get after resolving the problem into the x,y components and then putting the pieces back together is the right one? What if something is lost when you do it this way?
Note This is the unknown part of my argument. If the two components are really independant, then nothing is lost. To show that resolving the problem into components is valid, a proof is needed. When I was learning this topic, I was never shown a proof and in the subsequent time, I've never seen one. It was not obvious to me whether the proof was so trivial that none was needed or whether there was no proof and mathematicians were embarrassed to admit it.
Math departments, when I went through, had no problem delivering trivial proofs: at my university (Sydney University, 19656), the math department delivered proofs of 1+1=2 and talked at great lengths about the nature of "numberness". In light of their unwillingness to teach us any of the multitude of useful math tools, I regarded this as a declaration that the department had no interest in teaching its students anything and that it had no intention of fulfilling its contract with the tax payers, who provided their salaries, to do their job. The math department at Sydney University regarded themselves as unaccountable to society (they were unaccountable  the state government didn't care what they did with the money). It was a sham  I dropped math at the end of the 2nd year of university. There were only two good teachers in my courses  one I did my PhD with. The other, my first year chemistry lecturer, Peter Simpson, didn't get tenure. He was invited back for a reunion 40 yrs later by the students of my generation, showing that we were capable of appreciate his work, even if it went over the heads of the the chemistry department.
The remaining possibility then is that there is no proof. It's important when persuing a goal to know the assumptions you've made, in case when you get to the end, your destination isn't the one you expected. If there is no proof, then it should be declared in your list of assumptions. If the wings fall off your plane, you'll have to go back to find what went wrong and the first place to look is the list of things you can't prove correct.
In the meantime, in the absence of a proof that using orthogonal components is valid, let's list what we know about resolving problems into components. All of these conditions are necessary, but not sufficient for the validity of using components.
 It would be difficult to find a part of the technical world where calculations aren't based on components. If there were problems with the process, it's hard to imagine how they would have escaped notice. In general there's aren't other ways of handling the problems, so it's not like we can test for agreement using answers from methods that don't resolve the problem into components. However for the whole Middle (dark) Ages, people reverted to thinking that the world was flat. So it's possible that there are unnoticed problems in resolving components.
 If we accept the Cartesian system (positions are described by x,y), then we can resolve positions into any frame of reference. If we can resolve position, then by dividing by time, we can resolve velocity and acceleration too.
 We have to be able to show that orthogonal (ortho = right angles) directions are independant. When we say that we live in 3D space, it means that we need 3 (independant) pieces of information to specify our position in space. That we need 3 measurements; width, length and height, seems obvious. If the 3 pieces of information weren't independant, then we'd only need (say) 2.5 pieces of information, but noone has come up with a 2.5D space. It seems to be a fundamental part of math, that dimensions (as in 2D and 3D) are independant. Maybe dimensions are defined as being independant, and no proof is needed.
 The positions of an object in each dimension in 2D (or 3D) space are independant. You can change an object's position in one dimension without changing its position in any of the other dimensions. Maybe proving this is enough to prove the validity of resolving problems into components. If it is, then I'd like the proof stated somewhere, so that when the wings fall off my plane, I don't have to go back and find another method of calculation.
The whole matter is unclear and needs to be straightened out, at least for people like me. In the absence of better information, the rationale that I use is that resolving problems into components works. I'll admit this is not very satisfactory, but it's the best I can do.
Nov 2010: I asked a mathematician friend Chris Cosgrove, about resolving vectors into components, when I visited him on one of my hiking trips to Australia. Here's what I found.
The problem then becomes one of showing that you can represent vectors by components and that you can reverse the process; i.e. reassemble the original vector from its components. So the problem is about vectors.
Note  

What went wrong in the educational process is that the recipe for resolving a vector into (orthogonal) components is handed out to everyone in high school, but the explanation for why it works (and the procedures for dealing with nonorthogonal components) is not given till 2nd year in university (college). By then it's too late: 99% of the original cohort are doing something else, selling shoes, being lawyers, scientists and engineers, but with a crippled piece of knowledge rattling around their heads, which they'll abandon at first opportunity. I did math(s) at university, but I must not have been in the course that explained resolving vectors into components. 
To understand vectors, we need to know about basis vectors.
Basis vectors are linearly independant.
The two vectors i and a*i (where a is a real number coefficient) are dependant; you can make one from the other by multiplying by a nonzero coefficient.
The three vectors i,j, i+j (where i,j are any noncolinear vectors) are dependant  you can make any one of them from the other two. Show how to make each vector from the other two (you'll need 3 equations) ^{[35]} .
Show that the set of three vectors i,2j and 2i+2j (where i and j are noncolinear) are dependant, by making each of them from the other two (you'll need 3 equations) ^{[36]} .
The test for independance is whether you can make linear combinations of the vectors (sum them using nonzero coefficients) and not produce a zero vector. The set i,j,i+j are dependant because the linear combination i + j  (i+j) = 0. Using this test, show that the set of vectors i,2j,2i+2j are dependant ^{[37]} .
If vectors v_{1}...v_{n} are dependant, you can find a set of nonzero coefficients λ_{1}...λ_{n} which make the sum λ_{1}v_{1}+...λ_{n}v_{n}=0 .
Show by inspection that the noncolinear vectors i, j are independant ^{[38]} .
Now to the dimension of space. We live in 3D (3 space). How do you know? How would you check? People in the past have made worse mistakes than getting the dimension of space wrong. The answer is that if you can fill all of the space with linear combinations of 2 (or n) independant vectors, then the space has dimension 2 (or n). (This definition shifts the responsibility for defining dimensions back to the definition of independance of vectors). We can tell we live in 3 space, because we need linear combinations of 3 independant vectors to fill our space.
There are an infinite number of basis vectors in any space and an infinite number of basis vector sets can be formed from them (e.g. you can rigidly rotate any set of basis vectors, to produce another set of basis vectors). It is often convenient to pick basis vectors which are orthonormal (a telescoping of the words orthogonal==at right angles, normalized  dividing the vector by its length). You can always do this in a space where you can measure angles and lengths (e.g. Euclidean, Riemannian. You can't do this in affine space, which I know nothing about).
In 2D space, the x and y axes are orthogonal and are one basis set for 2D vectors. If you choose unit length versions of these vectors, then the unit vectors parallel to the x,yaxes are orthonormal.
Here's the pair of orthonormal basis vectors i,j parallel to the x,y axes.
y    ^j    i > x 
Here's another pair of orthonormal basis vectors u,v, at 45° to the x,y axes (the lines made of "/" and "\" are meant to be continuous).
y    \v  /u \  / \  / \  / \/  x 
We now understand basis vectors. The number of vectors in a basis set is the same as the dimension of the space. Let's say we've settled on our basis set (in our problem with the projectile, we choose the two basis vectors to be parallel and at right angles to the acceleration of gravity, i.e. along the y and x axis). The next step is to resolve a vector (e.g. velocity, force) into its components with respect to the basis set (abbreviated "resolve a vector into its components" or "resolve a vector with respect to its basis"). In the case of the projectile, we then do our calculations on the components. After we're done with that, we do the reverse, reconstructing the vector from its components. Here we'll just reconstruct the original vector from its components. The point is to show that resolution of the vector into components, followed by linear combination of the components multiplied by the basis vectors, reconstructs the original vector.
In the case of orthonormal basis vectors, to resolve a vector into it's components, you form the dot product with each of the basis vectors. Let's say the initial velocity of the projectile is u at an angle α to the horizontal.
Here's the dot product of the velocity vector and the horizontal basis vector i:
u_{horz} = u•i = u*cos(α)
Note  

The result is a (real) number, not a scalar and not a vector. 
A scalar is a number that doesn't depend on the coordinate system (e.g. mass, speed (not velocity) and viscosity are scalars). If the frame of reference was rotated, the components would change, so they aren't scalars. The components are not vectors; they are the dot product of two vectors and hence are real numbers.
Here's the dot product of the velocity vector and the vertical basis vector j:
u_{vert} = u•j = u*cos(90α) = u*sin(α)
Note  

The result is a (real) number, not a scalar and not a vector. 
Note  

In math(s), matrices are used for vector operations. You can use English (rather than matrices) to describe the simple manipulations we're doing here, but for anything more complicated (rotating a vector, finding eigen functions...) you'll need matrices. For high school work (e.g. the projectile problem), matrices are overkill. You can get the right answer without matrices (with a bit of shepherding by the teacher to make sure you use the right notation). 
Here's the matrix representation of the resolution of a vector u = ri+sj with respect to its basis set b=(i,j) into its components c
c = u  i  =  u.i  =  u*cos(alpha)   j   u.j   i*sin(alpha)  
The result is a column matrix (the column has two entries). Each of the components are real numbers. They are not scalars.
To reconstruct the original vector, you linearly combine the basis vectors, using the components as coefficients. Here's our reconstructed vector (let's call it r. It should be the same as our original u).
r = u*cos(α)*i + u*sin(α)*j
Here's the matrix representation of the previous equation. The column matrix of the components has to be transposed to a row matrix for the matrix operations to work. (Sometimes to get a matrix operation to work, you have to transpose the matrix  you can freely transpose a matrix from a row matrix to a column matrix and the reverse  it's the same matrix.)
r = c^{T}b
r = [ u*cos(alpha) u*sin(alpha) ]  i  = u*cos(alpha)*i + u*sin(alpha)*j  j  
The result is a 1*1 matrix (a single number). The number, having i,j in it, is a vector in 2 space. The vector has a horizontal component u*cos(α) and vertical component u*sin(α). What is this vector's magnitude?
r = √([u*cos(α)]^{2} + [u*sin(α)]^{2}) = u
What is this vector's argument (angle to the horizontal)?
angle_{r} = tan^{1}(ycoefficent/xcoefficient) = tan^{1}[sin(α)/cos(α)] = tan^{1}[tan(α)] = α
This is our original vector.
We've shown that resolving a vector into its components (at least when using an orthonormal basis for the vector space) and then reconstructing the vector from its components, returns the original vector. Thus we can analyse the projectile's (vector equations of) motion as components in the horizontal and vertical direction and then reconstruct the actual motion of the projectile from its components.
Note  

The components are real numbers, not vectors. Without attending the required 2nd year university course, I thought that components were vectors (after all, each component applies in a particular direction). After visiting Chris in Australia, on my return to USA, I polled some of my more technically educated friends and coworkers, each with multiple decades in their area of expertise, and asked whether they thought the components of vectors were real numbers, scalars, or vectors. It's apparent that not only the shoe salesmen missed out on the 2nd year university course on vectors, but many technical people have too. They think that the component is a vector, i.e. that ucos(α)*i is the component, rather than ucos(α). Admittedly, none of these people are using vectors in any more complicated way than they learned in high school. Presumably all of these people have been using the high school, nonmatrix representation of vectors for the subsequent decades. Since a vector is r = u*cos(α)*i + u*sin(α)*j it's reasonable to think of the terms being components. However the real representation of a vector is
Here the first (row) matrix is the components. 
Basis vectors don't have to be orthogonal: e.g. crystals, in general, don't have orthogonal axes. In the Triclinic Crystal System (http://en.wikipedia.org/wiki/Triclinic_crystal_system), the position of an atom in the unit cell is defined with respect to axes of different unit length and with cell angles α,β,γ≠90° (for some of the math involved in nonorthogonal basis sets, see Vector Spaces, http://fatman.cchem.berkeley.edu/xray/VectorSpaces.pdf).
You will likely start by describing your data/object in an orthogonal coordinate system (e.g. Cartesian). If you have your crystal in front of you, you can measure the lengths of the sides of the unit cell and the cell angles, giving you the nonorthogonal basis vector set. If you have some complicated mathematical construct, it may take a while to see that it would be simpler if moved to a nonorthogonal coordinate system. It might take a while to figure out what that coordinate system is.
Here we see how to resolve a 2D vector into nonorthogonal basis vectors. How many basis vectors do we need ^{[39]} ?
Note  

See the section "Example of Vector Components" in Basic Vector Operations (http://hyperphysics.phyastr.gsu.edu/hbase/vect.html) to show how to resolve a vector into nonorthogonal components. 
Note  

We need a coordinate system to represent the nonorthogonal basis vectors. Since we don't (yet) have a nonorthogonal coordinate system, we'll describe the problem, using the standard (orthogonal) cartesian coordinate system. (Once you can resolve vectors into nonorthogonal components, you have a nonorthogonal coordinate system.) 
Let i,j be the unit vectors along the x,y axes respectively.
In 2D, the unit vector along the xaxis (u=i) and the unit vector at an angle of 45° (v=(i+j)/√2) are (a pair of) basis vectors.
Note  

(u,v are normalised. There's no special advantage in normalizing basis vectors unless they are orthogonal.) 
y      /v (0.7,0.7)  /  / / u (1,0) > x 
How do you know they're basis vectors ^{[40]} ? Since u,v are basis vectors, you can make any 2D vector from a linear combination of u,v.
We've found these answers by guessing and thrashing around. We'd like a systematic way of finding the answers. Since any vector can be made from the basis vectors, the formula for any vector w, as a linear combination of the basis vectors u and v is
w = au +bv
where a,b are real number coefficients. Using this formula, repeat the example above resolving w = (0,1) into its components with respect to the basis set u=i+0j, v=i/√2+j/√2.
w is a vector being resolved with respect to its basis set u,v, thus
w = au +bv
substituting values in Cartesian space for w,u,v
(0,1) = a(1,0) + b(1/√2, 1/√2)
Since the numbers are in the orthogonal cartesian system, we can separate out (resolve) the coordinates into the x,y components giving us a pair of simultaneous equations in a,b.
resolving into x,y components
0 = a + b/√2 ;xcoordinate
1 = b/√2 ;ycoordinate
giving a,b = 1,√2. Thus w=u+√2v. We've resolved w with respect to its basis vectors. In the coordinate system with u,v as the axes, w has the location w=(1,√2). Confirm that w=(1,√2) in the u,v coordinate system is the same location in our 2 space as w=(0,1) in the i,j coordinate system.
Resolve the vector w=1/√2,1/√2 in the i,j coordinate system into its coordinates with respect to the same basis set above (i.e. u=i+0j, v=i/√2+j/√2) ^{[44]} .
There are 3 fields which we come across in our day to day living; gravitational, electric and magnetic.
Newton postulated the existance of a gravitational field. A field allowed a force (or object) to act at a distance, something for which there was no explanation at the time. Nowadays, everyone accepts the explanation that an object falls to earth because the earth attracts it. Before Newton, there was no accepting action at a distance. Even after Newton, the acceptance of action at a distance took a long time.
Here we're going to explore gravity and then later we'll return to fields.
Here's the modern form of the equation for the force of gravity between two masses m_{1} and m_{2} separated by a distance r
F = Gm_{1}m_{2}/r^{2}
where G is the Universal Gravitational Constant. What are the dimensions of G ^{[45]} ? The force of gravity is always +ve (attractive). Force is a vector, so the LHS of the equation is a vector. Is the RHS of the above equation a vector ^{[46]} ?
In the era that Newton formulated this equation, people were interested in celestial mechanics. The equation of more interest to people back then was
F = (GM_{earth})M_{moon}/r^{2}_{em}
where (GM_{earth}) was a constant of unknown value. Noone was particularly interested in G or M_{earth} separately. Even the product (GM_{earth}) could be sidestepped, if you just wanted the ratio of forces (or accelaration). This site Henry Cavendish: Weighing the Earth ("http://www.juliantrubin.com/bigten/cavendishg.html) explains
Newton was not particularly concerned to evaluate the constant of proportionality, G, for two reasons. First, a consistent unit of mass was not in widespread use at the time. Second, he judged that since the gravitational attraction was so weak between any pair of objects whose mass he could sensibly measure, being so overwhelmed by the attraction each feels toward the center of the earth, any measurement of G was impractical.
Here (in modern form) is the accelaration felt, by the the moon, due to the earth's gravity.
a_{moon} = (GM_{earth})/r^{2}_{em}
You can keep a body in circular motion (http://en.wikipedia.org/wiki/Circular_motion) (e.g. a weight on a string whirled around your body, by your arm held above your head; a car going around a corner; a planet in orbit around the sun), by applying a force (or accelaration) in the direction of the centre of the circle. The derivation of the equation for the force required and will be handled in another section. Here's the equation for the accelaration of a body in circular motion rotating at an angular velocity of ω radians/sec, about a center at distance r.
a = rω^{2}
Note  

a, ω and r are all vectors. We'll talk about cross and dot products later. 
Note  

New: 
In initial treatments (e.g. highschool), based on the assumption that you (the student) don't ever intend to make use of your education, the vector properties of the variables in this equation are ignored. a and r are obviously vectors. What direction are these vectors pointing (assume you're swinging a weight on a string above your head) ^{[47]} ?
The units of ω are radians/sec. What direction is the vector ω pointing? This isn't particularly obvious either. I'm not sure whether vectors are something convenient we made up, or are real and fall out of some math(s) independant of our existance. Assume that we just make up vectors, then we'd want the most useful direction for ω. If we used the tangent or the radius of the mass's movement as the direction of ω then we'd need to know the position of the mass at any instant to describe the direction of ω. Instead find a direction in 3space that specifies the direction of movement of the mass, but that doesn't change as the mass swings around your head ^{[48]} . Rotation is referred to the axis of rotation. The convention for the +ve direction is arbitary; you (or someone) decides whether clockwise or anticlockwise is +ve. The vector, describing the direction of rotation of a wheel about an axle, is along the axle. While it's obvious that an object rotating in a plane, at constant angular velocity, has an axis of rotation, a movement of an object through an angle, and leaving it there, is equally well described by the axis of the rotation. So the direction of a vector of units radians/sec and just plain radians are both along the axis of rotation.
If ω is a vector along the axis of rotation, what is ω^{2}? It's the dot product, which is a scalar.
Let's retrace Newton's calculation, as it would be done today. Newton knew that the moon was in a (nearly) circular orbit about the earth, and used the period of the moon's orbit and its distance (determined by parallax) to determine GM_{earth}.
Note  

In a reference I can no longer find, it said that Newton didn't have the term (GM_{earth}) in his calculations. He used the ratios of accelarations and the term GM_{earth} doesn't appear. 
The synodic period of the moon (the time for the moon to return to the same phase, e.g. to successive full or new moons) is 29.5 days. The siderial period of the moon (the time for the moon to return to the same position against the "fixed" stars) is 27.3 days. These two periods are close enough that you could use either for a back of the envelope calculation, but which one is the correct one ^{[49]} ? The distance to the moon varies between 363Mm (perigee; gee from geode meaning earth e.g. geology  the science of the earth; peri from the greek meaning around e.g. perimeter, peripatetic, periscope) and 405Mm (apogee; apo from the greek for away, apart, off, e.g. apology, apoenzyme, apostrophe, apostacy).
The moon's orbit is not circular, and not in the plane of the ecliptic (the inclination of the moon's orbit is 4.995.30°, see Orbit of the Moon http://en.wikipedia.org/wiki/Orbit_of_the_Moon), and the earth's rotation about it's axis is at 23.5° to the normal to the plane of the ecliptic, presumably all consequences the events which formed the moon.
It is assumed that a Mars sized object collided obliquely with the earth in the early days (50100My) of the solar system (see Moon's Formation http://en.wikipedia.org/wiki/Moon#Formation). For a dramatic image of the moon's formation see Giant Impact Hypothesis (http://en.wikipedia.org/wiki/Giant_impact_hypothesis). Here's a story and animations from youtube Origin of the Moon (http://www.youtube.com/watch?v=m8P5ujNwEwM). (I can't believe they made the woman scientist wear a large necklace, earrings, lipstick and makeup for the piece. They forgot the fingernail polish. They didn't require the male geologist to wear a suit out in the field; why does the woman have to put on a necklace in her office.) Our moon now is the original impactor together with material from the earth's mantle, giving the moon a lighter density (S.G.=3.35), with the denser earth being 5.4.
Because the moon's orbit is elliptical, its apparent size and its velocity through the sky varies through the month (the moon is largest and fastest when it's closest). However while the moon's size and velocity are changing, the rate of rotation about its axis is constant (the moon's axis of rotation is almost parallel to the earth's axis, being inclined at only 5.145° to the earth's axis, presumably a result of their common formation). The combination of constant angular velocity and varying speed means that the moon appear to rotate on its axis back and forth, a little bit, once each revolution about the earth. This rocking back and forth is called libration (from Libra, the balance). The rocking left to right is called longitudinal libration. With the moon being nearly in the plane of the ecliptic (23.5° to the equator), we see more of the moon's southern hemisphere when the moon has a northerly lattitude (and the reverse). The up and down libration is called latitudinal libration. The two motions (up and down/left and right) combine to give any point on the moon's surface a circular (or elliptical motion) during the month.
Look at the animation at the top right of this webpage Article 5 of 6 on Studying the Moon; GoodGood, Good Librations, Michael Packer . Identify the monthly changes that are due to
 the elliptical orbit of the moon ^{[50]}
the moon being nearly in the plane of the ecliptic, with its axis of rotation being nearly parallel to the earth's ^{[51]}
Due to libration, earth observers can see about 4/7 of the moon's surface. The English astronomer Patrick Moore, in the 1940's observed what he thought was the edge of a crater which he called Mare Orientale (Eastern Sea). Michael Packer's webpage has a photo of what Moore might have seen and gives an estimate for the amount of libration (longitudinal and latitudinal) that you see. Here's another webpage on Libration of the Moon (http://wwwistp.gsfc.nasa.gov/stargaze/Smoon4.htm).
Interestingly, when photos of the far side of the moon became available, (see further down on Michael Packer's webpage), Mare Orientale was found to be the largest crater on the moon with 3 concentric mountain rings around it.
Since we're talking about the far side of the moon (http://en.wikipedia.org/wiki/Far_side_of_the_Moon), it's intensely cratered, while the side facing the earth is covered with lunar maria. The low rate of cratering on the earth facing side is presumed due to shielding of the moon by the earth of incoming asteroids. The presence of maria on the earth facing side is thought to be due to a higher level of thermogenic radionucleids (with no explanation as to why they're on the earth facing side).
Note The far side of the moon (the side facing away from the earth) should not be confused with the dark side of the moon (the side facing away from the sun).
Let's get back to Newton's calculation. Use the accelaration of the moon in its (nearly) circular orbit around the earth to calculate (GM_{earth})
a = rω^{2} = (GM_{earth})/r^{2}
The angular velocity of the moon is 2π/27.3 radians/day = 2.66μradians/sec.
Assume that the moon's orbit is circular with a radius of 385Mm.
(GM_{earth}) = r^{3}_{em}ω^{2}_{moon} = 403 * 10^{6} m^{3}sec^{2}
Newton's great step (1686) was to realise that gravity was universal. i.e. the gravity that maintained the moon in orbit about the earth was the same gravity that made apples fall to the surface of the earth. (The story about the Newton's sudden understanding of gravity coming from being hit by a falling apple is apocryphal apo  away, cryphal  hidden). Newton's universal gravitation had to act at a distance, through what we now call a field. Using the value for (GM_{earth}) determined from the moon's orbit, what would be the acceleration at the surface of the earth?
a = g = (GM_{earth})/r^{2}_{earth}
What is r_{earth} ^{[52]} ?
Substituting the available values gives
g = 403 * 10^{6} / 6366^{2} = 9.944 m/sec^{2}
Note  

The value of the accelaration due to gravity (g) is about 9.8m/sec^{2} (we didn't use the right average distance to the moon). g varies about 0.5% over the surface of the earth (see Comparative gravities in various cities around the world, http://en.wikipedia.org/wiki/Gravity_of_Earth#Comparative_gravities_in_various_cities_around_the_world). Geologists use a gravimeter (http://en.wikipedia.org/wiki/Gravimeter) to detect changes in g as part of geophysical surveying (to detect changes in density of rocks underneath you, or the shape of the earth). (I used to call this a "gravitometer"; you can still find references by this name). The problem of the difference in g at any location, was of measuring a small change in a much larger number. A modern gravimeter (e.g. the A10 ballistic gravimeter http://www.lacosteromberg.com/pdf/A10Manual.pdf) allows a weight (with a corner reflector) to fall in a vacuum. The bottom of the weight is illuminated by a HeNe laser. The fringes of the return signal are counted in an iterferometer and timed with a rubidium clock. The accuracy is 10µgal (gal after Galileo) (what did Galileo do to get the gal named after him ^{[53]} ). 1gal=1cm/sec^{2}, so g=980gal. A superconducting gravitimeter ( Gravimeter http://en.wikipedia.org/wiki/Gravimeter) can detect changes of 1ngal, allowing you to detect the snow being shovelled off your roof. g changes 0.3mgal/m as you change altitude (gravitmetry http://en.wikipedia.org/wiki/Gravimetry). See Lucien LaCoste (http://en.wikipedia.org/wiki/Lucien_LaCoste) for the development of the zerolength spring, modern gravimeters and seismographs. Artificial satellites are convenient bodies being accelarated by the force of gravity in a much better vacuum than can be made on earth. Satellite gravimetry is being used to watch for shifts in icesheets e.g. http://earth.eo.esa.int/eosummerschool/envschool_2010/RR3.pdf (see the map on p47). 
Newton's calculation showed that the accelaration on the surface of the earth and the accelaration that kept the moon in its orbit around the earth was caused by the same gravity, i.e. gravity was universal. It was but a small step to realise that the same gravity kept the planets in orbit about the sun.
The relation stated above
(GM_{earth}) = r^{3}_{em}ω^{2}_{moon}can be used to derive Kepler's 3rd law (see Kepler's laws of planetary motion http://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion).
T^{2} α r^{3}
where α means "proportional to" and T is the period of the orbiting body. See if you can do this yourself. Newton was able to derive Kepler's laws from the inverse square field of gravity, a result that 400yrs later is still regarded as a tour de force.
FIX ME; derive Kepler's laws, talk about Tycho Brahe, Uraniborg and Kepler
People did eventually become interested in the value of G. The first step was taken in 17978 when Cavendish used the Michell torsion balance to detect the force between two masses. This should be called the MichellCavendish experiment, but is usually called the Cavendish Experiment (http://en.wikipedia.org/wiki/Cavendish_experiment). Michell, a geologist, thought of the experiment and designed the apparatus, but died before he could perform it. After his death, Michell's apparatus passed through the hands of several people before coming to Cavendish, who rebuilt the apparatus, and who gave full credit to Michell for the design of the experiment and the apparatus. The problem with the name then, is not with Cavendish, but is due to all of us who describe the experiment.
Good descriptions of the experiment are at
Cavendish put the apparatus in a sealed chamber and observed the deflections with a telescope. The juliantrubin site says
His work was done with such care that this value was not improved upon for over a century. ... Cavendish's extraordinary attention to detail and to the quantification of the errors in this experiment has lead many to describe this experiment as the first modern physics experiment.
and
Naturally, the value of the fundamental constant G has interested physicists for over 300 years and, except for the speed of light, it has the longest history of measurements. Almost all measurements of G have used variations of the torsion balance technique pioneered by Cavendish.
Thanks to the efforts of enthusiastic netizens we have these recreations of the MichellCavendish experiment
The details of the experiment (deflection, period of the torsion balance) weren't available to me as a student (presumably even my teachers didn't know them). I wondered how much deflection would be produced in a torsion balance, by bringing a two laboratory sized masses into proximity. Would the displacement of masses be microscopic or easily visible? While the force between magnets was obvious, I'd never detected any attraction between masses that I could lift with my hands. I couldn't imagine the experiment giving any deflection. Now due to the wonders of the internet and wikipedia, the details are now available for everyone. The period of the torsion balance was 7min; the deflection 4mm. A pair of large static lead balls (158kg each) attracted two smaller lead balls (730g each) suspended by the torsion element. The force of attraction on each of the two moving balls was 1.47 * 10^{7}N (about a grain of sand or 1:50*10^{6} of the weight of the small balls). I now see that the sensitivity of the balance requires a suspending wire capable of holding the weights, while requiring little force to twist it. Cavendish compared the attraction of the small balls to the large balls, with their attraction to the earth (their weight), thus giving the mass of the earth. Knowing the radius of the earth, Cavendish calculated the density of the earth.
This experiment is celebrated as the determination of G, but Cavendish determined the density of the earth (he declared that he had "weighed the earth"). Although the experiment also determined G, Cavendish had no need of or interest in this number. It wasn't till 75yrs later than anyone used Cavendish's density of the earth to calculate G (by Cornu and Baille; see the wikipedia article). The value of G determined by Cavendish's experiment differs by less than 1% from today's value.
According to wikipedia, the Cavendish experiment was inspired by systematic errors found in the survey of the MasonDixon line. Presumably the errors were due to the effects of gravity, but no details are given. The article by Hodges says that the experiment, being designed by a geologist, was to determine the density of the earth, something of interest to a geologist. The experiment was not designed to determine G or M_{earth}.
The webpage The MichellCavendish Experiment (http://www.public.iastate.edu/~lhodges/Michell.htm), by Laurent Hodges, gives a short biography of the under appreciated Michell. Another source on Michell is John Michell (http://en.wikipedia.org/wiki/John_Michell) wikipedia.
Michell's torsion balance was similar to Coulomb's, who measured the much stronger force of attraction between charged particles. Who was first, Coulomb or Michell? From Hodges:
It is not clear exactly when Michell devised his gravitational experiment, but perhaps it was in the early 1780s. Cavendish, writing in 1798, says it was "many years ago" although Michell did not complete his apparatus until "a short time before his death," which was in 1793. Coulomb investigated the electric force using a torsion balance, and is usually credited with having invented it, but Cavendish states in a footnote that Michell had described his torsion balance to Cavendish before the 1785 publication of Coulomb's results. Most references give Michell and Coulomb equal credit for having separately invented this balance.
Michell atticipated the existance of black holes (see dark star, http://en.wikipedia.org/wiki/Dark_star_%28Newtonian_mechanics%29).
Michell calculated that when the escape velocity at the surface of a star was equal to or greater than lightspeed, the generated light would be gravitationally trapped, so that the star would not be visible to a distant astronomer.
A better understanding of the interaction of light with gravity came later from Einstein's relativity.
Michell was the first to understand the cause of earthquakes, that they originated at an epicenter and showed various means to determine the location of the epicenter. As a result Michell is regarded as the father of seismology.
Michell was the first to understand the number of binary stars visible in the sky.
This is of personal significance to me. When I was an early teenager, I asked for a copy of Norton's Star Atlas. This was filled with terms and concepts beyond my understanding. Even worse, I had no way of filling in the missing information and resigned myself to the possibility of never understanding them. I noticed that astronomers had a great interest in double stars, for what reason I could not fathom. It wasn't till 40yrs later, on reading "Parallax" by Alan W. Hirshfeld (2001, Freeman and Co, ISBN 0716737116), that the reason for the interest was revealed.
Double (or binary) stars were assumed to be a random alignment of stars, rather than a pair of gravitationally bound stars. Astronomers, having no idea of the size of the universe, beyond that it was big, were interested in measuring the distance to the stars. Many pairs of stars consisted of a bright and a dim star. Presumably the bright star would be close and the dim star at a greater distance. Galileo proposed to use the method of parallax to measure the distance to the brighter star. You can see parallax if you look out a window onto distant objects and move your head side to side; the window will move, in the opposite direction to your head, with respect to the distant objects, confirming that the window is closer. Galileo realised that if the bright star were close enough, then as the earth orbited the sun each year, the bright star should be seen to move backwards and forwards with respect to the dim star, allowing you to determine the distance to the closer star. Astronomers set about finding binary star systems. Herschel, being the best telescope builder of his time, was way ahead, discovering in short time 1000 double stars. It was clear from early on that the distance of stars was large with respect to the diameter of the earth's orbit about the sun. Observing parallax would require luck (picking a double star system in which one star was close enough), precision equipment and many years of observing any particular system to build up enough data to show the annual wobble. The early results were negative or when they gave spurious positives, were revealed to be problems in the observations. The stars were just too far away.
The crushing blow (Hirshfeld, p188) came when Michell, doing a statistical analysis of the number of binary stars (in an era before statistics was well developed) found that there were many more binary stars than should be expected from random placement of stars in the sky. Michell found that most of the binary stars were gravitationally bound, orbiting each other, rather than apparent binaries. It turns out that binary (and ternary and quaternary) star systems are common. However you couldn't tell a gravitationally bound pair from an apparent pair just by looking. You had to track the systems, possibly for centuries, to see which ones were orbiting each other, to eliminate the gravitational binaries from your observing pool.
The parallax method for finding the distance to stars was abandonned. It wasn't till the 20th century, using telescopes with precise mechanisms, that parallax was detected, but then only for the closest stars.
Note  

I have little understanding of fields from a historical perspective. The webpage What is a field? (http://wwwistp.gsfc.nasa.gov/stargaze/StarFAQ17.htm#q294) gives some background, and refers readers to magnetic field lines (http://www.phy6.org/Education/whfldlns.html) and wave mechanics (http://wwwistp.gsfc.nasa.gov/stargaze/Q7.htm). The concept of a field arose to account for action at a distance. If a mass sitting in apparently empty space, was attracted to another remote mass, then there had to be something around the mass, originating in the remote mass, otherwise the mass couldn't detect the existance of the remote mass. 
This explanation for a field is paraphrased from "div grad curl and all that", h. m. schey (1992), Norton and Co, ISBN 0393962512. This book is available in several editions, and is referenced in several wikipedia webpages, which is how I found out about it. Schey introduces the concept of a field to describe electrostatics on p6. To maintain continuity with the previous section in this webpage on gravitation, I'll restate Schey's explanation in terms of gravity.
The force on a mass m_{0} from a mass M is given by
F=(Gm_{0}M/r^{2})˙û
where û is the unit vector from m_{0} to M.
The next point is the principle of superposition.
If a mass m_{1} exerts a force F_{1} on m_{0} and a mass m_{2} exerts a force F_{2} on m_{0}, then the resultant force on m_{0} is the (vector) sum , F_{1}+F_{2}.
The point is not that the resultant force is the (vector) sum of the two forces, for forces always add vectorially, but that the sum of the forces is only the vector sum of the forces; i.e. the two masses m_{1} and m_{2} act independantly; their effect is not modified by the presence each other (or of other mass(es)).
With the superposition of gravity established, we can introduce a vector function of position called the gravitational field
G(r)=F(r)/m_{0}
(This says that the field at r, is the force at r divided by m_{0})
For a mass m_{0} sitting on the surface of the earth, what is G(r) and what are its units (or dimensions) ^{[54]} ?
We say that the (strength of the) field of gravity is 9.8m/sec^{2}.
It is apparent that superposition holds for fields as well as forces. The force acting on mass m_{0} then is m_{0}G(r) (where now G(r) is the resultant gravitational field from any number of masses). What's the point of introducing the concept of a (gravitational) field (force per unit mass) which only differs trivially from the sum of the gravitational forces? Schey gives a page of explanation which amounts to "it's convenient for analysing systems with large numbers of, or continuous distributions of masses (or charges)". This is particularly so for electromagnetism. A field then is a bookkeeping device for large systems.
Does a field exist? A field was originally a construct for people uncomfortable with the concept of action at a distance. It didn't explain anything; all it did was to make up something to put in proximity to the affected mass. You could equally easily have said "the two masses interact through space" and your understanding of the situation would be identical; in both cases you didn't understand why they interacted, only that they did. In the meantime, a field became a convenient device for calculating the forces in large systems.
Field lines are used to visualise a field. By the definition of a field field lines show the direction of force at any point. A test particle, initially at rest, will move along a field line. The particles are a test mass, charged particle, or magnetic monopole for a gravitational, electric or magnetic field respectively. Since a magnetic monopole doesn't exist (except as a mathematical device), a magnetic field can be visualised by chains of magnetic dipoles e.g. compass needles or iron filings. Field lines are rarely used to show gravitational fields, perhaps because people already have a good idea of how a test mass would move in a gravitational field.
Here's a description of how to show a magnetic field (http://en.wikipedia.org/wiki/Magnetic_field)
Mapping out the strength and direction of the magnetic field is simple in principle. First, the strength and direction of the magnetic field is measured at a large number of locations. Each location is then marked with an arrow (called a vector) pointing in the direction of the local magnetic field with a length proportional to the strength of the magnetic field.
A simpler way to visualize the magnetic field is to connect the arrows to form magnetic field lines. Magnetic field lines make it much easier to visualize and understand the complex mathematical relationships underlying magnetic field. If done carefully, a field line diagram contains the same information as the vector field it represents. The magnetic field can be estimated at any point on a magnetic field line diagram (whether on a field line or not) using the direction and density of nearby magnetic field lines. A higher density of nearby field lines indicates a stronger magnetic field.
The direction of a magnetic field line can be revealed using a compass. The magnetic field points away from a magnet near its north pole and towards a magnet near its south pole. Magnetic field lines outside of a magnet point from the north pole to the south. This process works even for situations where there is no magnetic pole. A straight currentcarrying wire, for instance, produces a magnetic field that points neither towards nor away from the wire, but encircles it instead.
For examples of field lines see this webpage (above) and Field line (http://en.wikipedia.org/wiki/Field_line).
Electric fields and magnetic fields are dipolar; charges can be +ve or ve, magnetic poles can be N or S, so there can be attraction and repulsion. With gravity, there can only be attraction (there are no ve masses).
In a previous section, I told you that you could determine the time of flight of your projectile, by finding the time it took for the vertical component of the velocity to be the same, but of opposite sign to the initial vertical component (i.e. the projectile lands with the same speed as it was launched). I told you this extra piece of information, so you could get on with solving the problem. You could see that this was reasonable and likely, if not certainly true. It wasn't till I saw this section delivered in class, that I realised that you didn't have the information to show that this was true. You won't get this information till the next section, when we'll do the problem again (and properly).
In your further education, you won't always get a proof. However you need to know where in the stream of teaching, that a proof is needed (whether or not you got one, and whether or not the teacher will give it to you on later asking). You're building a castle of skills and knowledge. It will be based upon a series of proofs and dexterity derived from doing a large number of practice problems. If proofs are missing, you will find your castle built upon sand. It won't make any difference to the teacher if you don't get the proofs, but it's your life. If you don't get the proofs any other way, you should go figure them out yourself. If you know only the result, you can't go onto the next step and you're a technician; if you know the proof, you'll have enough stamina to explore the next step and you're an engineer.
Look for statements, like I made about the projectile's velocity (where I said that the projectile landed with with the same speed as on launch except going in the downward direction). Don't accept them; ask for proof.
With that in mind, I'm going to explain the condition for which the ball returns to the ground with velocity of the opposite sign. (I don't know the proof.)
A conservative force (http://en.wikipedia.org/wiki/Conservative_vector_field) (also called "field" and "vector field") is one in which the work done in moving a particle between two points is independant of the path taken. A corrolary of this, is that the particle on returning to its original position, will have its original energy. A nonconservative force is one in which the difference in energy is dependant on the path taken. A particle moving in a nonconservative field will have a different energy on returning to its original position.
With gravity being a conservative force (or field), then in the vertical component of the projectile problem, the energy of the projectile is dependant only on the difference in (vertical) distance from the starting point. When the particle returns to its starting point, it will have the same kinetic energy (½m.v^{2}). Because of the squaring, you can't tell whether the velocity has changed sign, but you know that the magnitude is the same. From experience and from v=u+at, you know that the velocity of the ball, when it returns to the ground, has the opposite sign. (In a later section we'll prove that the magnitude is the same, but here you're taking my word for it.)
Galileo explored the acceleration of gravity by rolling a ball down an inclined plane (http://www.mcm.edu/academic/galileo/ars/arshtml/mathofmotion1.html) showing that the distance covered by the ball was proportional to the square of the time (s∝t^{2}).
Note  

Galileo used an inclined plane to decrease the acceleration of the falling masses. The problem was that he didn't have a device fast enough, to time objects falling under the acceleration of gravity. Galileo hadn't invented the pendulum clock yet (which wouldn't have helped much, it didn't resolve time into small enough increments i.e. its resolution wasn't good enough) and modern clocks and stopwatches hadn't been invented. In the story I read as a kid, Galileo used his pulse to show that the period of a pendulum was independant of its amplitude. Often when you invent something new, like an inclined plane, the infrastructure to use or test it doesn't exist and you have to invent that too. For the inclined plane, Galileo was pushing the current timing technology, the klepsydra, a clock that drained water at constant rate. The weight of water collected indicated the time. (For Galileo's description of his clock see inclined plane http://galileo.rice.edu/lib/student_work/experiment95/inclined_plane.html). In one of the urls on "inclined plane", to capture the essence of Galileo's accomplishment, students use a computer to time the rolling ball. If Galileo had had a computer, he wouldn't have needed an inclined plane, with its decreased acceleration, and he could have let the balls drop under gravity. 
How far does a constantly accelarating mass fall. Here (again) is the graph of acceleration of a mass falling under gravity.
 ^    a .......................g   _______________________t time> 
Here (again) is the velocity (assumed to have v=0 at t=0; we're using downwards as the +ve frame of reference)
 o  o ^  o   o v  o  o  o o  t, secs > 
From experience (and reading) you know that s=f(t), for a constantly accelarating body, is a parabola. This graph shows the distance fallen as a function of time for a dropped body (drawn as 3 straight lines rather than a real parabola) (the downward direction is +ve).
 o  o ^  o   o s  o  o  o o  t, secs > 
We're now going to derive the equation for s=f(t) and show that it's a parabola.
By dimensions, the area under the v=f(t) graph is (change in) displacement. For constant velocity, by the definition of velocity, the dimensionless constant is 1, so the area under the graph is the actual displacement. The definition of velocity doesn't say anything about distance when the velocity is changing, so we have to figure this out ourselves. Let's see what we can find about the distance travelled (change in displacement) for the mass whose velocity is increasing linearly with time rather than constant.
In the (linear) v=f(t) graph above, let's say that the velocity at t=0 is v=0, while at the righthand end of the graph t=2sec, v=2m/sec. Find the displacement using the time and the average velocity.
v = final velocity u = initial velocity t = time ave_vel = (v + 0)/2 v = 2m/sec t = 2sec therefore s = ave_vel * t = ((v + 0)/2)*t = ((2 + 0)/2)*2 = 2m 
Let's find a general formula for s=f(t) using this average velocity method.
equations: ave_vel = (v + 0)/2 v = u + at (u = 0) therefore v = at s = vt #definition of velocity deriving average velocity=f(a,t) ave_vel = v/2 = at/2 #average vel = init vel + 1/2th increment due to gravity deriving s=f(a,t) s = ave_vel * t = (1/2)at^2 
Note  

For the moment, just consider situations where the object has u=0 at t=0. In this case s=½at^{2}. We'll derive the formula for u!=0 shortly. 
Galileo's experments had u=0, so the equation for his experiments is s=½at^{2}. We've found a formula that agrees with Galileo's data; s∝t^{2}. Using the average velocity to get our result sounds OK to most people, but to a mathematically educated person, deriving a result from the two end points, while ignoring the data in the middle (even if it is a straight line) is suspect. If the two end values were close, then a mathematician would say that the result is close to correct, but when one end is zero and the other end the highest value, you need another method to check your result.
Note  

For a linear v=f(t) the average velocity method we just used, gives the correct answer, but we haven't proven this yet. Just because a result looks right doesn't mean it is right; you have to prove it's right. For 2000yrs people thought that heavier objects fall faster than light ones, and that the earth is flat. 
Let's try another approach; find the area under the v=f(t) curve (a straight line).
 o  o// ^  o////   o////// v  o////////  o//////////  o//////////// o//////////////  t, secs > the graph is a triangle area = height * base/2 height = v (at time t) = at base = t area = (1/2)at^2 
The area has the dimension of distance. But now we have the problem of finding the dimensionless constant.
Let's try another approach. Chop the graph into small intervals. Make each interval short enough that the velocity can be regarded as constant. Then let the intervals become smaller and smaller. See if you get anything useful. First let's look at one interval. Values for the graph below might be t_{end} = 1sec, n = 10
  ^    _ v  /  //  //  //  0 t0 t_end t, secs > 
total time over which we're calculating distance = t_{end}
number of intervals = n
time per interval (the width of the slice) = t_{end}/n
time at interval shown = t_{0}
Now let's find the distance moved during the time slice of duration t_{end}/n at t = t_{0}.
equations:
s = vt #definition of velocity
v = at #definition of acceleration
values:
t_{slice} = t_{end}/n
velocity (regarded as constant) during interval starting at t_{0}
v = a.t_{0} # from definition of velocity
calculation:
s = v*t_{slice} # velocity can be regarded as constant
= (a*t_{0})*t_{slice}
= t_{0}(a.t_{end}/n)
Note  

The distance moved in the interval at t=t_{0} is t_{0}*constant. 
Sanity check 1: dimensions of RHS = a.t^{2} = l.
Sanity check 2: Check by substituting some numbers.
 t_{0} has values 0.0,0.1..1.0. Let's pick something in the middle, say t_{0} = 0.5.
 Let t_{end} = 1sec
 Let n = 10
 Let a = 1m/sec^{2}
The average method says that the distance moved in t = t_{end} secs is 0.5m ((0 + 1)/2 m/sec * 1sec = 0.5m). From the equation above, the distance moved in the time slice at t = 0.5secs is s = 0.5*1*1/10 = 0.05m . Since our timeslice is 1/10 of the total time, the distance we've calculated by the time slicing method is about what we'd expect from using the average method. (We'd accept anything here unless it was out by a factor of 100 or so.)
The particle moves t_{0}*(a.t_{end}/n) metres during the time slice (interval) at t = t_{0}, of duration t_{end}/n secs. Next we want to find the distance moved during each of the time slices and add them. To do this we can add all the t_{0} terms and multiply by the constant. After that we make the slices thinner and thinner and hopefully get something sensible for the total distance moved.
Here's our velocity graph showing all intervals.
 _  _// ^  _////   _////// v  _////////  _//////////  _//////////// //////////////  t, secs > 
(Don't get lost here. We're just finding the area of a triangle, only we're doing it in slices.)
Summing intervals for 0<t_{0}<t_{end}
t_{0} is a series of (real) numbers. For t_{end} = 1sec and n = 10, t_{0} = 0.0,0.1..1.0.
t_{0} = 0*(t_{end}/n),1*(t_{end}/n)..n*(t_{end}/n)
= (0..n)*(t_{end}/n)
The component (0..n) is an arithmetic progression. There are n+1 terms (remember the fencepost error problem). Sum the series using Gauss's method (where we sum the series twice):
0..n n..0  n..n (the sum has n+1 terms) sum of all terms (0..n) twice = n*(n+1) sum of all terms (0..n) once = n*(n+1)/2 Sum of all t0 terms: sum of (0..n)*(t_end/n) = n*(n+1)/2 * t_end/n = (n+1)*t_end/2 (note: as n increases, so does the sum of the t0 terms) duration of time interval: = t_end/n (note: as n increases, the duration of the interval decreases) sum of distances = sum of t0 terms * constant = (n+1)*t_end/2 * (a.t_end/n) = (1/2)a*t_end^2 *(n+1)/n Let n become large, when (n+1)/n becomes 1. s = (1/2)a*t_end^2 use the general symbol t (rather than t_end) for the time at the end of the period s = (1/2)at^2 
We summed the area of a triangle in slices, finding that the area was height*base/2. We already knew the area of the triangle, so this result should be no surprise. What we didn't know was how far a constantly accelarating mass moved. The definition of distance moved only applies for constant velocity. We have no definition for distance moved for an object which is changing velocity. By slicing time into small enough intervals that the velocity doesn't change during each interval, we found the distance moved. The distance is the area of the triangle, but we couldn't prove it with the average velocity method.
The notation we used is clunky and the derivation tortuous (at least the first time). However cutting the graph into small slices is how calculus works. You have to understand what happened in this example to understand calculus. Once the result and method becomes familiar, then you can use the simpler notation of calculus to get the same result. For the moment, you should use the slice method, to become familiar with slicing, then when you get to calculus, you'll just see calculus as a change in notation, rather than some big hairy deal.
Note  

In calculus, the process we just did (finding the area under a line) is done by integration. Here we found the area by summing a series. The sums of series are only known for a limited number of series, so the method we used isn't a general method for finding the area under an arbitary curve and isn't the method used in calculus. In calculus, integration relies on knowing that integration is the reverse of differentiation. Differentiation is simple and can be performed on any function, giving a table of functions and their differentials. The reverse process, of finding a function for any arbitary differential is not known. If the differential is in the table, then you can do the integration. If the differential isn't in the table, then you can't do the integration in analytic form (i.e. you can't produce an algebraic formula for it). You can always find a numerical value for the integral using numerical integration. There are no general formulae for integration; either the answer is known from differentiation or it's not known. People have been working on calculus for 400 yrs and it's almost certain that every function has been differentiated. So "not known" likely means "doesn't exist". 
Note  

from "e: The Story of a Number", Eli Maori, 1994 Princeton U. Press. ISBN 0691058547, pp6268. Before the advent of calculus, geometric methods had found the area under polynomials (i.e.y=x^{n}) by Fermat and under the hyperbola by Saint Vincent. 
Here's what we've accomplished in this section
Again, let's plot s=f(t) for an object moving under constant accelaration (it's a parabola). (The ascii version is 3 sections, each a straight line.)
 o  o ^  o   o s  o  o  o o  t, secs > 
Example: a ball is dropped on a planet with an accelaration due to gravity of 1m/sec^{2} (g on the Moon is 1.62m/sec^{2}). What is the total distance moved at the end of each second and the distance moved in each particular second ^{[55]} ? Notice that the difference in distances moved at the end of each interval is a linear series with the ratio 1,3,5,7,9.. What's a more familiar term for the difference in distances moved at the end of each interval ^{[56]} ?
At what time is the ball 18m below the dropping point ^{[57]} ? Again we used an equation continuous in t. The position of the ball is completely determined by the velocity at t=0 and the accelaration. If the ball is moving under an accelaration of 1m/sec^{2} and is not moving at t=0, then at both t=6sec, t=+6sec the ball will be 18m below the dropping point.
Experience tells us the the ball will be moving up at t=6sec and moving down at t=6sec. Calculate the velocity at both these times ^{[58]} .
What if the object has an initial velocity u before it starts accelarating? You sum the area under the v=f(t) graph the same way, in slices. This time the triangle is sitting on a pedestal. The area under the line (and hence the distance moved) is the sum of the area of the triangle (at the top of the diagram, filled with "/" characters) and the pedestal below (filled with "\" characters).
 o  o// ^  o////   o////// x  o ///////  o /////////  o /////////// x0 !o \\\\\\\\\\\\\\\ \\\\\\\\\\\\\\\ \\\\\\\\\\\\\\\ _______________ t, secs > 
You can calculate the distance moved yourself. Here's the answer.
s = ut + (1/2)at^2 
Previously (launch angle for projectile) we found the angle which maximized the distance of a projectile, although I had to inject extra information for you to solve the problem. With this new equation, the extra information is built in. Find the time of flight of a projectile launched at angle=ϑ, with initial velocity u (hint ^{[59]} ) ^{[60]} . We get the same answer as before. Notice we have an extra solution, t=0 (the equation is quadratic in t). We should see this solution in any properly handled problem. (We didn't see this solution in our first attempt.)
Let's look at where we are.
We're trying to find the velocity as a function of h. Once we find that, we can find the dimensionless constant for K.E.∝mv^{2}.
We have the equations
Let's eliminate t from these equations.
v = u + at (1) s = ut + (1/2)at^2 (2) from (1) t = (v  u)/a (3) substituting (3) in (2) s = (v  u)u/a + (1/2)a(v  u)^2/a^2 2as = (v  u).2u + (v  u)^2 = (v  u).(2u + v  u) = (v  u).(v + u) = v^2  u^2 the usual form of the equation v^2 = u^2 + 2as 
We now have the velocity of an object after accelerating through a distance s and we can calculate the dimensionless constant for K.E.
object falls through height = h P.E. lost = mgh equation: v^2 = u^2 + 2as initial condition: u = 0 on falling height = h velocity = sqrt(2as) = sqrt(2gh) therefore v^2 = 2gh K.E. = k.mv^2 #k is dimensionless constant = k.m(2gh) = 2k.mgh but K.E. gained = P.E. lost so mgh = 2k.mgh therefore k=(1/2) Thus K.E. = (1/2)mv^2 
Note  

FIXME  any connection between 1/2at^2 and 1/2mv^2? (both triangles?). The (1/2)mv^2 area comes from the graph y=at=v when x=t. I guess 1/2mv^2 is the area under the graph y=mv, when x=v. Where that comes from isn't obvious. 
The dimensionless constant for our K.E. formula is ½. We now have the formula for K.E.=½mv^{2}
You let a ball drop off a 10m high building. How fast is it going when it hits the ground ^{[61]} ?
Objects falling in air encounter aerodynamic resistance. The resistance depends on the area of the object and the coefficient of drag. A spherical object has a relatively high coefficient of drag, while a teardrop shaped object has a lower coefficient of drag. A skydiver going head first presents a low area to the air, while if falling horizontally presents a high area to the air. Because air presents a resisting force to objects falling through it, they cannot accelarate indefinitely  they stop accelerating when the resisting force is equal to the force of gravity. The velocity at this point is called terminal velocity (http://en.wikipedia.org/wiki/Terminal_velocity).
Here are some terminal velocities from Fluid Friction (http://hyperphysics.phyastr.gsu.edu/HBASE/airfri2.html)
Falling object Mass Area Terminal velocity Skydiver 75 kg 0.7 m2 60 m/s 134 mi/hr Baseball (3.66cm radius) 145 gm 42 cm2 33 m/s 74 mi/hr Golf ball (2.1 cm radius) 46 gm 14 cm2 32 m/s 72 mi/hr Hail stone (0.5 cm radius) 0.48 gm 0.79 cm2 14 m/s 31 mi/hr Raindrop (0.2 cm radius) 0.034 gm 0.13 cm2 9 m/s 20 mi/hr 
For a small enough object (water droplets in a cloud), the terminal velocity is smaller than any updrafts and the drops stay suspended indefinitely. From USGS Water Science (http://ga.water.usgs.gov/edu/watercyclecondensation.html) water droplets are about 10microns (µm) and using the calculator at the Fluid Friction site above, the terminal velocity is 0.6m/s.
How far would a ball have to drop in a vacuum to achieve the terminal speed of a skydiver in air ^{[62]} ? The height of the deck above the water of the (GW Bridge (http://en.wikipedia.org/wiki/George_Washington_Bridge) in the middle is 65m.
You throw a ball upwards at 14m/sec. What is its velocity 7.5m above the launch point ^{[63]} ? Notice that the ball 7.5m above you can be going up and can be going down. When the ball is 15m above you, what is its velocity ^{[64]} ? What happened here ^{[65]} ? What's the highest point in the trajectory ^{[66]} ?
Work is defined as F*s. This is just one formulation of energy/work/heat. It was the first. People wanted to measure the amount of work needed to pump water, bore canons, or grind flour and F*s gave it to them. The idea of energy (kinetic or potential) came later. The understanding that heat was energy came later yet. Since energy and work are the same, use the equation derived in this section to show that the work done to accelerate a mass in the horizontal direction, is equal to the change in kinetic energy. Assume the initial velocity is u, the final velocity is v, the particle moves a distance s while accelerating at a ^{[67]} .
On earth the record high jump (http://ihttp://en.wikipedia.org/wiki/High_jump) is about 2.5m. The men's pole vault record (http://en.wikipedia.org/wiki/Men's_pole_vault_world_record_progression) is about 6m. If these events were conducted on the moon (gravity 1/6th of earth's), what would the records be? First list your assumption(s) about these events (take as given that athletes perform fine in a vacuum and that air resistance doesn't affect the record). Here's my assumption(s) ^{[68]} . Assume that the record on earth is s, find the record on the moon. Insert numerical values the formula you get. Here's my answer ^{[69]} .
This is a standard question given to students who've just learned v^{2}=u^{2}+2as. The assumption you missed is that the s in the equation is the height of the centre of mass of the object at zero (vertical) velocity, while the record is the height of the bar cleared by the athlete. They are not the same. By agility (e.g. Fosbury Flop http://en.wikipedia.org/wiki/Fosbury_Flop), the athlete clears a bar higher than his/her centre of gravity. Assume a high jumper gets an extra 0.5m through agility, what would be the high jump record on the moon ^{[70]} ?
Often this question is posed as "if you can throw a ball up height x on earth, how high would it go on the moon". In this case the simple answer is correct.
On earth, the pole vaulter lands in a bed of rubber, to cushion the shock of falling from the bar at 6m. On the moon, how much greater is the shock of landing from a height of 36m ^{[71]} ?
Assume no agility in the high jump or pole vault. By what ratio does the time in the air (vacuum) change on moving the event to the moon ^{[72]} ?
If bipedal life exists on Jupiter (where the acceleration of gravity is 2.5 times that of earth), how much faster would its nervous system have to be, relative to ours, for it to be able to stay upright while walking ^{[73]} ?
Let's look at the 2D projectile problem again. Find the maximum height reached as a function of launch angle ϑ (the initial vertical velocity is u_{vert}=u.sin(ϑ)) ^{[74]} . Note that the maximum height occurs for ϑ=90°. Remember that the maximum distance occurs for ϑ=45°. What is is the maximum height reached for ϑ=45° compared to the height reached for ϑ=90° ^{[75]} .
The P.E. is a function of height. For a launch angle of 45° at the top of the arc what is the ratio P.E./K.E. ^{[76]} ?
For any launch angle ϑ, is the maximum height reached proportional to the ball's initial velocity or to the ball's initial energy (remember the initial energy is all kinetic)? Derive the height as a function of the initial parameter you choose ^{[77]} . You could have made a reasonable guess as to which it was, without knowing any of the equations for motion under constant acceleration. How would you have made your guess ^{[78]} ?
We found that the distance travelled for the projectile is s=u^{2}sin(2ϑ)/g. Is the distance travelled proportional to the initial velocity or to the initial energy of the ball? Derive the distance as a function of the appropriate parameter ^{[79]} . Could you have guessed whether distance was a function of initial velocity or initial energy, without knowing the equation for the distance ^{[80]} ?
We've derived the 3 equations which describe the motion of a particle under constant acceleration. The parameters for the equations are u,v,a,t,s. You need to remember these 3 equations.
v = u + at #doesn't involve s
s = ut + ½at^{2} #doesn't involve v
v^{2} = u^{2} + 2as # doesn't involve t
Each equation leaves out one of these parameters; e.g. in the first equation, the distance s is ignored. It will have some finite value, but you don't care what it is. You can calculate it from your input data using one of the other equations.
The first equation is derived from the definition of acceleration. The 2nd equation is derived from the first, for constant acceleration, by cutting time into small enough slices, that the velocity is constant, and then summing the distance moved in each slice. The 3rd equation is derived from the first two, by eliminating t.
In a situation where one variable is allowed to vary without constraint, the variable is called an unconstrained variable. Each of the 3 equations leaves a variable unconstrainted.
The equation v = u + at describes a situation where s is unconstrained. If u = 0 and we want to acheive the same v we can vary a,t such that the product a.t = constant. What effect do you imagine (from experience) changing a,t, while keeping the product a.t constant (say halving a while doubling t), will have on s?
The answer wasn't obvious to me. I had to look at the equations. From s = ut + ½at^{2}, we can see that s will double. From v^{2} = u^{2} + 2as, we get the same answer; the distance will double.
Unless you know something about the system, you can expect that unconstrained variables will change when other inputs are changed. There is a branch of calculus called "Calculus of Variations", which optimises the values of unconstrained variables, by varying the constrained variables. Example: can we optimise the profit from making cookies by varying costs of ingredients and the taste, texture of the cookies?
5th equation in "a"
In the set of 3 equations above, there isn't an equation which leaves u unconstrained. There is such an equation, but noone uses it. You can derive everything in it from the 3 equations above and the situation it describes isn't all that common. However for the exercise, derive the equation describing the relationship between v,a,t,s, which leaves u unconstrained, i.e. eliminate u ^{[81]} .
Example: the world record time for drag racing (driving a car in a straight line for ¼mile≅400m) is 4.428secs crossing the finish line at 541kmph. What's the acceleration (assumed constant) ^{[82]} ?
How is u is not involved here? Clearly in this example, we already know that u=0 (the drag racer will be disqualified if u != 0). We're after a set of circumstances where u is unconstrained.
Let's look at our example more closely. We've assumed that a is constant. This is unlikely with the power/torque curve of a gasoline/petrol engine. Let's calculate u from this information, using the other 3 equations. Here's my answer ^{[83]} . There are two different answers and neither is 0. Presumably the nonlinearity of acceleration accounts for this.
For the drag race record, let's look at the uncontrolled variable v in the more familiar equation s = ut + at^{2}; assume s = 400m, u = 0, a = 25m/sec^{2}, t = 4.4sec. What is the speed (velocity) at the end of the run ^{[84]} ? The real answer is 150m/sec. What constant acceleration would be needed to achieve a velocity of 150m/sec in 4.4sec ^{[85]} and what distance would be covered ^{[86]} ? The distance is short of 400m. Clearly the constant acceleration model doesn't fit here. However we're interested in the unconstrained variable u. Let's see what happens to u, assuming constant acceleration, if we change the acceleration and time in the 4th equation s = vt  ½at^{2}, with the convenient values v = 128m/sec, s = 512m.
FIXMETable 1. Effect of a,t on the unconstrained variable u while maintaining v = 128m/sec, s = 512m
t,sec  a, m/sec^{2}  u, m/sec 

1  768  896 
2  128  384 
4  indeterminant  indeterminant 
16  12  64 
32  7  96 
64  3.75  112 
As spotted by Zander, there is a 5th equation in which a is unconstrained. Derive the 5th equation.
^{[87]}Here we introduce vector notation: vectors are a better bookkeeping mechanism. All students of mechanics start with the cartesian system. For simple systems (1D where there are no vectors, or multidimensional systems resolved as components, which are assembly of 1D systems, where again there are no vectors), the cartesian system works quite well. However you do need to keep track of your frame of reference. Using the cartesian system, you'll find that similar problems are solved with different frames of references, leading to ad hoc'ery. You'll remember formulae with opposite signs in similar problems, but you won't really be clear as to why. When keeping track of directions and values (+ve,ve) by hand, the chances of mistakes are high. The vector system is more complicated to setup, but extends into 3D systems easily and reduces the sources of errors.
a scalar has magnitude but no direction; e.g. temperature, mass, volume, time, work, power, volts, amps, ohms. A scalar can be +ve or ve.
a vector has magnitude AND direction. Unlike a scalar, the magnitude of a vector is always +ve. In the frame of reference, the vector can point in the ve direction, but we don't say that the vector is ve. Instead we give its direction and magnitude. Familiar vectors are distance (when associated with a direction), velocity, accelaration (and gravitational field), force, electric field (volts/cm), and magnetic field (http://en.wikipedia.org/wiki/Magnetic_field) which aligns a compass (units are tesla in SI and gauss in cgs units).
Note  

(since magnetism is not talked about a lot...) The gauss is itself defined as one maxwell/cm^{2} (gauss http://en.wikipedia.org/wiki/Gauss_(unit)). The earth's magnetic field is 0.30.6gauss (about 1 gauss for a back of the envelope calculation) (3060μteslas). An MRI machine has a magnetic field of about 1T ( How strong are the magnets in an MRI machine? http://www.howstuffworks.com/question698.htm). The largest magnetic field produced on earth (in a laboratory) was 2800T (http://en.wikipedia.org/wiki/Magnetic_field). For more on magnetism see Hitchhiker's guide to magnetism http://www.irm.umn.edu/hg2m/hg2m_index.html. 
For the scalar of work, the direction of the vectors F (force) and s (distance) do not have to be colinear (i.e. they can be at an angle). Say a particle is confined to a rail allowing it to only move left or right. A force F acts on the particle at an angle ϑ. The work done on moving the particle a distance s is F*cos(ϑ)*s.
F . . . theta . > s 
You can think of the work as either
The work done F*cos(ϑ)*s, (where ϑ is the angle between the two vectors) is called the dot product of two vectors and written F⋅s. The dot operator, ⋅, multiplies the magnitude of the two vectors by cos(ϑ) and returns a scalar.
What work is done by a force F operating at right angles to the movement of a particle ^{[88]} ?
We're familiar with a force acting in the direction of movement (accelerating a car, a jet engine pushing a plane). In this case, the engine does work on the car/plane, raising its K.E. (or P.E.). Since energy is conserved, the engine has to loose energy. We normally think of an engine as being a generator of energy. This is a fiction; it's really converting potential (chemical) energy into kinetic energy (and heat). To show that an engine is a converter of energy, rather than a generator, we can leave it running for a while and eventually it will run out of potential energy and not be able to produce any more kinetic energy.
How does an engine arrange to loose energy? If the angle ϑ=180°, between the force and the movement, then work is ve. An engine produces a force in the opposite direction to its movement. A jet (or rocket) engine throws massive objects (molecules of gas) at high velocity out the back of the plane.
<front back> F>/ m> \o  <F /\  O O O O <s 
This diagram shows you on a frictionless rail car, throwing massive objects (m) off the back. This requires you to exert a force F on the massive objects. Since you are wearing friction soled shoes, Newton's 3rd law of motion says that your shoes will push on the bed of the rail car with a force F in the opposite direction. The engine (you) is exerting a force in the opposite direction to the movement, meaning the engine (you) is loosing energy.
You can throw massive objects off the back of the rail car in a vacuum. It will work even better than in the atmosphere (since there's no air resistance to moving the rail car). Goddard's (http://en.wikipedia.org/wiki/Robert_H._Goddard) proposal to fly to the moon with a rocket was greeted by derision by an editorial in the NYT
It expressed disbelief that Professor Goddard actually "does not know of the relation of action to reaction, and the need to have something better than a vacuum against which to react"
While the engine (you) is loosing energy, the railcar is gaining (kinetic) energy. The soles of your shoes are pushing on the railcar, in the direction it's moving. With ϑ=0, F⋅s is +ve. The important part of an engine, as far as the railcar, car, or rocket is concerned, is the motor mounts. (The car doesn't care what's pushing on the motor mounts.)
A plane/car can loose energy by braking  the force acts in the opposite direction to the direction of movement. Usually the energy is converted to heat, but could be used to run a generator to charge a battery, or converted to kinetic energy somewhere, to be later used to accelerate the car (this isn't practical at the moment). Generators in power stations convert hot steam to cool steam (loosing energy) and accelerate electrons (producing electric power).
In the cartesian frame of reference, P.E=mgh (remember the numerical value of g is ve). Why the ve sign? Let's think about the sign of P.E. of an elevated object. If a massive object is at height, it can do work when it looses height; it can raise another massive object, or it can generate sound and heat when it lands. The object at height has +ve P.E. When we raise the object to its height, we push upwards and the object moves upwards; i.e. we do +ve work on the object (using the dot product formalism, explain why is work +ve ^{[89]} ?) and the object at height must have more P.E. When the object falls, the force acting on the center of mass is pointing downwards, and the object moves downwards, so again ϑ=0 and the work done by the falling center of mass (converting P.E. to K.E.) is +ve.
Now that we have an idea of what's going on, let's change to the vector formalism.
force of gravity acting on center of mass F=mg. (g is a vector pointing downwards.)
In the cartesian frame of reference, the force is pointed in the ve direction. In the everyday world we can say that the value of the force is ve. However we can get into trouble unless we think clearly here. In the vector world, where vectors have magnitude and direction (but not sign), the force is just pointing in a particular direction. Scalars have signs, vectors have directions. We don't get anything with a sign until we have a scalar, such as work, which we get from the dot product of force with another vector (like direction). If you want to be sure of getting the sign right, you should let the vector math handle it for you.
o  F=+mg v ^  F=mg 
Let's say we want to raise the massive object  we'll have to push on it. We'll need a force infinitesimally larger than mg, but pointing in the opposite direction to mg (up). Our force then is F=mg (by Newton's 3rd Law of Motion). Since we're using vector notation, we don't need to know where g is pointing. Here g happens to be pointing downwards, but it could be up, down, or at some angle. (In daytoday conversation, we can say that the force is pointing in the ve direction, as long as we understand that we really mean it's pointing at 180° in the frame of reference.) (At intermediate steps, we can do checks, to make sure that we're doing something sensible.)
let's check the force of gravity on the object: it's F=+mg. g points down, so F=+mg points down too.
let's check the force we push on the object with: it's F=mg. g points down, so F=mg points up.
What work is done raising the object through a distance h? W=F⋅h=mg⋅h. In the vector notation, to do the dot product, the direction of the force and the direction of distance moved are not used, only the angle between them. F=mg points up (since g points down), and h points up. The angle between the force and the distance moved is ϑ=0°, so W=F⋅h is +ve i.e. we have to do (+ve) work to raise the massive object against the force of gravity.
Let's look at the the massive object. It's exerting a force F=mg. It's being moved a distance h by a force exerted from the outside. What work is the massive object doing on the external force? W=F⋅h=mg⋅h. Since g is facing downwards, and h is facing upwards, ϑ=180°, then cos(ϑ)=1 and the massive object has done a ve amount of work on the system pushing against it. i.e. the pushing system has less energy now.
What's the change in energy of the system? Change in P.E. of raised object + change in E. of the pushing system = (mg⋅h) + (mg⋅h) = 0.
Note  

Since most of the following sections have examples where vectors are colinear, the cartesian system will be used. However you can use vectors, in which case you should use vector equations like v^{2} = u^{2} + 2a⋅s. 
In the previous section we showed using the dot product, that work was scalar and used examples with P.E. Here we do the same thing with K.E. using the vector formalism and dot products.
W = K.E. = ½mv⋅v
v is a vector (it has magnitude and direction) and is always parallel to itself, so ϑ=0, and therefore v⋅v = v^{2}cos(0) = v^{2} . For K.E. you can ignore the fact that v is a vector and just square the magnitude of v (as you've always done).
A conservative (vector) field (http://en.wikipedia.org/wiki/Conservative_vector_field) is one in which the change in energy of a particle going between two points is independant of the path taken. A consequence is that the total energy of a particle is conserved, when moving under the influence of conservative forces. Energy is conserved as a loss of potential energy is converted to an equal quantity of kinetic energy or vice versa. It seems reasonable (although I can't prove it) that the reverse is also true; a body moving in a vector field whose energy is conserved, is moving in a conservative vector field. We're going to show that energy is conserved for a particle moving under the acceleration of gravity, even if this doesn't show that gravity is a conservative field (the heading for this section).
Note  

Other people have shown that gravity is a conservative field. I know that (gravity_is_a_conservative_field ⇒ energy_is_conserved_in_a_gravitational_field), but I can't show that (energy_is_conserved_in_a_gravitational_field ⇒ gravity_is_a_conservative_field). Since (energy_is_conserved_in_a_gravitational_field ⇔ gravity_is_a_conservative_field) then someone has already shown that (energy_is_conserved_in_a_gravitational_field ⇒ gravity_is_a_conservative_field) but I don't know how the basis of it. 
We'll first show that energy is conserved for the 1D problem (a ball moving only in the vertical plane), then do it for the 2D problem (the ball launched at an angle ϑ).
The potential energy (P.E.) is measured with respect to a reference position i.e. the difference in P.E. between its current position and some arbitary reference position. There is no absolute P.E. If you throw a ball up (by imparting kinetic energy (K.E.)), then the ball will ascend, converting K.E. to P.E., reaching a height when its velocity is 0. At this position it will have +ve P.E. (with respect to the throwing point, i.e. the ball can do work if it's allowed to fall through gravity back to the throwing point), but no K.E.
If instead you dropped the ball, it would gain K.E. and loose P.E. (i.e. the P.E. becomes ve and work needs to be done to return the ball to the dropping point).
Note (Note the presence or absence of dot product symbol ⋅ in these next formulae.) People often exchange frames of reference in midstream without being aware of it. The formula for P.E. is often stated (in nonvector notation) as P.E.=mgh, where h is the distance the particle can drop. Unstated is that the frame of reference is +ve downwards, automatically giving the correct answer of a +ve number. If you're being a bit more rigourous, you'll choose up as the +ve direction, and you'll use vectors. In this case P.E.=mg⋅. Just be aware of whether you're using vectors, or are in a 1D situation, where you're just using magnitudes and whether up or down is +ve.
For this problem, let
You'll calculate P.E. and K.E. From there you'll calculate total energy (E_{tot} = K.E.+ P.E.) of the ball after it leaves your hand.
Let's calculate E=f(h). After watching the ball's up and down motion, this would be a likely first approach.
P.E. is maximum for maximum height Here's the graph of P.E.=f(h).
1/2mu^2  o  o  o PE  o  o o  h 1/2.u^2/g 
What's the slope of the line ^{[91]} ?
Here's the graph of K.E.=f(h).
1/2mu^2 o  o  o KE  o  o  o  h 1/2.u^2/g 
What's the slope of the line ^{[93]} ?
For E=f(h), here's my answer for the total energy (E_{tot} = P.E. + K.E.) ^{[94]} . (With the slopes of the lines for K.E., P.E. being the ve of each other, we should expect that the sum of the two lines is a constant.) We find that no matter what the initial velocity (whether you throw the ball up or down, or let it drop), the total energy of the ball remains the same; the kinetic energy at the moment of launch. As height changes, K.E. is converted into P.E. (and back again). Energy is conserved.
We can plot a new diagram E = f(K.E.,P.E.). We know that K.E + P.E. = ½mu^{2}, so the graph with K.E. on the xaxis and P.E. on the yaxis is a straight line of what slope, and intersecting the P.E., K.E. axes where ^{[95]} ?
Here's the P.E./K.E. graph (caution  this is ascii art: the slope should be 45°)
1/2.mu^2 o  o  o PE  o  o  o  KE 1/2.mu^2 
What physical principle requires slope=1 ^{[96]} ?
You can make (KE,PE) diagrams for any system where KE and PE are interconverted e.g. oscillating systems (pendulum, mass on a spring). Because of conservation of energy, all diagrams are the same; i.e. there is only one (K.E.,P.E.) diagram. The only difference between the diagrams is the way the particle moves along the line (linear, 2nd order or sinusoidally as a function of some parameter e.g. time, height, velocity, angle...).
From thinking (we'll do the calculation shortly), what's the trajectory (locus) of the ball along this line as a function of h? i.e. what point on this graph corresponds to h=0 (start, end), h_{max} = ½u^{2}/g ^{[97]} ? If there was no ground to stop the ball at the end of the flight, what would be the location of the ball on this graph as it descended through the ground level ^{[98]} ?
Again just using thinking at the moment (we'll calculate the result shortly), what would be the speed of the locus along this line as a function of h? Is it linear with height or does it speed up in some places ^{[99]} ?
Mathematicians are quite happy to shift (translate) axes if it simplifies the analysis. The ball doesn't start at the origin of our graph, so if we want to describe how far its moved, we have to repeatedly say "from its original position". Let's translate the axes so that the ball's starting position is at the origin (we only need to translate the K.E. axis). Call the new coordinates (ΔKE,ΔPE) (the Δ is Delta), representing the change in K.E.,P.E. from their initial values. Here's our new graph
h=1/2.u^2/g o  1/2.mu^2 o  o  dPE o  o  h=0 o  1/2.mu^2 dKE 
Using K.E.=f(h), P.E.=f(h), find the coordinates of the ball on the (ΔKE,ΔPE) diagram as a function of height ^{[100]} ? Figure out (at least in principle, we'll do the calculation shortly) how to calculate the locus of the ball on the (ΔKE,ΔPE) graph as a function of height ^{[101]} . The line here is straight, simplifying our analysis. If the line was curved (e.g. a parabola), we would need some method of finding the distance moved along the curved line.
ΔKE is the change in K.E. from the launch value. The initial value is 0 (by definition). Since g is ve, ΔKE decreases linearly with increasing height, till at h = ½u^{2}/g it becomes ½mu^{2}. In the (K.E.,P.E.) and (ΔKE,ΔPE) diagrams, the ball moves uniformly with h. ΔKE then increases again, becoming 0 when the ball returns to the starting point on the (K.E.,P.E.) and (ΔKE,ΔPE) diagrams (note: this point on the diagram, corresponds in physical space to the end of the flight).
Here's a diagram of the projection of the locus of the ball, on the abscissa, as a function of height. The ball starts at the origin of this diagram, moves downwards (decreasing KE) and to the right (increasing height) till it reaches its heighest (physical) point at the bottom right. The ball then returns along the line going upwards (increasing KE) and to the left (decreasing height). The ball when it returns to the origin has reached the end of its flight. In the absence of ground, the ball will continue forever to the upper left, with KE increasing without limit.
below o  above ground o  ground o  o  abscissa (dKE) more KE o  o  h o start,end  o  o  o top less KE ground level 
Here's the velocity of the ball as it moves along the line.
e t below n o above ground n p ground  ooooooooooooooooooo mg   vel (d(dKE)/dh)   h    oooo +mg  s t t o a p r t ground level 
In the (ΔKE,ΔPE) diagram, the ball initially moves leftwards with constant velocity as a function of h, i.e. it has ve velocity of +mg. (Note: in physical space, the ball's velocity is also linear with respect to height.) When the ball reaches h_{max} = ½.u^{2}/g the ball reverses direction, now having +ve velocity in the (ΔKE,ΔPE) diagram. (Note: there is a discontinuity in velocity of the ball along the (ΔKE,ΔPE) diagram. There is no discontinuity in velocity in real space). After reaching the end of the flight, in the absence of ground, the ball continues to move along the (ΔKE,ΔPE) diagram to the left without limit.
Since the ball starts with P.E.=0, the P.E. and ΔPE axes are the same. ΔPE=P.E. is the change in P.E. from the launch value. The initial value is 0 (by definition), it increases linearly with height, till at h = ½u^{2}/g it becomes ½mu^{2}. ΔPE then decreases, becoming 0 when it returns to the starting point on the (K.E.,P.E.) and (ΔKE,ΔPE) diagrams (note: in physical space it's at the end of the flight).
There is a discontinuity in velocity of the projection along the ordinate, as for the projection of the abscissa.
There's a minor problem with using the distance from the origin, since taking the sqrt() can give us a ve result. Intuitively, if we regard the ball in the 2nd quadrant of the (ΔKE,ΔPE) diagram as being a +ve distance from the position at h=0, then we might say that the ball in the 4th quadrant has a ve distance from the position at h=0. However distances are always +ve; if the distance from NY to Boston is +ve, is the distance from NY to DC ve? Thus we can't use the distance along the line in the (ΔKE,ΔPE) diagram to say whether KE or PE has increased (or decreased). The plot of the distance of the ball from the starting point will have a discontinuity in slope at t_{end}. We could handle the matter manually and make the distance ve for h<0. It's simpler to use the projections on the axes (you'll have less ad. hoc'ery to explain).
Here's the distance diagram for the position of the ball on the (ΔKE,ΔPE) diagram as a function of h, using the convention that distance can only be +ve.
 x  x  o top distance  o  o o start,end  height 
The ball starts at the origin in the (ΔKE,ΔPE) diagram, moves to the top of the flight, then returns to the origin and then in the absence of ground, moves back towards the top position and then continues (the "x"s) without limit.
Here we derive the K.E. and P.E. as a function of time. The equations for E=f(t) are more complicated than for E=f(h) (P.E. is a simple function of height).
You've already determined t for P.E._{max} (it's half the time for the flight i.e. half the time between the two points with P.E. = 0). Derive t for P.E._{max} using calculus to determine the time when P.E. is stationary. Here's it is ^{[106]} . You've already done this, but again substitute the value of t for P.E._{max} into P.E. = f(t) to find the value of P.E._{max} ^{[107]} ? Confirm that this is a reasonable value for P.E.
Here's the graph P.E.=f(t).
 (u/g,1/2.mu^2)  o o  o o P.E. o o  o o o o  t 
You already know that the minimum K.E. occurs at t = u/g. Use calculus to rederive the same result: Differentiate K.E. wrt t to find t for stationary K.E. (assume the stationary value is a minimum). Then find the value for K.E._{min} ^{[109]} .
Here's the graph K.E.=f(t).
 o o (1/2.mu^2)  o o K.E. o (u/g,0) o  o o oo t 
For E=f(t), here's my answer for the total energy (E_{tot} = P.E. + K.E.) ^{[110]} .
We now have K.E. and P.E. as functions of time and can plot the (K.E.,P.E.) diagram.
This is the same diagram as was derived for E=f(h), but this time marked with the location according to time. Here's the trajectory (locus) of the ball along this line as a function of t, showing t=start, t=u/g (half the time of the flight at the top of the parabola), t_{end}
1/2mu^2 o t=1/2  o  o PE  o  o  o t=start,end  KE 1/2mu^2 
Since we're looking at the same diagram as for the E=f(h) section, the ball moves similarly, starting at bottom right, then moves up the line to top left (at the top of the flight) and then returns to the bottom right (at the end of the flight).
(remember it's the same diagram as we saw with E=f(h)) If there was no ground to stop the ball at the end of the flight, what would be the location of the ball on this graph as it descended through the ground level ^{[111]} ?
Just using thinking at the moment (we'll calculate the result shortly), what would be the speed of the locus along this line as a function of t? Is it linear with time or does it speed up in some places ^{[112]} ?
Let's translate axes as we did for E=f(h) so that the ball starts at the origin. Here's our new graph (now with the locus of the ball parameterised in t, rather than h).
o t=1/2  1/2.mu^2 o  o  dPE o  o  o t=start,end  1/2.mu^2 dKE 
Using K.E.=f(t), P.E.=f(t), find the coordinates of the ball on the (ΔKE,ΔPE) diagram as a function of time ^{[113]} ? Calculate the locus of the ball along the line as a function of time, using the same methods as for E=f(h).
ΔKE is the change in K.E. from the launch value. The initial value is 0 (by definition), it decreases in a parabola with time, till at t = u/g it becomes ½mu^{2}. In the (K.E.,P.E.) and (ΔKE,ΔPE) diagrams, the ball moves slowly when it's near the top left point. ΔKE then increases again, becoming 0 when the ball returns to the starting point on the (K.E.,P.E.) and (ΔKE,ΔPE) diagrams (note: this point on the diagram, corresponds in physical space to the end of the flight). At large t, ΔKE increases with t^{2}.
Here's the diagram of the projection, on the abscissa, of the locus of the ball in the (ΔKE,ΔPE) diagram as a function of time.
  below ground proj.  o abscissa o  o +ve dKE  o  o  o start t o end  o o  o o ve dKE  o o top less KE above ground  
Since the ball starts with P.E.=0, the P.E. and ΔPE axes are the same. ΔPE=P.E. is the change in P.E. from the launch value. The initial value of ΔPE is 0 (by definition), it increases in a parabola with time, till at t = u/g it becomes ½mu^{2}. ΔPE then decreases, becoming 0 when the ball returns to the starting point on the (K.E.,P.E.) and (ΔKE,ΔPE) diagrams (note: in physical space it's at the end of the flight). If there was no ground to stop the ball, at large t ΔPE decreases with t^{2}.
There is the same problem using distance along the line as previously (for the (ΔKE,ΔPE) diagram for E=f(t)) in that the distance formula gives +ve and ve distances, but the convention is that distance is always +ve.
Now let's calculate E = f(v).
The result isn't quite as informative as the equations for E=f(t), E=f(h). Let's see what we can find from this equation. For what v is P.E. = 0 ^{[118]} ? After our previous work, these answers are obvious, but they may not be if we'd just started characterising a new system (the 2nd answer, v = u, wasn't obvious in our earlier attempts to characterise the problem). When is P.E. a maximum ^{[119]} ? Well then you may ask, when (time wise) is v^{2} a minimum? We need one or more of our 3 equations to find this. However we do know the minimum possible value for v^{2} (it's 0), giving P.E._{max} = ½mu^{2} . In this situation it's possible to convert (all) the initial K.E. to P.E. (this isn't true in the 2D case).
What is the name of the curve P.E. = f(v) (it has the equation P.E. = ½m(u^{2}  v^{2}) ) ^{[120]} ?
Is the parabola symmetric about the v = 0 axis ^{[121]} ? Use the quadratic formula to show that parabolas without any first order terms are symmetric about the yaxis ^{[122]} . Is the P.E. parabola concave downwards or upwards? Is the P.E. at v = 0 a maximum or minumum ^{[123]} ?
Here's the parabolas for P.E., K.E.
x oo x PE=o x o  o x KE=x xo  ox ox  xo o x  x o oxxo v> 
Since both P.E. and K.E. parabolas have the same constant (±½m) infront of the v^{2} term, they are mirror images through a horizontal line (what is the height of the horizontal line ^{[125]} ?).
Note  

These graphs being parabolas, you may think you're looking at graphs of the flight of the ball. As it turns out, parabolas describe the flight path and E=f(t  h  v) because the equations are 2nd order, not because the various parabolas represent the same thing. Note that that abscissa is v not s or t and that the point v = 0 is in the centre of the graph. On this graph, the LHS has v ve. When/where in the flight is v ve, +ve^{[126]} ? 
In the E = f(v) graph, the ball starts at the RHS and moves to the left, hitting the ground at the mirror image spot on the LHS.
Note  

Since we have a 1D system, there are no such things as vectors and v is a scalar in a cartesian frame of reference. 
For E=f(v), here's my answer for the total energy (E_{tot} = P.E. + K.E.) ^{[127]} . We now have K.E. and P.E. as functions of velocity and can plot the (ΔK.E.,ΔP.E.) diagram.
This is the same diagram as was derived for E=f(h),E=f(t), but here it's marked with locus of the ball as a function of v. What's the trajectory (locus) of the ball along this line as a function of v? i.e. what point on this graph corresponds to v=u (start of flight), v=0 (top of the arc), v=u (end of flight) ^{[128]} ?
Just using thinking at the moment (we'll calculate the result shortly), what would be the speed of the locus along this line as a function of t? Is it linear with time or does it speed up in some places ^{[129]} ?
Let's translate axes as we did for E=f(h) so that the ball starts at the origin. Here's our graph (now with the locus of the ball parameterised in t, rather than h).
o v=0  1/2.mu^2 o  o  dPE o  o  o v=u,u  1/2.mu^2 dKE 
Using K.E.=f(v), P.E.=f(v), find the coordinates of the ball on the (ΔKE,ΔPE) diagram as a function of time ^{[130]} ?
As a function of v
ΔKE is the change in K.E. from the launch value. v_{t=0} = u and decreases as the flight progresses. The initial value of ΔKE is 0 (by definition), it changes in a parabola as a function of velocity, initially decreasing till it reaches the vertex of the parabola at v=0 when ΔKE = ½mu^{2}. At that point the locus is moving with zero velocity. Then as the ball descends in physical space, the locus starts moving again to return to the starting/ending point with v = u, At large v, ΔKE increases with v^{2}.
Here's diagram of the projection, on the abscissa, of the locus of the ball in the (ΔKE,ΔPE) diagram as a function of velocity.
below ground o  o proj.  o abscissa o  o more KE  o   end o v o start o  o o  o less KE o o top above ground 
Compare this graph with the one for the projection on the abscissa for E=f(t). The locus of the ball starts on the right, moves through the v=0 axis and the after reaching ground level heads off to the upper left.
Since the ball starts with P.E.=0, the P.E. and ΔPE axes are the same. ΔPE=P.E. is the change in P.E. from the launch value. The initial value is 0 (by definition), it increases in a parabola with time, till at v = 0 it becomes ½mu^{2}. ΔPE then decreases, becoming 0 when the ball returns to the starting point on the (ΔKE,ΔPE) diagram. At large v ΔPE decreases with v^{2}.
In the (ΔKE,ΔPE) diagram the locus moves linearly with h, but is 2nd order with t and v. There's nothing particularly deep about this. The (ΔKE,ΔPE) line is straight so the projections on the ΔKE, or ΔPE axis can be used to follow the movement of the locus of the ball on the diagram. The movement of the locus then is determined by the behaviour of PE (or equivalently KE) as a function of the parameters h,t,v. We can find the dependency of the locus on the parameters by knowing that PE∝h and from our 3 equations we have h∝t^{2}, h∝v^{2}.
Here's what we've found from the 1D trajectory problem (equations use the dot product even though we're in 1D).
General info:
Note: the average velocity on the ascending part of the arc is h_{max}/t_{top} = u/2. This agrees with what we know about the flight: it starts with velocity=u, has a constant decelleration and ends with velocity=0.
Energy=f(h).
Energy=f(t)
Energy=f(v)
Note: E_{tot} always is ½mu^{2}. We've shown three different ways that gravity is a conservative field in 1D.
Now let's do the 2D problem. Make the following change to the 1D problem
We can calculate E_{tot} = P.E. + K.E. as a function of (among other things) time and of height.
Here we calculate P.E. = f(t).
Calculate the P.E. as a function of the variable t (and the initialising constants u,ϑ) ^{[134]} . We see that P.E. is 2nd order with respect to t (i.e. P.E. = f(t^{2})).
Note  

When we say a function is nth order with respect to a variable, this means that the highest order term in the variable is to the nth power. This statement allows you to determine how fast the function grows (lower order terms are insignificant as the variable goes to ±∞). 
What is the name of the curve which is 2nd order in a parameter ^{[135]} ?
Note  

When you get an equation you haven't seen before, you should do some sanity checks on it. The only easy check (by eyeball, i.e. without resorting to pen and paper), for the function we have, is that for large t (+ve or ve), (when the highest order term in t, here t^{2}, dominates), P.E. is ve (the projectile is below the starting height). Here we go to pen and paper, to show that P.E. is +ve for the expected values of t. 
For what t is P.E. = 0? ^{[136]} ? We know from our original problem that the time of flight of the ball is t = 2u.sin(ϑ)/g. By plugging the start and end times into our formula for P.E., we see that P.E. = 0. This is as we'd expect; we defined our reference frame to have P.E. = 0 at the start. Since the start and end of the flight is at the same height, the P.E. at the end will also be 0.
Next check the calculated value for P.E. at the top of the arc. For this you need to know t at the vertex of the parabola (instead of knowing, the time honoured method of guessing is often just as fast). Since a parabola is symmetrical about its vertex, the top of the flight will occur half way (in time and distance) through the flight, at t_{top} = u.sin(ϑ)/g (remember g is ve, so t is +ve). Use this value for t_{top} to find the P.E. at the top of the arc ^{[137]} . We find that the P.E. at the top of the arc is the vertical component of the initial K.E. (The horizontal component of the initial K.E. doesn't affect the vertical component of the problem.)
Our sanity checks show that P.E. as a function of time
This is not proof that our equation is correct (however it's a good indication). If we get into trouble further down the line, these checks indicate that we can first look elsewhere for the cause.
The vertex of the parabola is offset from the yaxis (the vertex occurs at +ve time, rather than at t = 0), making the equation for the P.E. a little complicated.
Note  

You can characterise the parabola using calculus. Still you need to be able to do it the simple way too (like we do here). Later we'll use calculus to find out more about the flight of the projectile. 
You can simplify the equation of the parabola by translating the axes (in this case, we'll shift the xaxis). You need to know the value of t at the vertex (the top of the flight), which we've guessed occurs half way, at t_{top} = u.sin(ϑ)/g. To put the vertex on the yaxis (i.e. at x=0), translate (offset) the xaxis by guessed_center = u.sin(ϑ)/g (remember this has a +ve value). To do the offset, create a new time variable T and make the following substitution:
T = t  guessed_center = t + u.sin(ϑ)/g
∴ t = T  u.sin(ϑ)/g.
Substitute this formula for t into the formula for P.E., giving us a parabola with the new independant variable (the translated version of time) T, rather than the original t ^{[138]} .
Do a sanity check; check that all terms have the dimensions of energy. This parabola is of the form y + a = bx^{2} (actually P.E. + a = bT^{2}). With a ve sign infront of the T^{2} term, is the parabola concave down or up ^{[139]} ? The center of the arc (top of the flight) occurs for what T ^{[140]} ? Substitute this value of T into our formula for P.E., to check that we get the same result for the P.E. at the top of the arc, as we did before we translated the xaxis. Find the value of T that gives P.E. = 0 ^{[141]} and confirm that it's some value you expect (and check its dimensions).
If we wanted to simplify the parabola further, we could translate the yaxis too. (This doesn't add much clarity in our case, but in other situations it could be useful.) Pick a new energy scale for the yaxis and call it J.E., with a value
J.E. = P.E. + ½mu^{2}sin^{2}(ϑ)
i.e. P.E. = J.E.  ½mu^{2}sin^{2}(ϑ)
Note  

J.E. and P.E. go in the same direction (both have constants with the same sign), so the direction of the parabola (concave downwards) doesn't change. 
Substitute our new value for P.E. into the parabola ^{[142]} . Now we have a standard y=ax^{2} parabola. Confirm that terms on both sides have the dimensions of energy (you should be able to do this by inspection). With the ve sign infront of the T^{2} term, the parabola is still concave downwards.
Note  

We could further simplify the parabola by defining a new variable Y = J.E./½mg^{2}T^{2} (or even Y = J.E./½mg^{2}T^{2}). The numbers on the yaxis now are dimensionless. Mathematicians/physicists like using scaled/tranlated forms of equation(s), as the problem can be studied in a form that works for any initial conditions (the details for any particular flight are ignored). The scaling/translation factors are kept somewhere else (on another piece of paper, or in a struct in a program) and only consulted when you need the real world numbers for your answer. You have more chances of making mistakes if you change the dimensions of a problem. In the case here, it's likely that you would use a form that has dimensions of energy (rather than dimensionless). 
We've found the formula for P.E. as a function of time. Now let's find the K.E. for the projectile as a function of time. We'll find the two components.
Find the vertical component of the K.E. as a function of time ^{[143]} .
Find the horizontal component of K.E. as a function of time ^{[144]} .
Find the total K.E. as a function of time ^{[145]} . Let's do some sanity checks on our formula for K.E. K.E. is ½mv^{2} so it should always be 0 or +ve. Thus K.E. should always be non ve. Whether this is true is not obvious from the formula (remember g is ve). We could assume that we've got it right and press on regardless and return here if we find something inexplicable further down the road. In general this is not a good idea. Since we're learning and we mightn't recognize an inexplicable equation (there are lots of ways of things going wrong, not all of which you'll pick when they happen), let's check before going on. The K.E. is 2nd order in t (it's a parabola). By inspection you can see that at large t (+ve or ve), when the terms in t^{2} dominate, that K.E. is +ve. If the curve is always non ve, then we should check that it doesn't cross the xaxis for some value of t.
How do we check that a parabola doesn't cross the xaxis ^{[146]} ? Do the calculation to tell if the parabola crosses the xaxis ^{[147]} . The maximum value for sin^{2}(ϑ) is 1 (for ϑ=90°). The value for our term (inside the √ in the formula for the roots of the quadratic equation) then is either ve or 0. The 0 value indicates two roots at the same point on the xaxis, i.e. the parabola is concave upwards and touches the xaxis at one point only (and never has a ve value). At this value of ϑ (ϑ=90°) the projectile is fired directly upwards. The projectile reaches v = 0, with K.E = 0 and P.E. at its maximum. For any other ϑ, there are no real roots and the K.E. is +ve throughout the whole flight (i.e. the projectile is always moving). Our sanity check shows that K.E.=0 only for ϑ=90°.
We've shown that the P.E. parabola is concave downwards, while the K.E. parabola is concave upwards. Next we're going to add P.E.+K.E. What does the law of conservation of energy predict for the sum of these two energies (parabolas) as a function of t ^{[148]} ?
Add P.E.+K.E. (as functions of time) ^{[149]} . Throughout the whole flight, the energy of the projectile is constant at the initial K.E. This is both blindingly obvious (if you know about the conservation of energy) and profound at the same time. Nowhere in our equations (1D, 2D) did we build in the conservation of energy; we just allowed our ball to be accelarated by gravity. By doing so, we found that gravity is a conservative field.
Compare K.E._{tot} for arbitary ϑ with ϑ=90° i.e. find K.E._{tot, ϑ}  K.E._{tot,ϑ=90°} ^{[150]} . For 0≤ϑ<90°, this is +ve. This result shows that for flights with a nonvertical launch that the ball has a higher K.E. as a function of time. (When ϑ=90° at the top of the flight, the ball has no K.E.; When ϑ=45° at the peak the K.E. is ½E_{tot}.) Confirm that you get an equivalent result for P.E._{tot,ϑ=90°}  P.E._{tot, ϑ} ^{[151]} P.E. decreases with decreasing ϑ. The difference in P.E. as a function of ϑ is mg(1sin(ϑ)ut #g is ve
The locus of the ball in the (ΔKE,ΔPE) diagram starts at the origin. For ϑ=90° the (ΔK.E.,ΔP.E.) diagram is the same as for the 1D case. For ϑ<90° the ball doesn't convert all its KE to PE at the top of the arc (it's moving horizontally at the top of the arc). For ϑ=45deg;, at the top of the arc, the ball only reaches half the height and ball has equal amounts of K.E. and P.E.. Here's the (ΔK.E.,ΔP.E.) for E=f(t) in 2D for ϑ=45°. Note that the locus of the ball never reaches ΔPE=½mu^{2}. In the absence of ground, the ball continues to the lower right (not shown).
 1/2.mu^2   dPE top o  o  o t=start,end  1/2.mu^2 dKE 
For ϑ<90° the (ΔKE,ΔPE) diagram extends a shorter distance into the 2nd quadrant. As well the time to reach the top of the arc is less. Do these two factors cancel, i.e. does the ball move along the line in the (ΔKE,ΔPE) diagram at the same velocity as for the vertical flight? Here we find the velocity of the ball along its locus in the (ΔK.E.,ΔP.E.) diagram as a function of time for any ϑ.
Before launching into a detailed mathematical analysis, it's often useful to calculate numbers for some convenient location(s) on the line. This may tell you in short time if you're barking up the wrong tree. From previous work we know
Time to the top of the arc t_{top} = u.sin(ϑ)/g . Time is multiplied by sin(ϑ). For ϑ=45°, t' = (t_{ϑ}/t_{ϑ=90°}) = 0.707 .
Note  

the prime mark on t, t' denotes a reduced (and dimensionless) number. It's the value at some point divided by some obvious scaling factor, such as the maximum. 
Here's the P.E. at the top of the flight (because there is no offset on the ΔPE axis, P.E.=ΔPE)
ΔPE = P.E. = mg(ut.sin(ϑ) + ½gt^{2}) . For t_{top} = u.sin(ϑ)/g , ΔPE = P.E. = ½m(u.sin(ϑ))^{2}. ΔPE is multiplied by sin^{2}(ϑ) (or as we found above, the velocity is decreased by mgu(1sin(ϑ))). For ϑ=45°, P.E.' = P.E._{ϑ}/P.E._{ϑ=90°} = 0.5 .
For ϑ=45° the height of the top of the flight is multiplied by 0.5, while the time to get there is multiplied by 0.707. The velocity of the locus of the ball in the (ΔKE,ΔPE) diagram is multiplied by a factor of sin(ϑ).
For any ϑ the P.E. (or ΔPE) is multiplied by sin^{2}(ϑ). When dividing (ΔPE/t), we're left with a factor of sin(ϑ) i.e. the velocity of the locus of the ball is multiplied by sin(ϑ).
Let's find a more general derivation of this result. The velocity along the line as a function of time is
d(P.E.)/dt = mg(u.sin(ϑ) +gt)
The velocity of the locus of the ball compared with the locus for ϑ=90° is smaller by mgu(1sin(ϑ)).
In the previous section we calculated the (total) energy as a function of time. Here we do it as a function of height. We'll get the same answer (energy is conserved).
Calculate the total energy of the projectile at any h ^{[156]} .
Note  

The mgh terms cancel in P.E. and K.E._{vert}. In the K.E. terms, the horz and vert components of initial velocity add (via Pythagoras) to give the initial velocity. 
Note  

Doing this problem as function of height turned out to be simpler than doing it as a function of time, although I didn't realise this till after I'd done both methods. Sometimes any approach will do. Sometimes you want energy as a function of time and you'll need the time approach. 
Here we calculate E=f(v).
Starting with P.E. = mg⋅h and knowing h = f(v), derive P.E. = f(v) (you can use the 1D version for hints) ^{[157]} . While this answer is correct, were asked for P.E.=f(v), not P.E.=f(v_{vert}). Change this result into a function of the form E=f(v) ^{[158]} . There's no dependence of P.E. on ϑ (this may be useful in some circumstances).
Find the components of K.E. in the x,y plane. Here's K.E._{horz} (it's a constant) ^{[159]} K.E._{vert} ^{[160]} .
What's K.E._{tot}=K.E._{vert}+K.E._{horz} ^{[161]} ? (NEW FIXME FIXED  one term has a u and the other a v. Show steps to result.) We could have written this result by inspection (i.e. without finding the components of K.E.). Again there's no dependency on ϑ. The dependency on ϑ has been subsumed into v, i.e. v=f(ϑ). (We usually don't have a lot of flexibility in how our equations work out; we might have got equations in E=f(v) that had ϑ as a variable, as we did with E=f(t), but as it turns out we didn't.)
Here's E_{tot} ^{[162]} .
Although we've now analysing a 2D problem, we're still treating the variables (e.g. v) as scalars. This works because we're dividing the 2D problem into component 1D (scalar) problems and adding our results, the scalar quantities P.E., K.E. However while v_{horz}, v_{vert} are scalars, v is a vector. In 2D, v doesn't become ve at the top of the flight; only the angle (also confusingly called the argument) becomes ve.
For ϑ=45°, what's v at the start of the flight? It's (x,y)=(u/√2,u/√2) (cartesian coodinates) and (r,ϑ)=(u,π/4) (polar coordinates). What about the top of the arc? It's (x,y)=(u/√2,0) (cartesian coordinates; the 0 indicating that the ball has no vertical velocity), or (r,ϑ)=(u/√2,0) (polar coordinates; the 0 here indicating that velocity vector is pointing along the xaxis). (Coincidently, with the vector pointing along the xaxis, the cartesian and polar forms are the same).
Here's the diagram of the location of the vector v in cartesian coordinates during the flight for ϑ=45° (note: v_{horz} is constant).
 o start  o  o  o  o v_vert otop  v_horz o  o  o  o  o end u.cos(theta) 
NEW FIXME FIXED note: In the above diagram, the points are evenly spaced (it was the easiest way to draw them). If the points were drawn at constant time intervals, what would be the spacing ^{[163]} ? The locus of the vector moves at constant rate of g, as a function of time, moving parallel to the yaxis.
Note  

We've found the velocity of the velocity vector (it's constant). The phrase "velocity of the velocity vector" is confusing in casual conversation. What you mean is that the velocity is changing at a constant rate or the time rate of change of the velocity vector is constant. The underlying physical cause is that the ball is being given a constant acceleration (or equivalently the velocity is changing at constant rate, or the position is 2nd order with respect to time). 
Let's explore the symmetry of the graph above. The graph is symmetrical about the xaxis (more correctly  on reflection through the xaxis): this may be obvious, but if we're going to say that it's symmetrical, we have to be able to prove it (or more accurately show what the symmetry is). Declaring that it's obvious is not good enough (it was obvious to some people that the earth is flat). To say, without further backup, that the graph is symmetrical, might be interpreted as handwaving.
handwaving: the practice of glibly dismissing an area of (your) ignorance by confidently asserting some fact. This mostly happens in seminars (or verbal discussions), when the speaker will say (waving a hand in the air) something like "it's been known for thousands of years that the earth is flat" or "everyone needs to drink 4 glasses of water a day" and continue as if this is true.
The truth (or otherwise) of the asserted fact is not the issue (the fact may or may not be true). If an unsupported statement can't be verified by the majority of well informed audience members in real time, then they will be lost to the talk trying to work out whether the fact is true or not. The critical part of handwaving is that it's done to cover up an absence of data. You may be incompetent and not realise that data is needed; you may not have done the work and be trying to intimidate or bluff the audience (outside the scientific world, most people are terrified of announcing their ignorance on a topic, by putting up their hand and asking for an explanation of something the speaker has declared to be obvious). In a scientific presentation, handwaving is an insult to the audience and only 1 step above lying (you will not be invited back). Handwaving is normal in newspapers, from politicians and in any area where decisions are not based on fact. (e.g. "these photos show that Saddam Hussein has weapons of mass destruction").
This is handwaving and is not an acceptable statement
 "The line is symmetrical about the xaxis"
This is an acceptable statement
 "You can prove that the line is symmetrical, with respect to time, on reflection through the xaxis"
Presumably the proof is simple algebra and can be handled by the majority of the audience (be prepared to do the proof after the talk for any interested audience member).
The position of the ball, as a function of time, is a parabolic arc; the position is symmetrical about the vertex of the parabola. We found this by making the substitution t = T  u.sin(ϑ)/g when the graph of the position of the ball as a function of the new time variable T was symmetrical about the (translated) yaxis (i.e. y = f(T) = f(T)). This is symmetry by reflection through the yaxis.
The velocity of the ball in the 1D problem is symmetrical about the xaxis. At t = 0 the ball has +ve velocity, at t = u/g the ball has 0 velocity, and at t = 2u/g, the ball has ve velocity. Here's the graph again
o  o  o v o  t o  o  o 
The graph is symmetrical on rotation of 180° about the point (t,v)=(u/g,0)
In the 2D (v_{horz},v_{vert}) graph above, the locus moves uniformly with time, so we can assume that the graph is symmetrical (on reflection) about the xaxis. We know the time for which v_{vert} = 0 (it's t = ug).
You can see the velocity in polar (r,ϑ) coordinates, by drawing a vector from the origin, to each of the points in the cartesian representation of the velocity vector. To determine the (r,ϑ), people usually eyeball the cartesian vector diagram and leave it at that. Here's the (r,ϑ) diagram.
pi/2  theta  pi/4 start o v=u  o  r pi/4  end o  o pi/2  
The intercept on the raxis is u/√2 The angle never reaches π/2 (i.e. the ball never goes directly downwards).
FIXME  find the equation for this line.What is the equation for the line ^{[164]} ? This is where graphing software comes in handy. We can calculate some values relatively easy. Use these for the coordinates (r,Φ) = ( √( u^{2} + 2ugt.sin(ϑ) + (gt)^{2} ) , tan^{1}(tan(ϑ) + gt/u.cos(ϑ)) ) For ϑ=45° calculate the coordinates for the start ^{[165]} , top ( t = u.sin(ϑ)/g ) ^{[166]} , end of flight ( t = 2u.sin(ϑ)/g ) ^{[167]} and for large t ^{[168]} .
To plot P.E.=f(v), we need a 3rd dimension. To plot K.E. and P.E. on separate axes to look for conservation of energy, we'd need a 4th dimension. Trying to visualize these functions will be difficult, and I can't see that any great insight will come from it, so I'm not going to attempt it. (Admittedly, the cartesian graph of the locus of v is a straight line, so you could plot K.E. and P.E. in 3D.)
We've done E=f(t). Now let's do E=f(x) (horizontal distance).
Rather than deriving the equations from scratch, let's use x = u.cos(ϑ).t to derive our equations for E=f(x) from the equations in E=f(t). Here's h=f(t).
h = ut.sin(ϑ) + ½gt^{2}
Substituting t = x/(u.cos(ϑ)) we get
h = x.tan(ϑ) + ½g(x/u.cos(ϑ))^{2}
Do a sanity check. The throw is 2u^{2}sin(ϑ)cos(ϑ)/g . The distance to maximum height is half this at x = u^{2}sin(ϑ)cos(ϑ)/g . Check that you get the expected value for h at x=0 ^{[169]} , x for max height ^{[170]} , and x_{end} ^{[171]} .
As another sanity check, we should expect some strange behaviour in h as a function of ϑ for ϑ→90° (when tan(ϑ)→∞, cos(ϑ)→0 and x→0). Try substituting these values into the h=f(x) ^{[172]} .
With our equation for height having passed some sanity checks, find P.E.=f(x) ^{[173]} .
In the equation above for h=f(x), there are terms in g.x. Both of g,x are vectors and they are orthogonal. The dot product of two orthogonal vectors is 0. Quite what is happening there I don't understand (you can't scalar multiply two vectors).
Here's K.E._{tot}=f(t)
K.E._{tot} = ½m(u^{2} + 2u.sin(ϑ).gt + (gt)^{2})
Find K.E._{tot}=f(x) ^{[174]} . We need to do lots of sanity checks on anything looking like this (terms in tan(), cos()). Do a dimensional check on any terms you don't recognise as being energy. Check K.E._{tot} at the start (x_{start} = 0) ^{[175]} , top of the flight (x_{top} = u^{2}sin(ϑ)cos(ϑ)/g) ^{[176]} and end of the flight (x_{end} = 2u^{2}sin(ϑ)cos(ϑ)/g) ^{[177]} .
Now find E_{tot}=f(x) ^{[178]} .
The equations for E=f(x) are more complicated that for any of the other variables we've analysed. The problem of following an accelarating body is determined by initial conditions (g, u, ϑ) and is driven by t. None of our 3 equations are driven by horizontal distance. To make horizontal distance our independant variable, we have to use scaling constants (u, ϑ) which complicates our analysis.
Sometimes to get a better understanding we have to pick a difficult independant parameter, and we'll have to deal with complicated equations. In the case of commercial passenger jet aircraft, their speed (about 900kmph) has not changed since their introduction in the 1950s. You'd think with advances in technology in the last 50yrs, that faster planes would have been produced by now. It turns out that one of the major costs of running an airline is jet fuel. It also turns out that the fuel usage (in l/km) of planes has a minimum at about 900kmph, and another at Mach1.5 (where the Concorde flies). All commercial jet aircraft fly at these two speeds  it's matter of physics and aerodynamics. All that advances in techology have done is to make planes more economical to fly, when they fly at their most economical speed, but nothing anyone can do will shift that speed. You don't see the minimum in the fuel economy until you plot fuel usage as a function of velocity. Airlines don't plot speed on the yaxis of their graphs; they plot fuel usage.
Let's explore around the problem, to expand the radius of problem space that we understand. Mathematicians like doing this and you'll get pats on the back for finding some new result.
What is the energy imparted to the ground by the projectile at impact (compared to the initial energy) ^{[179]} ? Why is the P.E. the same at the beginning and end of the flight ^{[180]} ?.
For what ϑ is the time of flight longest (you've already done the calculation which finds this, to find the angle for the furthest throw, but you didn't record the time). FIXME  we've done it ^{[181]} ? What is the time of flight for the trajectory with maximum throw, relative to the longest flight. ^{[182]} ?
FIXME  put graph PE, KE (again) for theta = 90degHere's the parabolic flight of our projectile for ϑ=45° (maximum throw), with the xaxis showing distance. The yaxis shows P.E. For didactic purposes, I've shown the flight before the moment of throwing and after the end of the flight, allowing the projectile to pass effortlessly through the ground. The P.E. reaches half the initial K.E. The K.E. drops to half the initial K.E. (the vertices of the two parabolas touch; which isn't clear in the ascii art). The total energy (P.E+K.E.) is shown by e.
x x exeeeeeeeeeeeeeeeeexe x x KE=x  x x  x x  x x  x x  o o PE=o  o o  o o  o o o 45deg o oo o o x > 
Here's the diagram for the projectile shot more vertically (say ϑ=60°); the P.E. parabola extends up further (the ball goes higher), while the K.E. parabola goes down deeper (the ball goes slower) (not a great pair of parabolas I'll admit).
x x exeeeeeexe x x KE=x x oo x xo ox  x x  x x  o o PE=o  o o ox xo o xx o o o ooo o o x > 
Draw a similar diagram for the ball shot more horizontally (say ϑ=30°) ^{[183]} .
The diagrams above (and experience) allow us to compare the slope of the h=f(x) parabolas at the start and end, for different launch angles. For ϑ = 0°, the initial slope is 0, while at high ϑ (e.g. ϑ=89°) the initial slope is highest. We say that the initial (and final) slope is a monotonic function of the launch angle.
monotonic: a function whose value is always increasing (monotonically increasing) or decreasing (monotonically decreasing); i.e. the slope of the line does not change sign. A straight line is a monotonic function. The height of the projectile as a function of time or distance is not monotonic; it increases and then decreases (it's a parabola). The length of the throw (the distance) is not a monotonic function of launch angle; starting at ϑ=0°, the throw increases till ϑ=45° and then decreases again. The initial slope of the projectile is a monotonically increasing function of ϑ; the final slope of the projectile is a monotonically decreasing function of ϑ.
Since P.E. is a constant multiple of height (P.E. = mgh) we expect (and we'll show below) that the relative initial slopes of the P.E.=f(x) curves (as above) will be in the same order; i.e. the initial slope of the P.E.=f(x) parabola is lowest for ϑ=30° which is less than the slope for ϑ=45° which is less than the slope for the ϑ=60° shot. FIXME this is obvious. initial slope is the same thing as initial angle FIXME use the same variable for distance (x,s...)
We've plotted height (or P.E.) on the yaxis, with distance on the xaxis; i.e. h=f(s). For the xaxis, we could equally well have plotted t; i.e. h=f(t). For any particular flight (i.e. ϑ fixed), there's no practical difference in the two graphs. However the two families of graphs (h=f(s), h=f(t)) is different when ϑ is varied.
Here's h=f(s) for 3 different values of ϑ (say 30°, 45°, 60°). This graph is what we'd see if we were standing to the side of the thrower. The distance starts at 0 for ϑ=0°, increases till ϑ=45° and then decreases again to 0 for ϑ=90°.
 6  6 ^ 6 6 44  6 6 4 h  33 4 3 63 4  s> 
The family of parabolas turns back on itself (i.e. the horizontal distance thrown becomes shorter) at ϑ=45°.
For the same flights, the h=f(t) graphs don't fold back on themselves when ϑ=45°. Here's the h=f(t) graphs for 3 different values of ϑ (say 30°, 45°, 60°). Unlike the h=f(s) parabolas, you wouldn't ever physically see this family of curves; you'd have to record the data and then plot it on a graph. The ϑ=60° is the longest flight (in time) of the three shown, but as you see from the previous graph, it has less throw than the ϑ=45° flight.
 6 6  6 6 ^  6 44 6   64 4 6 h  33 4 6 3 3 4 6  t> 
From inspection of the diagrams, the variation in initial slope as a function of ϑ is smaller in the h=f(t) graph. Derive the slope of the h=f(t) graph (i.e. find the first derivative) ^{[184]} . Sanity check; t_{dh/dt=0} = u.sin(ϑ)/g. (This is the top of the flight.)
The initial slope is the slope at t = 0 and has the value
dh/dt_{t=0} = u.sin(ϑ)
NEW FIXME FIXED  why isn't this tan(theta).The ball is launched at angle=ϑ. The slope of a line at angle=ϑ is tan(ϑ). Why is the initial slope of our graph = u.sin(ϑ) and not = tan(ϑ)?
The slope that tan(ϑ) is measuring is dy/dx (=dh/dx). The slope that u.sin(ϑ) is measuring is dh/dt. We can derive dh/dx from the equation we found in the section on E=f(x)
h = x.tan(ϑ) + ½g(x/u.cos(ϑ))^{2}
dh/dx = tan(ϑ) + gx/(u.cos(ϑ))^{2}
dh/dx_{x=0} = tan(ϑ) and we get the expected slope=tan(ϑ).
As we derived our E=f(x) equations by scaling the E=f(t) equations with x = u.cos(ϑ).t, we can similarly scale the slope. Here we derive dh/dx from dh/dt. The chain rule for differentiation says that da/dc=(da/db)(db/dc). In our case dy/dx=(dy/dt)(dt/dx).
Starting with dh/dt_{t=0} = u.cos(ϑ) and dt/dx = 1/(dx/dt)
dh/dx = (dh/dt)(dt/dx) = u.sin(ϑ)/u.cos(ϑ) = tan(ϑ)
Note It took several centuries before the theory of calculus was put on a solid theoretical basis. Miraculously people found that calculus just worked, even if you didn't really understand why. (Usually when you're trying to get something new to work, it doesn't work at all, and much effort is required to get it to work even badly.) By the time I studied it, there were several centuries of methods that worked, and about 50yrs of rigour that either seemed irrelevant to me, or was just too torturous to be bothered with. I just wanted to use calculus; later if I was interested I'd take the time to find out why.
The chain rule says that you can multiply differentials as if the numerator and denominator of the differential term are variables. The people who used this rule were scientists and engineers and presumably were using results from the early days of calculus. There is another school of calculus (the mathematicians) who say that dy/dx is not dy divided by dx in the limit dx→0, but that d/dx is an operator operating on y and there is no numerator or denominator to even think about manipulating; i.e. you can't use the chain rule (even though it worked). Presumably the mathematicians were being more formal and rigourous and there was a way to do the same thing formally and rigourously. However noone explained it to me. A painful part of my university education was being required to use the chain rule in one context, but being marked wrong if I used it in another context.
Here are the distance and time formulae as a function of ϑ
Here's these two sin functions plotted (yaxis has s = distance thrown, t = time of flight)
 ss ^  s s   s s s  s s  0 theta 90  t ^  t   t t  t t  0 theta 90 
In the s=f(ϑ) graph, there are two values of ϑ for any s. To show the two valued property of the flight, here's another way of presenting it, this time as h=f(s).
We've previously found height = f(t), as an intermediate step to finding P.E.= f(t), but we didn't explicitely note the equation. (re)Derive h = f(t) here ^{[185]} .
Instead of the height as a function of time, we now want the slope of the flight as a function of time. We need calculus for this, so let's do some calculus. We'll do some noodling around as well. Find the derivative of height with respect to time, i.e. dh/dt. If you know calculus, do it yourself, otherwise here's the answer ^{[186]} . Check dimensions (what should they be ^{[187]} ?). You've seen the time derivate of height before, when you looked for u_{vert} (they're the same thing). Then you derived the result from v = u + at. (This formula itself is the time differential of s = ut + ½at^{2} ).
At t = 0 what is the slope of the height ^{[188]} ? The initial slope is the initial u_{vert}.
The derivative (dh/dt) is 0 at the top of the flight (why is this true ^{[189]} ?) We should get a familiar expression by equating dh/dt = 0. What is it ^{[190]} ? We've (re)found the time required to reach the top of the arc (which you'll remember is half the time of the flight).
Let's find the formula for the slope of the P.E.,x (P.E, distance) parabola. First let's find the slope of the h,x (height, distance) parabola. This parabola has an initial angle of ϑ, so by definition slope=tan(ϑ).
The formula for the P.E. parabola is P.E. = mg(ut.sin(ϑ) + ½gt^{2})
We've already found, for the flight that throws the projectile the longest horizontal distance (ϑ=45°), that at the top of the flight P.E./K.E. = 1. This result is for one point only in one particular trajectory (the longest horizontal distance). On finding such a result, mathematicians like to see if they can generalise the result (to expand the radius of understanding the problem); e.g. is there a simple formula for P.E./K.E. for any point in the flight for any launch angle. It's possible that the formulas will be nasty in the general case, in which case you can always throw them at a computer to get numerical results. If you get one of these nasty formulas, the you back off and look for subsets of the problem for which there are simple answers; you might find several of those (or only one). Let's see what we can find. Write down the formula for P.E./K.E. for any ϑ,u,t. ^{[191]} .
This formulae for P.E./K.E. is a little unwieldy. Remember that we already know that P.E.=K.E at the top of the arc when ϑ=45°. We're looking (hoping) for something equally simple here. You could try asking a simpler question. Notice that K.E. is dependant on three variables u, ϑ, t. You should look for (simpler) answers that depend on a smaller number of variables. Remember that there's really only one parabola (after translating/scaling the axes). (Similarly there's only one straight line, one circle...). The route to a simpler question/answer isn't obvious at this stage, and may not be until you get to the end by brute force. So let's perservere and see what happens.
Here's the summary of what we've found out about the 2D flight of a ball. (remember g is ve).
General:
equations as a function of time
P.E._{ϑ}  P.E._{ϑ=90°} =  mg(ut.sin(ϑ) + ½g^{2}) + mg(ut + ½gt^{2}) = mg(1sin(ϑ)ut
remember g is ve. For any ϑ<90°, P.E. is lower.
K.E._{tot,ϑ}  K.E._{tot,ϑ=90°} = (½m(u^{2} + 2u.sin(ϑ).gt + (gt)^{2}))  (½m(u^{2} + 2u.gt + (gt)^{2})) = mg(1sin(ϑ))ut
remember g is ve. For any ϑ<90°, K.E. is higher.
equations as a function of height
Equations as a function of velocity
Equations as a function of distance
Note  

these are just examples of things you can calculate  you don't need to remember them. When you need them, you calculate them on the spot from these more fundamental equations, which you should remember (or do enough examples that you remember them). v = u + at s = ut + ½at^{2} v^{2} = u^{2} + 2as 
Note  

html doesn't have the full range of mathematical operators. I'll use "*" for multiply (numbers), "•" or "." for dot product and "⨯" for cross product. 
Previously we've looked at the dot product of two vectors u,v being defined as u*v*cos(α), where α is the angle between the two vectors. This number is a scalar (its value is independant of the coordinate system). The dot product is maximum when the two vectors are parallel.
Here's the dot product P=u^{T}v in matrix form, where u=u_{1}i+u_{2}j is the vector resolved into an orthogonal basis.
DP = [ u1*i u2*j ] v1*i  = (u1*i).(v1*i) + (u1*i).(v2*j) + (u2*j).(v1*i) + (u2*j).(v2*j)  v2*j  
Now i•j=0 and i•i=1 etc, giving
DP = u1*v1+u2*v2
Pick a pair of vectors at right angles (e.g. the unit vectors along the x and y axes and confirm that their dot product is 0 ^{[192]} . Do the same for two identical unit vectors (e.g. the unit vector at 45° to the xaxis) and show that their dot product is 1 ^{[193]} .
Find the formula in the u=a*i+b*j format for the unit vector u at 45° to the xaxis ^{[194]} and for the unit vector v. at 60° to the xaxis ^{[195]} . Use the dot product formula to show that the angle between these two vectors is 15° ^{[196]} . Make a trivial change to the 60° vector, to find the unit vector at 30° to the xaxis. Show that this new vector makes an angle of 15° with the 45° vector.
Note that if the angle of the vector is α, then u_{1}=cos(α) and u_{2}=sin(α). The dot product of two unit vectors of angle α,β is then
DP = cos(α)cos(β)+sin(α)sin(β)=cos(αβ)
This is also the formula for cos(αβ), showing that the dot product gives the cos() of the difference in the angle.
Note  

When trying to remember trig identities like the one above
(rewritten in conventional form here)
cos(αβ) = cos(α)cos(β)+sin(α)sin(β) I had a hard enough time remembering the combination of cos's and sin's on the right hand side. On top of that I had to remember that a ve sign on the LHS became a +ve sign on the RHS. I now see that the difference in two angles (αβ) is going to be smaller than either of the angles (α, β). Since the cos of small angles (cos(αβ)) is going to be larger than the cos of big angles (either of cos(α), cos(β)) then cos(αβ) is going to have to be larger; i.e. the sum of something and not the difference. 
There is another vector product, the cross product (http://en.wikipedia.org/wiki/Cross_product). (Both the dot and cross product originated with Lagrange in 1773.) The formula (in nonvector form) for the cross product of u,v intersecting at an angle α is
CP = u ⨯ v = u*v*sin(α)
This is maximum when the two vectors are orthogonal.
If we have diagrams of forces for which we want the cross product, we can do it in two logically identical ways.
 u alpha/ /  /  v/ a 
the normal way
w = u*v*sin(α)
find the component of v orthogonal to u (shown as a in the diagram above)
a = v * cos (90α) = v * sin(α)
w = u * a = u * v * sin(α)
You'll probably use the normal way if you do a lot if it. If you only do this occassionally, then the 2nd way is the least likely to make mistakes.
While the dot product results in a scalar, the cross product results in a vector, orthogonal to the plane common to the two original vectors. Thus you need to be in (at least) 3 space to have a cross product. (For many problems, as you will see in examples below, you can ignore the 3rd dimension.)
An example of cross produce is the force on an electron moving at right angles to a magnetic field (the electron feels no force if moving parallel to the magnetic field). In the diagram below, assume the poles of a large horseshoe magnet are above and below the plane of the paper. The lines of the magnet field then are perpendicular to the paper and you're seeing the lines end on. (If you're not sure what a magnetic field line is, a magnetic monopole initially at rest, will accelarate along the lines of the magnetic field.)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . e . . . . . . . . <v . . . . . . . . 
Let's assume you can have a region of constant magnetic field, adjacent to a region of no magnetic field. (You can't in practice have an infinitely steep gradient; there will be a region with a magnetic gradient between them, but this doesn't affect our argument. You could instead turn the field on when the electron reaches a predetermined spot.) An electron entering on the right and moving left at velocity v initially experiences no force. On entering the magnetic field it will experience a force at right angles to its velocity (in the diagram, this is upwards in the plane of the paper). This force, called the Lorentz force (http://en.wikipedia.org/wiki/Magnetic_field#Force_on_a_charged_particle) is at right angles to both the magnetic field and the velocity.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ^ . . . . . . .  F . . . . . . . e . . . . . . <v . . . . . . . . 
F = q * v ⨯ B
where B is the magnetic field, v is the velocity and q is the charge on the electron.
What will be the path of the electron after it enters the magnetic field (unless you already know the answer, it won't be obvious. Draw the electron at a location in the magnetic field, see where it will go next, move the electron there, see what will happen..., rinse, lather, repeat) ^{[197]} ?
A couple of things to note here
A simple machine (http://en.wikipedia.org/wiki/Simple_machine) is one that changes the direction of a force. There are 6 simple machines as defined by the engineers of the Renaissance
The complete dynamic theory of simple machines was worked out Galileo, in 1600 in Le Meccaniche ("On Mechanics"), who was the first to understand that simple machines do not create energy, only transform it.
Th section on levers is after the section on cross products, as the calculations for levers involve a cross product. The cross product wasn't invented till 1773 by LaGrange (see Cross Product http://en.wikipedia.org/wiki/Cross_product#History) and the ancient Greeks managed to understand levers without it. If you're happy to understand levers at the level of the ancient Greeks, then you don't need to know about the cross product either. However if you want to use any of the math developed after 1773, then levers are a convenient place to start using the cross product.
Pulleys and hydraulics are simple machines which don't use cross products.
Statics is the study of the forces within objects at rest e.g. levers and trusses (e.g. bridges). Kinematics is the study of objects that accelarate (or at least move and collide). The critical difference between the two topics is that in statics, the resultant force on any point in the object it 0, i.e. the point is not accelarating. In kinematics objects accelerate because the resultant force is nonzero.
If the object is not accelerating, it doesn't matter whether it's stationary or moving. In Newtonian mechanics there is no difference between an object moving at constant velocity and one that's stationary; you can chose a frame of reference moving with the object being studied, and the previously moving object is now stationary. This does not change the forces acting on the body.
Note  

stationary: unchanging in condition; time invariant. Here this means not moving for a long period. This is different to not moving which can be for an instant (as well as a long time). A ball thrown vertically will not be moving for an instant at the top of its flight. Unfortunately, in calculus the top of a parabola, e.g. the flight of a ball, is called a stationary point. We'll just have to be careful what we mean when we say stationary. Here it means time invariant. In calculus it's a point with the first derivative being zero. 
The important thing to decide is whether the object is or is not accelerating. If it's not accelerating, you analyse it the same way whether it's moving or stationary.
In this section on simple machines, in all cases the object will be stationary. This means that the object is also not accelerating. In your analysis to find the forces involved, in just about every step you will be required to make a statement that the resultant force on an element, hinge, joint, lever, bar is zero because the element is not accelerating. How do you know the element is not accelerating? Because it's stationary. That the object is stationary is your data; the result which is the important part is that it's not accelerating, as it let's you know that there's no resultant force on the object.
When I was taught statics, you said "the resultant is 0 because the point is stationary". This is true, but not rigourous. (The resultant could be zero if the object is moving.) You should say "the resultant is 0, because the point is not accelerating. We know the point is not accelerating because it's stationary".
In the work world, noone's going to ask you to justify that the resultant on a point is 0; the whole object is stationary and not accelerating. If you're doing a statics problem for homework or for an exam question, you'll have to find out what sort of explanation is required ("it's stationary" may be enough).
First look at the wiki page Pulley (http://en.wikipedia.org/wiki/Pulley) for examples of pulleys. A pulley sits on a rotating shaft and usually are connected to other pulleys by a belt. In a car engine the crankshaft provides power via pulleys to the camshaft, water pump, alternator, power steering pump and air conditioning compressor. When a system of pulleys of different diameter are connected by a belt running at constant velocity, the forces transmitted from one pulley to another are determined by moments.
Here we'll analyse a particular compound pulley, the block and tackle (http://en.wikipedia.org/wiki/Block_and_tackle).
Pulleys are fixed (class 1) or movable (class 2). I've not heard either term (fixed/movable),(class 1/class 2). I expect the fixity or movability of a pulley is obvious and no terminolgy is needed. I will use the terms fixed and movable, because wikipedia does, and because it simplifies discussion below. Once you understand pulleys, you probably won't need the terms either.
One part of a block and tackle is the rope. Let's look at the behaviour of a piece of rope. Here's a vertical piece in space, with a force at the bottom (maybe you're pulling on it, maybe a weight is hanging off it). Let's have a mass M attached to the bottom of the rope.
    M  Mg v 
This piece of rope is accelerating downwards (accel=g). Let's apply a force F=Mg upwards at the top of the rope (you grab the rope with your hand).
^  F=Mg     M  Mg v 
This piece of rope is not accelerating. The resultant force on it is 0. What is the tension in the rope? Is it F=Mg or F=2*Mg? The answer is either blindingly obvious or a great puzzle. How would you justify your answer (hint) ^{[198]} ?
How do we know the scale is correct and the scale hasn't been predivided (or premultiplied) by 2 to enhance our weighing experience ^{[199]} ?
What we've found is that when one end of the rope has a force F on it, the rope accelerates. When the same force is pulling on both ends, the rope has a resultant force of 0 on it (and doesn't accelerate) and the tension in the rope is F.
Note  

We didn't prove anything in the mathematical sense here. We calibrated our scale with volumes of water and we defined tension to be that force which a spring scale sees. 
In a fixed pulley, the axle, while free to rotate, cannot move in space. Here's a fixed pulley, fixed by attatching it to a ceiling.
/////////////   ( O )     F  v M  Mg v 
Because of the limitations of ASCII art, I'm going to ask you to accept that this is a diagram of a pulley on an axle, with a rope over it, supporting a mass M and that a counterbalancing force F is pulling on the other end (i.e. M is not accelerating  it's stationary). When you pull on the other end of the rope (the end with the force), you raise the mass.
Let's look at what's happening to the mass. Here it is in space.
 M  Mg v 
This mass accelerates downwards. We know that the mass on the pulley drawn above doesn't accelerate (it's stationary), so the resultant force on the mass is 0. The force which counterbalances Mg comes from the rope pulling upwards with F=Mg.
^ F=Mg  M  Mg v 
If the rope is pulling up, then it should accelerate (here's the rope, imagine you're pulling up on it with F=Mg).
 ^  Mg  
However the end of the rope doesn't accelerate, it has zero resultant force on it, because the mass is pulling down on it. Here's the end of the rope (not accelerating, it's stationary).
 ^  Mg   M Mg  v 
A pulley changes the direction of a force. The rope coming down on the right side of the pully transmits a force F=Mg
/////////////   ( O )     F v 
see lever (http://en.wikipedia.org/wiki/Lever)
The earliest remaining writings regarding levers date from the 3rd century BC and were provided by Archimedes. "Give me a place to stand, and I shall move the earth with a lever". Archimedes formally stated the correct mathematical principle of levers. It is assumed that in ancient Egypt, levers were used (with the help of slaves) to move and uplift obelisks weighing more than 100 tons.
First we need some nomenclature and then we'll return to levers
moment: (see torque http://en.wikipedia.org/wiki/Torque)  the tendency of a force to rotate an object about an axis, fulcrum, or pivot. Just as a force is a push or a pull, a moment is a twist.
The moment (in nonvector form) of a force F rotating an arm of length d about a pivot (or pivot point) O
F v O d 
is M=F*d.
Note  

The moment is always referenced to the pivot point. You say "the moment about O is X nm". 
Here's the handle of a screwdriver (or the steering wheel of a car, or the helm of a ship) being twisted clockwise. The faces are d apart (and the distance to the center of the screwdriver is r=d/2). The resultant force on the screwdriver is 0 (it does not move sideways or up and down on the paper). The twisting effect is called the moment.
For the screwdriver, there are two forces acting and the moment (in nonvector form) is M=2*Fr=Fd.
F>    / \  r  0  d  \ /    <F 
However both F and d are vectors. We need a vector multiplication operator then. Which is going to provide the most twist; F at right angles (normal) or parallel to d ^{[200]} ? Which vector operation do we need then ^{[201]} ? Is M a vector or a scalar ^{[202]} ? M, being the cross product of F and d, points along the shaft.
The nonvector equation for moment is better written as M = F*sin(90°)*d. If you've agreed not to use vectors and the angle is known to be 90°, then M = F*d. The vector form of the equation is M=F⨯d.
Note  

Both work (a scalar) and moment (a vector) are the product of a force times distance. They both have the same dimensions (ml^{2}t^{2}) and units (newtonmetres). For the moment, you can regard this as a coincidence. There is no work done when putting a moment on a screwdriver and there is no twisting involved in elevating a mass against the field of gravity. (There is no reason why different quantities can't have the same dimensions.) However when a moment moves through a distance, work is done (we'll get to this). 
We don't know yet which direction along the shaft is +ve for the moment vector. It turns out that the answer is arbitary and is agreed upon by convention. We could have labelled any of the directions in the 2D cartesian axes as +ve. We arbitarily chose the right xaxis and the up yaxis as +ve. If the European languages were written right to left and from the bottom of the page to the top, we might have a different convention now. Similarly when we extend the 2D cartesian system to 3D, by adding a zaxis, we have to chose a +ve direction. For this class we have simple problems and we will chose a convenient +ve direction for each problem. The greater world and complicated problems need a convention (you can work on remembering this convention when you start needing it).
The convention chosen is the righthand screw rule or right hand grip rule (http://en.wikipedia.org/wiki/Righthand_rule). Look at the photo labelled "Use of right hand"; this uses the 3rd (middle) finger as the vector product. Compare this with the diagram labelled "Finding the direction of the cross product by the righthand rule" in cross product (http://en.wikipedia.org/wiki/Cross_product) where the cross product is on the thumb. Both diagrams give the same result (this works as long as you start at any finger and work around your thumb, first and middle finger in order have the order a,b,c in the numerical order of your fingers).
While the dot product is the same if you swap the order of the two vectors (because cos(α)=cos(α)), swapping the order of the two vectors in a cross product changes the sign of the result (a⨯b=a⨯b, because sin(α)=sin(α)).
There are plenty of opportunities for mistakes in handedness. Make sure that you and whoever you communicate with have agreed on your convention(s).
Here's a handedness mistake from the first relay satellite Telstar (http://books.google.com/books?id=npI5NsFG8ngC&pg=PA176&lpg=PA176&dq=telstar+transatlantic+circular+polarisation&source=bl&ots=mf7A3thbIS&sig=k2KWq8bSTC6GyVXXPWtFzN7z4Fw&hl=en&ei=njoCTbXqNcGqlAeqr6X0CQ&sa=X&oi=book_result&ct=result&resnum=5&ved=0CCkQ6AEwBA#v=onepage&q=telstar%20transatlantic%20circular%20polarisation&f=false).
This one has a handedness incompatibility in laying one of the early transatlantic undersea cables not knowing left from right (http://www.independent.co.uk/news/science/notknowingleftfromrightmalcolmcornwallrecountstwooccasionswhenwronghandednessandmurphyslawcombinedtofrustrateadvancesintransatlanticcommunications1508602.html). It includes a nontechnical write up of the Telstar handedness problem.
A vector product requires a 3D world. You cannot have a vector product in a 2D problem; there's nowhere for the product to go. Despite this, you can be taught lots about circular motion, moments, couples and torque, while ignoring this fact. This is pidgin physics and it isn't easy to extend; to progress, the first thing you'll have to do is go back a redo all your knowledge about circular motion. (An equivalent in English class would be to only use present tense and only read books in present tense.)
Let's say you'd spent your whole life thinking of simple (e.g. 2D) problems, you hadn't heard of the cross product, and you'd just learned how to calculate moments using the nonvector method. Then imagine you're told that a moment was a vector. Your first question is "where does it point?". You might not immediately conclude that it was along the axis of rotation. It might be hard to think of a sensible answer. Once you're heard that the moment vector is parallel to the axis of rotation, your reaction might be "well who would have ever guessed", a shrug and to turn the page. If (like me) you didn't see, first time, the direction of the moment vector, then you should stop to recognise that you and nature have a different idea about the world. You need to figure out what nature is telling you about the world we live in, that you don't understand. You've been in this spot many times before; you had to learn that physical objects are hard and to avoid collisions with them. Rather than trying to remember the answer (that the moment vector is normal to the force and the distance, and parallel to the axis of rotation), you should try to rationalise it so that when a similar problem comes up, you might be able to guess the answer (like Wessel did). Here's how I rationalised the direction of the moment vector ^{[203]} .
Of course it's much simpler to ignore the vector aspects of all of this. The only thing you loose is learning something useful.
Uncharacteristically, the wiki entry couple (http://en.wikipedia.org/wiki/Couple_%29) is not helpful. It gives a definition like this, which depends circularly on knowing "moment".
couple: a system of forces, with the resultant=0, but with moment≠0.
This definition won't mean much till you understand how to differentiate a translational force from a rotational force (next section).
I don't have a good definition of couple either. Instead I'll start with examples. Here's you twisting something; a turnstyle, screwdriver, steering wheel, screwtop jar, tap or die (http://en.wikipedia.org/wiki/Tap_and_die). The bar has a pivot and two arms of length d/2
Note  

In the following examples, a clockwise moment is +ve. 
^ F  d/2 d/2 O  F v 
The couple is the two forces separated by the distance d. By the definition of moment, the moment of the couple about the pivot point (in nonvector format) is
M = F*d/2 + F*d/2 = F*d (in vector format M=F⨯d)
What's the moment about a pivot point at the end of the bar (labelled "A")?
^ F  d O A  F v 
The left hand force has no moment about A, while the right hand force has a moment of F*d. Thus the moment is
M = F*0 + F*d = F*d (in vector format M = F⨯d)
What's the moment or the two forces about the pivot point "B" off the end bar at a distance d?
^ d F  d O B  F v 
M = F*d + F*2d = F*d (in vector format M = F⨯d)
Hmm. Seeing a pattern yet? Maybe it only happens along the line joining the points where the two forces act.
Note  

If the force isn't normal to the distance vector,
we can add a factor sin(ϑ)
(M=F*sin(ϑ)*d)
or remember the formula for cross product
M=F⨯d.

In the diagram below (the triangle is equilateral to make calculations easy) you might be trying to turn a wheelbarrow.
C / \ / \ / \ ^ / \ F / d \   F v 
What's the moment of the two forces about "C" ^{[204]} ?
A generalised version of this derivation will get the same moment for a pivot point anywhere in the x,y plane. The pivot point is often half way between the point of application of the two forces. However the moment is independant of the pivot point (see the couple is the same no matter how you measure it http://pages.uoregon.edu/struct/courseware/461/461_lectures/461_lecture12/461_lecture12.html). The moment is determined only by the force(s) and the distance between them. The effect of the moment will depend on the location of the pivot, but the size of the moment has nothing to do with the location of the pivot.
This point is explained in the wiki page couple (http://en.wikipedia.org/wiki/Couple_%28mechanics%29)
The moment of a force is only defined with respect to a certain point P (it is said to be the "moment about P"), and in general when P is changed, the moment changes. However, the moment (torque) of a couple is independent of the reference point P: Any point will give the same moment. In other words, a torque vector, unlike any other moment vector, is a "free vector". This fact is called Varignon's Second Moment Theorem.
A couple is then a pair of equal (and opposite) forces about a pivot point. By a quirk of math(s), the location of the pivot is irrelevant. As will become clearer in the next section, a couple is a system with moment, but no resultant translational force.
Let's broaden our undestanding of couples. Here's a standard couple, this time with the forces at 30° to the normal to the pivot arm.
^ F \ d/2 d/2 O \ F v 
Calculate the moment ^{[205]} . The moment is half that of our previous examples (the force isn't normal to the distance vector anymore).
This one is a little different
^ F / d/2 d/2 O \ F v 
Using vector addition of forces, turn this into a diagram showing a translational force and a moment. Calculate the moment and the translational force ^{[206]} . Reassuringly, the moment of the couple and the translational force are independant of the set of components used to find the resultants.
The wiki page forces and couples (http://en.wikipedia.org/wiki/Couple_%28mechanics%29#Forces_and_couples) explains the combination of translational forces and moments. The forces in these diagrams are equivalent.
 F'>  d      O == O <F == M=F*d O <F    d   O' <F O' <F'  
In the left diagram, the point "O" might be the center of gravity, the pivot point, or some arbitary point. The body is being pushed (translationally) by F The force F is also exerting a moment M=F*d on the body about "O".
In the middle diagram we move the translational force to the point O (the resultant force on the body cannot be changed). We restore the moment with a couple of the same moment. When calculating the couple, it doesn't matter where along the bar we apply the equivalent forces F', as long as we get the correct moment (all couples of the same moment are equivalent). Here I've placed the forces F' at the ends of the bar. In this case with the moment M = F*d being replaced by the couple consisting of the two forces F' at a distance d from O, then F*d = 2*F'*d and hence F'=F/2.
In the right diagram, since it doesn't matter where the forces of the couple act, there's no need to draw the forces of the couple; you just indicate the moment. Due to the limitations of ASCII art, I've used M=F*d to show a moment at O. In a normal diagram the M would be accompanied by a 270° part of a circle with an arrowhead, around the O .
Here's the relevent section from the wiki page.
A single force acting at any point O' of a rigid body can be replaced by an equal and parallel force F acting at any given point O and a couple with forces parallel to F whose moment is M = F*d, with d being the separation of O and O'. Conversely, a couple and a force in the plane of the couple can be replaced by a single force, appropriately located.
Any couple can be replaced by another in the same plane of the same direction and moment, having any desired force or any desired arm.
A couple then is a pair of moments from equal and opposite forces. I've never heard the term couple used, except in text books. You'll usually say "the moment on the screwdriver is..." rather than pointing to the diagram showing the screwdriver and saying "the moment on the couple is...".
The textbook Elementary Treatise on Natural Philosophy (see p14, section 28) (http://books.google.com/books?id=u8NHAAAAIAAJ&lpg=PA14&dq=single+force+couple&hl=en&ei=IXsmTcLLOsaAlAfxtrihAQ&sa=X&oi=book_result&ct=result&resnum=9&ved=0CFQQ6AEwCA#v=onepage&q=single%20force%20couple&f=false) explains that any system of non coincident forces can be resolved to a single force and a couple whose axis is normal to the force (remember moments are vectors, we'll return to this later). (This book was published in 1880 when Science was called "Natural Philosophy" i.e. "natural knowledge").
This book calls the system here (which we studied above) a wrench. A wrench has both translation and moment.
 d  O  d <F 
Note  

We will use the equivalence of a wrench with a translational force and a moment to clarify our understanding of the king post bridge in a section below. 
Let's test our knowledge on some examples. Here's a ASCII art circular steering wheel of diameter d.
  / \   O  d \ /    
Let's say you want to turn right and you only use one hand, your right hand. You'll do this
  / \ F   O   d \ / v    
We know that the steering wheel doesn't accelerate downwards. Complete the above diagram, calculate the moment and replace the downward force from your right hand with a couple using both hands ^{[207]} .
Here's the definition of couple again, from the wiki page. Now that you know the difference between moment and a translational force, it should make sense.
couple: a system of forces, with the resultant=0, but with moment≠0.
A space craft has pairs of small attitude adjustment jets (I think usually called reaction engines) in 3 axes to maintain orientation. It would be convenient if the center of gravity of the space craft were located half way between the engines, but due to the many constraints of design, this is often not possible. As well interplanetary space probes are often not symmetrical (e.g. Voyager http://www.jpl.nasa.gov/missions/missiondetails.cfm?mission=Voyager) requiring non symmetrical placement of the jets. Be prepared to orient a spacecraft using jets located at arbitary positions with respect to the center of gravity.
The Gemini 8 mission, led by Neil Armstrong, almost came to disaster when one of the attitude jets became stuck in the on position, leading to the spacecraft spinning at 1rev/sec. This would lead the astronauts to black out and eventually die in space. The astronauts didn't have time to figure out the problem and handled it by turning on jets one at a time till the spinning slowed down. Eventually the fuel was exhausted and the errant jet was no longer able to thrust. This is described in Gemini 8 Emergency. Armstrong's ability to think in an emergengy and contain the problem lead to him being in the crew of the first manned trip to the moon.
torque: in most cases a synonym for moment.
Moment is used in physics, where quite complicated systems are possible. I've only heard of moment in connection with forces in static systems. Torque is used in engineering and is usually applied to simple systems. I've only ever heard of torque in connection with engines.
This next block is of interest to people who work on cars. You can skip it if you like
A torque wrench (http://en.wikipedia.org/wiki/Torque_wrench) is used to tighten bolts to specified torque. The pictures on this website show you a beam torque wrench (cheaper and less precise, but good enough for a car and for home mechanics) and a click type torque wrench (expensive and precise, used by professional mechanics). When tightening multiple bolts on a piece of metal on a car, you want them the same tightness. Steel is hard and tough. When tightening bolts in steel, you can usually tighten them as tight as seems sensible for the size bolt and you never needed a torque wrench. With much softer aluminium now being used in cars, the bolts are tightened to a lower torque (regular strength will tear out the threads) and you need a torque wrench so you won't overdo it. When I was younger and torque wrenches were expensive, the only bolts you tightened with a torque wrench were the head bolts. One friend had a click type torque wrench (the only ones available then) and I paid him $5 each time I borrowed it (which was fair enough). Nowadays, beam torque wrenches are cheap enough that any home mechanic can have one, and car manuals have torques for every bolt. I think this is because of lawyers, not engineers.
If you don't know what a cross type lug nut wrench (wheel brace) is, look it up on the internet. In the old days with steel rims on cars, you tightened the lug nuts on your wheels as tight as you could. Using this technique, all nuts were about the same tightness, so one wouldn't come undone on you. The torque turns out to be about 160ftlbs. If your lug nut wrench has arms 2' across, what force will you have to apply to the ends of the arms ^{[208]} ? Alternately you can use your whole body weight on one arm of the lug nut wrench. In contrast, aluminium alloy wheels are tightened to 80ftlbs. You can't easily do this by hand as you have to remember to pull short, and it's hard to do this reproducably. You need a torque wrench (or devices like torque sticks in combination with an air driven impact wrench, that do it for you).
Torque is used to quantify the strength of an engine/motor. A motor/engine outputs its power through a rotating shaft. You measure the torque as a function of load, speed (rpm) and throttle settings. The engine needs a load such as a brake rotor attatched to the shaft. The brake stator is mounted on an external frame through a torque arm with a scale to see how much twist is being applied to the brake (see torque vs horsepower http://www.revsearch.com/dynamometer/torque_vs_horsepower.html and the photo at brake http://www.revsearch.com/dynamometer/waterbrake.html). This machine is called a dynamometer (colloquially known as a dyno).
Power is also used to quantify the strength of an engine/motor. Here's how we calculate power. Our engine is generating a torque T. The diagram shows the circular (as circular as you can do with ascii art) flywheel around the shaft. Let's say we have a small pad on one part of the circumference of the flywheel (here shown at 3 o'clock), which rubs against a friction surface (a brake, not shown, which surrounds the whole flywheel) thus delivering power.
  / \_  r  o _  \ /  
The force against the brake is
F=T/r
^ T F r 
Let's say we know nothing about vectors (as happens in high school and anyone before Lagrange in 1773). If the wheel revolves once, how much work is done ^{[209]} ?
Vector version FIXME
^{[210]}If the shaft is rotating at f sec^{1} and producing torque=T, what is the formula for the power output in terms of T,f ^{[211]} ?
This next section is practical details about car engines. You can skip it if you're not interested (or read it out of class time).
Here's a torque and power curve. From the graph, the maximum power, 241kW is produced at 6400rpm. What torque is the engine producing ^{[212]} ?
On a dyno, the engine's power output (e.g. heat delivered to a brake) is not measured, but is calculated from the torque and rpm.
A regular passenger car is never tested on a dyno; the mechanics make sure all the parts are at the manufacturer's settings. If your car runs about right, and gets the expected fuel consumption, then your car is tuned about right. Racing cars are tuned on a dyno for maximum output. There is enough money at stake in racing cars, that the owners can afford a dyno and mechanics spending days optimizing the engine to extract the last 1% of power. These tests are done on a whole car dyno. The car is clamped down with the driving wheels turning rollers connected to the brake. Here's a Ferrari on a dyno (http://www.speedlab1.com/FERRARI_288_GTO_028_b.jpg). Hoses remove the exhaust gases and fans on the front of the car blow air over the radiator.
You don't need a brake with a torque arm to measure the torque of your cars' engine. The engine is mounted to the car's frame through rubber blocks. As torque is applied to the engine, the engine heels over. You can see this with the car stationary (at least if you have an automatic transmission): lift up the hood/bonnet, have someone stand on hard the brakes and put the car in gear. Stand to the side of the car (never stand in front of a car with a running engine and a person in the driver's seat) and watch the engine roll over as power is applied. Depending on the state of the rubber mounts, the engine will move by about 30°. You can calibrate the torque by pushing on the engine (at least in principle, you'll need 100200lbs) and recording the tilt of the engine. Presumably later, as you drive along, you can see the amount the engine heels over and know your torque (you might need the bonnet/hood off or a TV camera inside the engine with a scale showing the angle the engine has rotated).
If the torque curve is flat (i.e. flat with engine rpm), then the power produced is proportional to the rpm. Engine designers have some freedom to change the shape of the torque curve. Here's sets of torque curves (http://www.calsci.com/motorcycleinfo/Horsepower.html). Chart 4 shows 5 different motorcycle engines. Some engines have flat torque curves, for some the torque increases with engine speed, and for some the torque decreases with engine speed. Chart 1 shows two different torque curves for the same engine.
Here's the torque curve (along with HP curve) (http://www.carkeys.co.uk/images/Technical/powertorque.jpg) for a gasoline/petrol engine taken from power and torque explained. Note that the torque curve is relatively flat and the power output is linear (at least in the mid range).
Note Since P=T*rpm, if the torque is proportional to rpm (as it is in this curve at low rpm) then power should be 2nd order (rather than linear) with rpm. The power curve should be parabolic at low rpm. Gasoline/petrol/diesel engines doesn't work at low speed, requiring a multigear transmission and a slipping clutch (or automatic transmission) to allow the car to start from a stop.
Here's the torque curve for a DC motor (http://lancet.mit.edu/motors/colorTS1.jpg) from DC motor torque curve tutorial (http://lancet.mit.edu/motors/motors3.html). Look at the webpage to see that you understand why the power maximum comes at half maximum speed (hint ^{[213]} ). Note that this sort of motor can operate at zero speed and doesn't need a transmission to operate a car/train.
Here's the torque curve for an AC motor (http://cipco.apogee.net/mnd/mfcttor.asp). It also operates at zero speed.
Do you want a flat, peaked or sloping torque curve? It appears to be a complicated topic, which I don't understand. The regular driver wants a tractable engine that lasts a long time, needs a minimum of servicing and doesn't require changing gears a lot. A racing car driver wants all the speed they can get, is prepared prepared to fuss over the engine, strip it down to the last nut and bolt after every race and to do whatever is neccessary in the gear changing department.
Torque is responsible for the force that's applied to the road. The force that accelerates the car is the difference between the force from the engine and the aerodynamic resistance of the air. If you want to accelerate more at low speed, you want a torque curve that decreases with rpm. If you want to accelerate more at high speed, you want the reverse. If you want a minimum of changing gears, you want a flat torque curve. Of course you don't need a gearbox/transmission if you use an electric motor  it can produce torque when stopped and can accelerate heavy objects; trains, earth moving equipement and busses from a dead start.
A lever is a rigid bar, with pivot (fulcrum), a load and a force applied to it. There are 3 classes of levers, defined by the order/arrangement of these three places. Descriptions are at How to identify the three types of levers (http://www.ehow.com/how_2174950_identifythreetypeslevers.html) and lever: definition (http://www.answers.com/topic/lever).
A 1st class lever has the fulcrum between the load and the force. An example is a seesaw
mg F   v v  /\  
Other examples are a pry bar (crow bar), and a pair of scissors (being a pair of 1st class levers working together).
2nd and 3rd class levers have the fulcrum at the end.
A 2nd class lever has the load in the middle and the force at the end. An example is a wheel barrow.
mg  v  O ^  F 
A nutcracker is a pair of 2nd class levers working together.
A 3rd class lever has the force in the middle. An example is the human mandible
mg  v  O ^  F 
Another example is a pair of tweezers. Other examples of 3rd class levers are almost anything used to hit something else, e.g. a baseball or cricket bat. At the moment of striking, the hand is the fulcrum, the momentum of the centre of mass of the bat is the force and the ball/nail is the load. Another 3rd class lever is the forearm lifting an object in your hand, using the contraction of the biceps (lever http://www.daviddarling.info/encyclopedia/L/lever.html).
Forces and loads are interchangeable, making the difference between a 2nd class and a 3rd class lever somewhat pedantic.
Note  

You will rarely hear the class terminology applied to levers. You'll get it in your first lesson on levers and never hear about it again. I had trouble finding explanations on the web. You should be familiar with the terminology incase you ever bump into it again, but it's most likely that the location of the fulcrum, load and force will be obvious and not need to be labelled by a class name. You apply the principle of moments to a lever in the same way no matter what its class, so mathematically there's no difference in the levers. 
Identify the location of the fulcrum, load and force in
Here's a lever and fulcrum (acting as a pivot), with two masses M,m, at distance l_{M} and l_{m} from the fulcrum.
M m  /\  
Moments require you to know force. Here's the diagram showing the forces.
Mg mg   v v  /\  
We need a convention for the +ve direction of moment. We could use the right hand screw rule from the cross product, but there's too many ways you can get this wrong. Instead, in a 2D problem like this, people usually chose clockwise as the +ve direction and ignore cross products (we just multiply; the moment vectors all come out of the page from the fulcrum, so we only have one orientation of vector to keep track of).
Here's the moment for mass M
moment_{M} = Mg*l_{M}
Here's the moment for mass m
moment_{m} = mg*l_{m}
Note  

don't forget g. Moments are about force, not mass. In this problem, g cancells out in the end, so you'll get the right answer if you leave out g. 
The lever will be in equilibrium if the resultant moment about the pivot is 0.
Note  

We're dealing with a branch of mechanics called statics (as against kinematics, i.e. bouncing balls and projectiles). Most problems in moments involve static systems, or can be restated in terms of a static system. For a list of the branches of mechanics see Mechanics: Statics, Kinematics (http://www.splung.com/content/sid/2). Statics deals with systems in equilibrium (i.e. stationary), which in this context means not accelerating. In proofs, people will say "this element is stationary, so...". Since everyone following the proof knows that the structure is stationary, you are only restating the obvious, to show the audience your chain of reasoning. Being stationary implies not accelerating. However even if the element is moving, it still can be not accelerating, so just saying that it's stationary is sloppy communication. To be rigourous, you should say "it's not accelerating" or "it's in equilibrium" and then say "we know this because it's stationary". You should not just say "it's stationary", although this is accepted by just about everyone. The important part is that the element is not accelarating. If it's stationary, then it's not accelerating. If you state that it's stationary, then there is one step of inference to the property you're interested in (that it's not accelerating). It's cleaner to say straight off that the element is not accelerating. I am probably being superpedantic here. 
Since the lever is in equilibrium (it's not accelerating) the sum of the moments must be 0, so (with g cancelling)
mg*l_{m}  Mg*l_{M} = 0
M*l_{M} = m*l_{m}
(note: units aren't specified here) If M=4, l_{M}=1, l_{m}=4, what is m for the lever to be inequilibrium ^{[218]} ? With a lever and fulcrum, you can lift a heavy weight.
Let's say instead of using mass m to balance the lever, you push down with your hand using force mg. Let's say you move your hand down a small distance s. How much work have you done ^{[219]} ? Using the symbolic values l_{m},l_{M}, how much work does mass M do ^{[220]} ? What is the total amount of work done ^{[221]} ? Sure you and all the other Egyptian slaves had to work to lift up the obelisk on the other end of the lever, but physicists will tell you that you can get it back again by lowering the obelisk back to the ground. You still have to do the same amount of work, using a lever, to lift the object through some predetermined distance, as you would have if you'd picked it up directly. With a lever you get to do the work using less force.
Here's our lever again. The diagram doesn't tell us all that's going on (we didn't need to know it all to handle the problem above).
Mg mg   v v  /\  
The lever has a force of Mg+mg pushing down on it, yet it isn't moving. Why not ^{[222]} ?
Let's draw just the lever to show our new found information.
Mg mg   v v  lM ^ lm  (M+m)g 
Note  

the resultant force on the lever in the vertical plane is zero. (there are no horizontal forces). The lever isn't accelerating (actually it isn't moving). 
We used our knowledge that the lever was in equilibrium, to state that the sum of the moments about the fulcrum was 0. What is the sum of the moments about point immediately under mass M ^{[223]} ? We get the same result if we look at the moments about the point under mass m. We can apply the same process to any point in a system in (or outside) a system in equilibrium and find that the sum of moments about it are 0.
We've looked at the forces and moments on the lever. Let's look at the forces and moments for other elements in the lever problem.
Mass M has a force Mg acting on it, but it isn't accelerating. Why isn't mass M moving/accelerating ^{[224]} ?
Here's what's going on at each mass
___    M  ___  Mg v ^  Mg  
This diagram is a common way of showing the force of gravity Mg downward on the mass, balanced by a force Mg upward from the lever.
Here's our updated lever diagram
Mg mg   v v ^ ^ Mg mg  lM ^ lm  (M+m)g /\  
Now we have the fulcrum pushing on the lever, but nothing resisting the push. Clearly the lever is pushing back with a force (M+m)g. Here's our new diagram.
Mg mg   v v ^ ^ Mg mg  lM  lm v(M+m)g ^  (M+m)g /\  
We can go on to look at the forces between the fulcrum and the ground, but here we've covered all the elements of the lever. The elements are stationary and all have resultant forces of zero. The masses are acted on by gravity and resisted by the lever. The lever is pushing out on the masses in one direction and with the same force on the fulcrum in the other. The fulcrum is bearing the weight of the two masses and is pushing back with the opposite force.
Note  

Beware diagrams with unbalanced forces. They indicate that something should be accelerating. 
Presumably Egyptian slaves not only build pyramids with levers, they shovelled dirt too. Here's a 1m long shovel in action. You push down on the handle with force F_{1}, and pull up on the shaft with force F_{2}, at a distance s along the shaft from the handle. Find F_{1},F_{2} as a function of s.
F1 mg   v 1m v  s ^  F2 
Start by taking moments about everywhere and anywhere. e.g. taking moments about the handle (i.e. the spot where F_{1} is applied).
F_{2}*s = mg*1 = mg
∴ F_{2} = mg/s (1)
Taking moments about the spot on the shaft where F_{2} is applied.
F_{1}*s = mg*(1s)
∴ F_{1} = mg(1s)/s (2)
We now know F_{1},F_{2} in terms of mg. As you move your hand on the shaft towards the handle, what happens to F_{1},F_{2} ^{[225]} ?
We could have taken moments about the pan of the shovel
F_{1}*1 = F_{2}*(1s)
∴ F_{1}/F_{2} = (1s) (3)
This doesn't give any new information. We can derive this same relationship from (1) and (2) above.
By noticing that the shovel is not accelerating in the vertical direction, find a relationship between F_{1}, F_{2}, mg ^{[226]} . Even though F_{1},F_{2} become infinity large as s→0, the difference between them is always mg.
See if you can derive the same relationship between F_{1}, F_{2}, mg using Eqn (1,2,3) above ^{[227]} . We should expect to get the same result by any method in this case moments, i.e. not rotationally accelerating and by not accelerating vertically.
When using the standard short gardening spade/shovel, most people have one hand about 3/4 the way down the shaft. If your hands are too far apart, the angle between your arms becomes large and your arms hold extra tension because of the angle. A shovel is really this
/ \ / \ /F1 \F2 mg / \  v s ^ v  1m 
with an angle ϑ between your arms. If the tension in your arms is F_{1},F_{2} then the force applied to the shovel is F_{1}cos(ϑ/2),F_{2}cos(ϑ/2).
Note  

These notes are a place holder. This topic will be moved into section on circular motion, after angular momentum has been discussed. 
A bat or sword when swung at a target are 3rd class levers. The batter when striking the ball can feel a kick backwards or forwards, depending on where the ball strikes the bat. There is a spot on the bat where no (or little) kick is felt, and this is often called the sweet spot. There are several effects which contribute forces at the handle, all of which have zero effect at different locations, so there is no single sweet spot. Manufacturers make claims about the size of the sweet spot, all of which are difficult to verify. For explanations of the effects contributing to the sweet spot see What and where is the sweet spot (http://paws.kettering.edu/~drussell/batsnew/sweetspot.html), baseball bat sweet spot (http://paws.kettering.edu/~drussell/batsnew/bendsweet.html), finding the sweet spot (http://www.exploratorium.edu/baseball/sweetspot.html) which shows a simple procedure to demonstrate the sweet spot.
Here we calculate the center of percussion (http://en.wikipedia.org/wiki/Center_of_percussion), a major contributor to the sweet spot effect and a spot whose location can be calculated exactly.
Calculate the center of gravity and the effective pendulum length of a uniform beam.
A basic element of engineering is the truss (http://en.wikipedia.org/wiki/Truss), which is any structure made of triangles (see pictures on this webpage for examples). A truss is rigid without requiring rigidity at the triangle's vertices (called nodes in trussspeak). The elements of a truss can be joined by freely moving fasteners e.g. swivels, pins (a pin here can be thick, not the ones used for sewing for an image of a pin  about half way down  Statics http://www.ecourses.ou.edu/cgibin/ebook.cgi?doc=&topic=st&chap_sec=05.1&page=theory), nuts and bolts (which hold the elements together but allow rotation about the axis of the bolt) or hinges. geodesic domes (http://en.wikipedia.org/wiki/Geodesic_dome) can be made of trusses, as are capsids for some viruses (T4 phage (http://www.bacteriophagetherapy.info/ECF409468E2F48909CA6D390A26E39C1/What%20is%20it%20all%20about.html). The regular solids, the tetrahedron, hexahedron (cube), octahedron, dodecahedron and the icosahedron have been known since at least Plato's time and are called the platonic solids (http://en.wikipedia.org/wiki/Platonic_solid). A subset of the platonic solids are trusses: the tetrahedron, hexahedron, octahedron and the icosahedron. The octahedron is used in the octet truss, invented recently (about the 1950's) by Buckminster Fuller, used to support wide roofs and crane jibs (see Space_frame. http://en.wikipedia.org/wiki/Space_frame).
Note  

The official name for a hinge in physics/engineering/math is a revolute joint (http://en.wikipedia.org/wiki/Revolute_joint), i.e. a joint which revolves. I've never heard of it. Have a look on the webpage, just to make sure you know the term. There's no point in talking about hinges when you really mean revolute joint. 
Let's look at our first truss; a simplified version of a window box (holds flowers) mounted on an outside wall, a kitchen cabinet mounted on an inside wall, a TV antenna mounted on a chimney, or shelves mounted on a wall (where brackets take the role of the sloping element). Although usually a rectangle in its cross section, the forces on a window box can be analysed as if it were a truss. The window box supports a mass M. The two arms (elements) of the box are rigid, light and are hinged (free to rotate in the plane of the paper) at A,B,M. The arm BM makes an angle ϑ with the wall.
 Mg Aoo v _ /  / wall  /  /  / Bo  
If we're attaching a window box to a wall, we have to know the forces that will be applied to the wall by the window box. Here we find the forces exerted on the wall at the mount points A,B.
We can disassemble a truss at the nodes as long as we replace the hinge with the forces it provides. This simplifies analysis of the truss by making it easier to view.
Let's look at the mount point B. First we decide on the directions that are going to be +ve at each point.
Note  

In the cartesian projectile problem, we arbitarily chose +ve x,y directions before analysing the problem. In a simple truss problem, you can see by eye which elements are pushing on the revolute joints and which are pulling. You label each force separately, so that the +ve force is pushing. (You can chose anything you like, this is just a simple scheme that makes checking easier.) (If we're wrong, our forces will come out with ve values. No big deal. In this case we relabel the forces with +ve in the other direction, proclaim cheerily that "obviously the forces go this way" and redo the problem changing the corresponding signs.) 
From our years of experience in building window boxes, we expect the wall has to both push out on BM and hold weight (we don't know how much yet) in the directions shown below.
bv / ^ o B'  Bo> bh  
Note  

Points B, B' are the same point. They are the two parts of the hinge, here shown disassembled. 
Since the node B is not accelerating (it's stationary), then the arm BM must push on the node B with equal and opposite forces. Here's the forces, which push on the hinge, being provided by the element BM.
bh / < o B'   bv Bo v  
We've now found the forces at node B. (We could have one diagram showing the 4 forces, but it's crowded. It's your choice whether you have the four forces in one diagram or two.) Note: the resultant force on node B is zero (B is not accelerating), i.e. the two b_{v}s cancel and the two b_{h}s cancel.
Now let's look at the stationary element BM unhinged at B (but still hinged at M). and find the direction of the resultant of b_{v},b_{h}. From simple mechanical principles (i.e. without needing any math(s)), can you tell the direction of the resultant ^{[228]} ?
Now we'll find the same result using math(s).
M  o  /  /  bv / lBM  ^ /  /  o>bh  B 
We know that B is not accelarating about the pivot point M (the element BM is stationary), so the total moment about M for the element BM is 0. Let's take clockwise as our +ve direction for moments. The angle between b_{v} and BM is ϑ. The length of the arm BM (l_{BM}) is the same for both moments, so in the end, it will cancel out, but we'll include it here for completeness.
The (clockwise) moment of b_{v} about M is
moment_{bv} = b_{v}sin(ϑ)*l_{BM}
Alternately, the force b_{v} makes an angle ϑ with the element BM. Thus the component of b_{v} at right angles to BM is b_{v}cos(90ϑ) = b_{v}sin(ϑ) .
The (clockwise) moment of b_{h} about M is
moment_{bh} = b_{v}sin(90ϑ)*l_{BM} = b_{v}cos(ϑ)*l_{BM}
Alternately, the component of b_{h} at right angles to BM is b_{h}cos(ϑ).
Now we add moments (or equate forces) and find
b_{v}sin(ϑ)=b_{h}cos(ϑ)
b_{h}/b_{v}=tan(ϑ)
Let's find the direction of the resultant R. Here are the two components added.
bh > ^ /  / bv / R  / / 
The angle of R is tan^{1}(b_{h}/b_{v}) , which is ϑ. ϑ is also the angle of the arm BM. The resultant then is along the arm BM.
This result is well known and seen so often, that you'll never be asked to prove it in a problem (except perhaps in the first class, where you meet this result). You will likely rederive it many times, until you learn to recognise it.
Now we have the ratio b_{h}/b_{v}, but not the individual values yet. We have to find the forces at all the nodes and then set up simultaneous equations to solve for the individual forces. (For simple cases like the one triangle window box, some of the equations will have only one variable.)
Let's look at node A. Following the method we used with node B, find what you can about a_{v} and a_{h}. In particular what's a_{v} ^{[229]} ? (Trusses with horizontal and vertical elements are helpful in this way.)
We don't have to wait to analyse all nodes in the truss before assigning values to some of the forces. Let's see what we can do knowing a_{v}=0 Let's analyse the vertical components of the problem (slightly disassembling the truss).
 Mg  A oo v  /  / wall bh /  ^ /  / B o  
This diagram shows only the vertical forces. (a_{v}=0 and is not shown.)
Note  

The two vertical forces exert a clockwise moment on the truss, but since the truss is not accelerating (it's stationary), this moment must be cancelled by an equal and opposite moment. The remaining forces are the horizontal forces, so these must produce an anticlockwise moment. We'll determine the strength of the horizontal forces shortly. 
Since the truss is not accelerating in the vertical plane (it's stationary), then
b_{v}=Mg
Since we know b_{v} and we know ϑ we can determine b_{h}
b_{h}/b_{v}=tan(ϑ)
b_{v}=Mg
∴ b_{h}=Mgtan(ϑ)
What's the force in the arm BM? (You'll need to know this to design an element which can support the weight of the flower box.) (Hint; it's the resultant of b_{h} and b_{v}.)
Using vector addition (in the equations below b_{h}, b_{v} should have hats over the "b", instead they are in bold type)
R = b_{h} + b_{v} = Mg√(1+tan^{2}(ϑ)) = Mg sec(ϑ)
Note  

sec()=1/cos(). From the identity sec^{2}(ϑ)=1+tan^{2}(ϑ) , sec() is a little more than tan(). 
Is the element BM under compression or tension ^{[230]} ?
Whether then element is under compression or tension effects the material you use. Bricks and concrete can take compression, but not tension. (Bricks are much better than concrete, but then laying bricks is expensive.) (see compression tension strength of common materials http://www.engineeringtoolbox.com/compressiontensionstrengthd_1352.html for common materials under compression.) Steel is the material of choice for tension (see Elastic Properties and Young Modulus for some Materials http://www.engineeringtoolbox.com/youngmodulusd_417.html). Elements under tension don't have to be rigid, you can use (steel) cables instead.
The sec() term indicates that the arm BM has to support a multiple of the supported mass M. What multiple does the arm have to support for ϑ=0,30,45,60,75° ^{[231]} ? Note: the tension in the element rises rapidly for ϑ>45°.
The remaining force to be determined is a_{h}. What does a_{h} do that allows us to calculate its value ^{[232]} ?
Here's the diagram showing the moments about revolute joint B.
 <ah Mg  A oo v  /  / wall  /  /  / B o  
Here's the results from looking at the contributors to the moments about B.
Mg sin(ϑ) = a_{h} cos(ϑ)
a_{h} = Mg tan(ϑ)
Since a_{v}=0, then a_{h} is the force in element AM. Is AM in tension or compression (you should be able to tell this without doing any math(s)) ^{[233]} ? You can use a rope for AM.
Add to your previous table, the multiple of the supported weight in element AM for ϑ=0,30,45,60,75° ^{[234]} ? For large angles of ϑ, hinge A will be torn out of the wall.
We've found the horizontal forces in different steps in the problem. Here they are combined in one diagram
 <ah=Mg*tan(theta) A oo  /  / wall  /  /  / B o >bh=Mg*tan(theta)  
Note  

The resultant horizontal force on the window box is zero. Why is this a requirement for a correct solution ^{[235]} ? 
The wall pushes out on the bottom and pulls in on the top.
Here's the summary of forces in the window box truss.
quantity/ value force ah Mg tan(theta) = tension in horizontal arm AM av 0 bh Mg tan(theta) bv Mg BM Mg sec(theta) = compression in sloping arm BM 
The considerations involved are
The success of the design rests on ϑ being low. A real window box will be made out of wood, which is strong enough to support flowers, pots or soil. The weak point will be the mounts which may be screws into wood or brick. The bottom set of screws have to support the weight of the window box. This is a small problem. If the screws can go in far enough, they'll support the weight; the weight will be supported by the long screws bending. If the wall is brick, then it doesn't matter that the screws are in relatively weak cement.
The real problem is that as ϑ increases, the horizontal forces increase dramatically. The bottom anchors will push on the wall, which will be OK, since the wall has a large area, but the top anchors will just be a set of screws, which will easily pull out, whether the wall is brick or wood. Even for small ϑ, while it's relatively easy to support weight, it's not easy to resist the same force pulling screws out. The pulling force is being counteracted only by the area of the screw threads.
Conclusion: window boxes are fine for small loads, as long as ϑ is small. If you have weight, it can't be supported by a vertical wall; the weight should be supported by the ground.
Abseiling (in German, English and Australian) and rapelling (in French and US) is a technique for descending a cliff using a rope (http://en.wikipedia.org/wiki/Abseiling).
The forces involved in abseiling can be analysed using an upside down window box truss. The abseiler's center of mass (M) is held by the rope (under tension), while the abseiler's legs are under compression (feet push against the cliff face). Again let ϑ be the angle of the rope with the wall (cliff).
Bo \  \  \ cliff \ _ \. O   \ . / Ao  M . ..  
Some way down the descent (when ϑ is small), the abseiler's weight is held by a harness on the torso and little force is felt by the feet against the cliff. However the start of the abseil (called "getting on the abseil") can be difficult. The rope may be tied to a tree at waist height and the abseiler has to back out over the cliff edge. If the cliff edge is square and there are no footholds to step down the cliff partially supported by the rope, then the abseiler is faced with the prospect of changing from standing vertically, supported by their feet, to lying horizontally with their feet against the cliff, being held by a nearly horizontal rope, with large (tan(ϑ)) tension. To get around this, the abseiler is faced with the options of crawling over the edge (ungainly) or jumping off the edge (and crashing back onto the cliff, dangerous). Some effort goes into chosing a good place to abseil. Often you have little choice. If the edge is overhung, there's not much you can do.
See truss bridge (http://en.wikipedia.org/wiki/Truss_bridge) and Bridge Basics, a Spotter's guide to Bridge Design (http://pghbridges.com)
Trusses have been used extensively for bridges and in particular for rail bridges. (Road bridges have been replaced by reinforced concrete. They aren't building a lot of new rail bridges, so I'm not sure whether a truss or concrete would be used for a new rail bridge.)
One of the problems with early rail bridges was that they collapsed after repeated crossings by trains. It took a while for engineers to appreciate the effects of microcracks on the brittleness of the steel of the era. A bridge well within specs at the time of its construction, would later collapse while a train was running across it. Gustave Eiffel (http://en.wikipedia.org/wiki/Gustave_Eiffel), although mainly remembered for the eponymous tower, was responsible for developing iron into structural steel (http://en.wikipedia.org/wiki/Steel) (iron with 0.22% carbon). The tower was designed for the 1889 World's fair in Paris to showcase French prowess in structural steel, and was meant to be torn down at the end of the fair. The tower was a great hit with the populace and was never demolished. (It must have been greatly overengineered to have lasted 110yrs past its design lifetime.)
Early train and rail development was done by the Scots. It was clear that rail would never be adopted for goods and passengers while bridges collapsed under passing trains. The most dramatic collapse was the Tay Bridge (http://en.wikipedia.org/wiki/Tay_Rail_Bridge#The_first_Tay_Bridge), killing all 75 people on board, including the bridge designer's son. The official enquiry into the bridge collapse (see Thomas Bouch, http://en.wikipedia.org/wiki/Thomas_Bouch) found problems with the design and construction. However, the bridge was never designed to handle large cross winds and the immediate cause of the Tay Bridge collapse was not any design fault, but sending a train across the bridge in a storm with a force 10 to force 11 cross wind. Here's photo of a ship at sea in a force 10 wind and in a force 11 wind force 11 wind from the webpage on the Beaufort Scale (http://en.wikipedia.org/wiki/Beaufort_scale). (While you're there, have a look at a sea in a force 12 wind.) The station master who sent the train over the bridge would have had no idea about structures and wind loading and probably said "a little rain and wind never stopped a train around here". The train that hurtled into the Tay had 6 carriages. Let's do a back of the envelope calculation of the lateral forces on the bridge from the windload on the train. I don't have much of an idea of the length of a train carriage back then, but lets say it was 50' long and 12' high; a cross section of 600sq'. From wind load formula (http://www.askmehelpdesk.com/physics/windloadformula211780.html) a cross sectional area of 477sq' at 50mph (force 11 wind) has a wind loading of 3760lbs. The 6 carriages of 600sq' + engine (about the same) then have a loading of 33000lbs (15 Tons). I don't know if this is a lot of lateral force for a bridge, but it was more than the Tay bridge could handle.
To demonstrate their ability to build strong and safe bridges, the Scots then built the Forth of Firth bridge (http://en.wikipedia.org/wiki/Forth_Bridge) a beautiful and grossly overengineered bridge. When I was a kid I had a book on engineering marvels of the world. Included was the Hoover Dam (which I've seen) and the Forth of Firth Bridge. When I visited Scotland, I had a friend drive me out to see the bridge. I was disappointed by its small size (and what's more, it was only single track). I guess everything looks big when you're 6 yrs old. Still although not as big as I expected, it was beautiful.
Let's design a king post bridge, a simple truss bridge (http://en.wikipedia.org/wiki/Truss_bridge) with two trusses.
. .  . .  . .  . .  .  
The bridge is supported at its ends. Let's put a train on it at the centre (the weight of the train is drawn under the bridge for convenience).
B . .  . .  . .  . . D . A  C /\ Mg /\  v  
The sloping elements make an angle ϑ with the rail bed. All elements are light, rigid and have revolute joints. The rail bed (the horizontal member) is two elements joined at the center D. Find forces at the nodes and the forces (tension/compression) in the elements.
Start by looking at what you know; which elements are under tension, which are under compression and which are holding the weight of the train? To test if an element is under tension, see if you can (in your mind's eye) replace it with a rope (or see if the element will stretch/lengthen, when you put the train on the bridge). If the element is under compression, you can't replace it with a rope, and the element will shorten when you put the train on the bridge. Try to do the problem yourself from here. If you get stuck, look at my solution below.
Element BD is under tension (it's holding the train). Elements AB, BC are under compression (as BD is pulled down, the sloping elements will be compressed  will get shorter). With a weight in the middle of AC, the elements making up AC will be under tension.
Element BD is holding at least some of the weight of the train. How much of the weight of the train is it holding (i.e. how much weight are AD and DC holding)?
AD and DC are horizontal (and under tension), so they can't be holding any of the weight of the train. (Do you know why? We proved it when we analysed the horizontal element of the flower box.) BD then is holding all the weight of the train. The tension in BD then is Mg. Let's look at the node B. (Since the two triangles are symmetrical, we only need look at one of them.)
B bh>. . . ^ Mg . .  v . . bv  . . theta D . A CEach sloping element will be pushing up on B with force b_{v} and inwards on B with force b_{h}. The resultant of b_{v}, b_{h} runs along AB, making an angle ϑ with the horizontal. We can now determine b_{v}, b_{h}.
2b_{v} = Mg
∴ b_{v} = Mg/2
We proved this next statement in the window box problem. There it took a bit of work; here we can write it out by inspection
b_{v}/b_{h} = tan(ϑ)
∴ b_{h} = b_{v}cot(ϑ) = ½Mgcot(ϑ)
What's the resultant force in AB
R_{AB} = ½Mg√(1+cot^{2}(ϑ)) = ½Mg cosec(ϑ)
This is the same problem as found in the window box, except here I've let ϑ be the angle with the horizontal, rather than vertical (as with the window box), so the trigonometric function is cosec(ϑ) rather than sec(ϑ).
From experience/guessing, what effect does ϑ have on the compression in elements AB, BC ^{[236]} ? What is the compression in AB, BC for ϑ=90° (the bridge will be infinitely high with AB, BC vertical) ^{[237]} ?
Bridge designers have ϑ about 45°. What is the ratio of increase in compression in each sloping element if ϑ decreases from 90° to 45° ^{[238]} ?
The last part is to find the forces on the horizontal elements AD,DC. In the next diagram, we start at node B and find the forces at node A.
<bh o B B o>bh . . . <bh o A A o>bhNode B is pushing horizontally on element AB with force b_{h}. Node B is not accelerating (it's stationary).
Note I'm repeating this phrase about nodes not accelerating for didactic purposes. No working engineer or person practiced in statics would bother; they already know that it's a statics problem. Element AB is pushing horizontally on node B with force b_{h} (we've already calculated this force: it's ½Mgcot(ϑ)). Since element AB is not accelerating (it's stationary), then element AB must also be pushing outward on node A with the same force. Since node A is not accelerating (it's stationary), then some force (which can only come from element AD) is pulling inward on node A with the same force.
In the next diagram, we look at node A. Since node A is not accelerating (it's stationary), and it's being pushed horizontally outward by force b_{h} from element AB, then node A must be pushing horizontally inwards on element AB. This force comes from element AD. In turn node A is pulling on AD with force b_{h} putting AD under tension.
A . o>bh <bh o DNow let's look at element AD. AD is not accelerating (is stationary) so there must be an opposite force on it as well (coming from node D)
A . o>bh bh> <bh <bh oo o A D DThe tension in AD, DC is b_{h}.
Note If you think keeping track of all these forces is complicated, it is. It helps to remember whether an element or node is being pulled or pushed. This will not be obvious (at least initially) in a large system. However you won't be dealing with large systems unless you've had practice at small systems. By the time you get to a large system, you'll be able to spot the elements under tension and compression by eye. Heres the forces and tensions/compressions in the king post bridge
quantity/ value force node A,C horizontal Mg cot(theta)/2 vertical Mg/2 node D horizontal Mg cot(theta)/2 vertical Mg node B horizontal Mg cot(theta)/2 vertical Mg sloping elements AB, BC compression Mg cosec(theta)/2 vertical element BD tension Mg horizontal elements AD,DC tension Mg cot(theta)/2By inspection (and symmetry) the weight on each of the pylons (at A, C) is ½Mg.
Note  

The tops of the frame (at B) push on each other with force=b_{h}. The bottoms of the frame (at D) pull on each other with the same force, but in the opposite direction. 
In each half (each triangle) the horizontal force at the top and the horizontal force at the bottom are equal (and opposite). This means that the horizontal force on each half of the bridge is 0. This wasn't one of our constraints or conditions. It just fell out of our analysis. (It wasn't something I was expecting either.) When something simple (i.e. the forces being equal, and the sum being 0) like this falls out of your equations, without you expecting it, you should check that you understand why. (This might be a moment like when you found the direction of the moment vector.) There's likely to be something simple (or what might be the same thing, something profound) behind it.
Why is the resultant horizontal force on each half 0? Or why are the forces at the top and bottom of each half equal? Or if they're equal, what is nature telling you? (The answer wasn't obvious to me. I thought about it for a couple of weeks. Give yourself enough time think about it before looking at my answer.) ^{[239]} .
We didn't use moments to analyse the king post bridge. What's the king post bridge doing in a section on moments? Moments were needed to derive the direction of the resultant in a truss element (we did this with the window box). If you accept this result, then the king post bridge (and probably all static problems) doesn't depend on moments for its solution. The king post bridge is just the next step after learning about moments.
Note that all the weight of the train (at least when it's in the middle of the bridge) is supported by the vertical beam, which then moves the weight through the sloping elements to the pylons at the ends. No weight is born by the horizontal elements which make up the road bed. This is at least in principle. When the train is 1/4 or 3/4 across the bridge, the horizontal elements will bear load. As well the road bed has width and even when the train is in the middle of the bridge, the road bed has to support weight in the sidetoside direction.
To get some practice detecting compression and tension in truss elements look at the Howe truss and the Pratt truss. Both are multiple king post trusses. The Howe truss is a horizontal extension of the king post truss. Excluding the two end sections, the Pratt truss is the vertical inverse of the Howe truss.
Assume a train is at the center of both bridges. For the 4 center (square) sections, which elements are in tension and which in compression? (imagine pushing down on the top of the diagram with your hand; which elements lengthen and which elements shorten?)
Here's the Howe truss
..... .  .  .  .  .  . .  .  .  .  .  . .  .  .  .  .  . .  .  .  .  .  . ...... 
^{[240]} .
Here's the Pratt truss
.... .  .  .  .  .  . .  .  .  .  .  . .  .  .  .  .  . .  .  .  .  .  . ...... 
^{[241]} .
Here's the summary (T=tension, C=compression)
element Howe Pratt roadbed T T vertical T C sloping C T top C C 
Because a bridge crosses a previously uncrossable spot, it's not always possible to build the bridge in one piece. Sometimes a bridge is built from both sides and the two halves lowered to allow joining in the middle. This happened with the construction of the Sydney Harbour Bridge (http://www.columbia.edu/cu/gsapp/BT/BSI/ARCH/img0069.jpg). The weight of the two halves of the bridge was supported by cables bolted into rock at the ends of the bridge (in the photo, seen as the dark bar at the end of each arch, which continues the arch down to the ground). Here's the ceremony where the two arches were closed by a pin (http://www.cultureandrecreation.gov.au/articles/harbourbridge/images/closingofarchweb.jpg). It's not likely that you'll use this procedure with a king post bridge, however you should be ready to do it for a bridge with many trusses, so let's practice with the simple king post bridge.
Here's the king post bridge as two halves. Although there won't be a train on either half, presumably workmen will need tools and equipment to build the bridge, so the centre ends of each half of the bridge are shown carrying a force of ½Mg. To counterbalance the moment caused by the weight of equipment, forces at B1,B2 hold the two halves apart (forces provided by cables)
B1 B2 <bh . . bh> .   . .   . .   . . D1 D2 . A   C /\ Mg/2 Mg/2 /\  v v  
You can do this problem on your own if you like. For the moment, see if this version of the bridge is any different to the whole bridge. A quick check is to look for tension/compression in the elements. Do you see anything different ^{[242]} ? Compare how the two halves (left, right) of the bridge are stopped from rotating into each other in the two versions of the bridge (if it's not obvious what I'm asking, pretend you're explaining the bridge to the railway officials and say something profound or erudite in the engineering sense about the difference). Here's what I see ^{[243]} .
If you separate the two halves of the king post design, you get a queen post truss (http://en.wikipedia.org/wiki/Truss_bridge#Queenpost_truss) and queen post truss (http://en.wikipedia.org/wiki/Queen_post). The queen post is used in framing the roof of houses. From bridge basics (http://pghbridges.com/basics.htm)
the queen post truss adds a horizontal top chord (to the king post truss) to achieve a longer span, but the center panel tends to be less rigid due to its lack of diagonal bracing.
FIXME  do length of arc integral to find length of parabola. As check do length of arc integral of straight line, semicircle
Note  

I had fun here. This isn't central to what I'm teaching. You can read this in your own time if you like and ask questions later (see classical mechanics http://en.wikipedia.org/wiki/Classical_mechanics). 
We've found that mgh=m.(l/t^{2}).l and mv^{2} = m.(l/t)^{2} have the dimensions of energy. What about m.l^{2}/t^{2} and m.(l^{2}/t).t^{1} both of which have the dimension of E?
I'm just playing games here (but that's where discoveries are made). Let's see if we can think of anything physical for these new quantities.
m.l^{2}/t^{2}: The can be rearranged to (m/t^{2}).l^{2} . The first term (the rate of change of mass/second) has the same dimensions as surface tension or the spring constant.
Note  

Surface tension is the force required to pull a length of wire, parallel to a liquid's surface, away from the surface. The spring constant is the force required to change the length of a spring by l. Both have dimensions F/l=ma/l=ml/(t^{2}l)=m/t^{2} . 
The second term is area.
In the case of surface tension, as you pull the wire away from the liquid, a sheet of liquid forms between the wire and the surface of the bulk liquid. Energy, proportional to the newly created surface area, is required to pull the wire away from the liquid.
In the case of the spring constant, the force is proportional to the extension of the spring. Work then is proportional to the square of the extension of the spring.
One of the laws of physics is that if something can exist, it must exist. You may not be able to guess what it looks like, or you may not know where to look for it, but that doesn't mean it doesn't exist. In this case of looking for new quantities, only the rate of change of mass (m/t) isn't in use (hasn't been discovered). If these quantities aren't known yet, you can guarantee that one day someone will find them.
Just as it takes energy to lift a hiker (and pack) or elevator against the force of gravity, energy is released (and can be used) when massive objects fall with the force of gravity. An example is hydroelectric power; falling water pushes the blades of a turbine and produces electricity. In earlier times a small dam in a river was used to power a mill to grind flour.
The tallest hydroelectric dam in the world is the 300m Nurek dam at Vakhsh, Tajikistan (http://en.wikipedia.org/wiki/Nurek_Dam) (I couldn't find a good photo.) (here's a list of high dams http://www.infoplease.com/ipa/A0113468.html).
Let's calculate the amount of work that water can do in a hydropower station (assume turbines and generators efficiently turn this energy into electricity  they do a pretty good job, I think it's >90%).
First let's calculate the amount of energy that a drop of water of mass m picks up when it free falls through a height h (let's assume the water doesn't meet any air resistance).
Note  

The height of water, in daytoday running of the dam, is called the head. For a physics problem, we'll use the generic term "height". 
water level     h   .  .  . o drop of falling water __x__ turbine blades x 
The force acting on the mass is
F = ma the acceleration of gravity is g, so F = mg 
The amount of work done to accelerate the water drop is F*s
W = F*s = mgh 
Note  

This amount of work is seen in many problems; the amount of energy that a mass gains when falling in a gravitational field is mgh. It's easy to derive so you don't neccessarily have to remember it, but don't be surprised when you keep seeing it. 
How much power do we get? Assume that mass m passes through the turbine every t seconds, then the power available from the dam is
P = W/t = mgh/t watts 
If we want more power we can increase the mass flow (m/t) or the height of the dam h. Since in some places, the amount of water is finite, we can get more power from the same amount of water just by building a higher dam. The higher dam isn't always free. Sometimes the land around is too flat to have a deeper/higher dam; sometimes the extra weight of water causes seismic problems (earthquakes); building a higher wall is expensive and requires a good understanding of the geology near the wall. Still if you can double the height of the wall, you double the amount of power you can get from the same mass of water. (Dam operators like to keep the dams as near full as possible for this reason. This requirement conflicts with another purpose of dams; flood mitigation, where the dam level is required to be low enough to absorb any conceivable flood.)
In hydrodams, water isn't free falling at high speed into a turbine; it's waiting almost stationary at the bottom of the dam under enormous pressure at the turbine intake. Let's see how much work we can get out of a mass m of water, at the bottom of a dam of height h. Let's say the mass of water occupies a rectangular box of horizontal area A and height dh.
water level     h    ___  / A / dh  /____/ / ____/ __x__ turbine blades x 
The force acting on the area A is the (mass of water above it)*g. We know the volume of water above the box (it's A*h). To get the mass, we multiply volume*density. Density is usually represented by the symbol ρ (pronounced "rho" with units mass/volume). What is the dimensions of ρ ^{[244]} ?
Note  

The density of water, by original definition (the definition of the gram was the mass of 1cc of water), is 1gm/cm^{3}, 1kg/l, 1tonne/m^{3}. 
What is the force acting on the area A?
F = Mg ; M = mass above rectangular volume calculating M M = h.A.rho thus F = h.A.rho.g 
The mass of water in our rectangular box is m=vol*density=dh.A.ρ. Let's say our mass m moves a distance dh without encountering any backpressure (i.e. the only force acting on it is the pushing force described above) by moving a distance dh.
Note  

in the falling drop calculation we put a similar restraint on the problem, that there be no backwards force on the drop as it fell. 
How much work is done by the volume of water as it moves through the distance dh.
W = F*s = h.A.rho.g.dh = (A.rho.dh).g.h the mass of the rectangular volume is m=A.rho.dh thus W = mgh 
This work is only done when the water moves from the pressure at the bottom of the dam at the intake of the turbine, to a region of no pressure  i.e. at the turbine outlet. The water sitting a short distance back from the intakes, with equal pressure on all sides, isn't capable of doing any work at all (and has no reason to move in any direction).
In reality the water doesn't move from high to low pressure in the distance dh. It moves through the multiple sets of turbine blades, along a gradient of decreasing pressure, releasing a bit of energy at a time, till it reaches the outlet at 0 pressure, when it can do no more work. The total amount of work is determined by the difference in pressure from turbine intake to outlet.
Note  

Various devices extract work from a fluid under pressure

We've found that no matter whether we let gravity accelerate the water by falling the height of the dam, or let gravity push on the water at the bottom of the dam (through the mass of water above it), we get the same amount of energy from the water. It turns out that energy is energy and it doesn't matter which way we use it (to first convert to motion or to pressure), it's still the same amount of energy. If this were not true, then it would be possible to construct a perpetual motion machine, where two sets of turbines, one operating through water pressure and another set, operating through water velocity were operated backtoback, with the difference in energy being available for work.
Let's say we want to boil a cup (250ml≅250g) of coffee in a microwave oven using hydroelectric power. How much water has to pass through the turbines to do this? To simplify the problem, assume there is no loss in the turbines, transmission lines or microwave oven (a real microwave oven is about 60% efficient). Assume that our coffee water starts at 0° (in reality we'd start at 2025°). Let's get our power from Nurek.
Energy needed to heat 250ml water from 0100deg Work needed to heat water for coffee W = m.C.dT W = 250*4.2*100 = 105kJ Work available from mass m of water at Nurek W = mgh = m*9.8*300J = 2.94m kJ equating these two quantities m = 105/2.94 kg = 35.7kg = 142 cups (of 250g) 
Thus using power from Nurek, to boil our cup of coffee, we'd have to let 142 cups of water flow through the turbines.
Lets say instead we used power from the 233m Hoover dam at Boulder Az. (http://en.wikipedia.org/wiki/Hoover_Dam). How much water would need to be used to boil our cup of coffee now ^{[245]} ?
How about using hydropower from the 168m Grand Coulee Dam on the mighty Columbia R (http://en.wikipedia.org/wiki/Grand_Coulee_Dam) ^{[246]} ?
Our local park Eno River (http://www.enoriver.org/eno/Shop/Journals/MillJournal/Geos.htm) has a dam once used for a mill (the water wheel is still there). I can't find the height of the dam, but let's say it's 10m (tops). How much water would have to flow through the Eno River mill to boil my cup of coffee ^{[247]} ?
Even in the best of circumstances (no power loss), and getting our power from Nurek, the highest dam in the world, we need 142 cups of water going through the turbines to boil a cup of water. How much is 142 cups? My kitchen faucet (tap), running full blast, fills my coffee cup in 3 secs. It takes 2mins to boil my coffee. During that time, even if I left the kitchen tap running full blast, for the time I was using the microwave oven, I would only be letting 40 cups of water go down the drain.
The US has plenty of water (maybe), but most of the world doesn't have enough water to turn on faucets full blast everytime you use electricity. Most hydroelectric dams in the US are quite low (50100m?)
Note  

looking up hydrodams in NC with google, I find information about the area of the lake behind the dam, but rarely is the height of the dam listed. The Apalachia Dam http://en.wikipedia.org/wiki/Apalachia_Dam is 46m high. The Idol Dam on the Yadkin R (http://www.digitalforsyth.org/photos/818) and Idol Dam (http://www.digitalforsyth.org/photos/836) looks to be no higher than 5m. 
The Hoover dam was a technological marvel of its era (I remember being impressed with it in my childhood, and took the first opportunity on moving to the US to visit it). It sits astride the mighty Colorado River, which appears as a mountain of water as it flows through the Grand Canyon. However the demands on the Colorado are such that none of it reaches the ocean anymore  it's all been diverted for agriculture. The Hoover Dam was planned in an era of unprecedented rainfall (although the builders, without historical records didn't know this). The amount of water available to be flushed to boil cups of coffee is much less than anyone planned. It's possible that the lake behind the Hoover Dam will be empty in our lifetimes. Still there's lots of water in the Columbia R for the Grand Coulee dam, isn't there? Well sure, but no anadromous fish (http://en.wikipedia.org/wiki/Anadromous#Classification) live in the river anymore (they can't get past the dam to spawn) and the way of life of the native americans went with it. Putting high dams and large lakes in the middle of a river is an environmental and habitat catastrophy (see Environmental_and_cultural_consequences http://en.wikipedia.org/wiki/Grand_Coulee_Dam#Environmental_and_cultural_consequences) that might have been acceptable 100yrs ago, when the world's population was much less, but it isn't anymore.
Hydropower has worked well for Norway, which has lots of snow, and deep valleys to produce the head (height) needed to generate the power. Hydropower would be fine if it was just a matter of sticking a turbine at the outlet of a dam that was already needed for something else, but the environmental costs of dams are such that now there are many successful projects that have removed dams and allowed the fish to return to their habitat.
What height would the 250ml of water in my coffee cup have to fall to have its temperature raised 100° on impact ^{[248]} ? What velocity would this 250ml body of water have on impact in my coffee cup (assuming no atmospheric friction and a constant acceleration of gravity) ^{[249]} ?
A little side trip as to why public transport is important.
In 1977 Caesare Marchetti published an analysis of the commute times from the Neolithic to the present (FIXME). He found that humans are prepared to commute upto 1.5hrs. As modes of transport have changed (foot, horse, car) only the size of the conurbanisations have changed not the commute time. Thus the fallacy that this generation's transport gets you to work quicker than the last generation's transport: no  you wind up living further from work in a bigger city.
In america, there are people who make their living making/selling cars. With a car you have "freedom" they say.
FIXME You need 50100yrs and stable zoning laws to redesign/build a city around a train line (cf Perth).
Example: The TGV does 500kmph and uses power at a rate of 12MW. Assuming that all tractive power goes to overcome aerodynamic resistance (a reasonable assumption), what is the aerodynamic force resisting the TGV ^{[250]} ? What mass, under the attraction of gravity, would need that amount of force to resist it's movement ^{[251]} ? The nose of the TGV must be able to withstand the weight of 6 US cars.
Heat is energy; it has the same dimensions as energy (this took a while to understand too). Heat can be used to lift objects against gravity. The energy in objects falling from a height is converted to heat when they impact.
Temperature is a familiar concept and has been so for many millenia. In winter it's cold and water turns to ice. In summer it's hot. We all know that when you add heat to an object that it becomes hotter and when you remove heat it becomes cooler. When objects of different temperature are in contact, heat flows from the hotter to the colder object, till they are at the same temperature. There is a strong connection between heat and temperature. We all have a good idea and agree upon temperature.
But what is temperature? Temperature is a different beast altogether from heat. In SI (but not cgs, or mks), temperature has its own dimension T (I don't understand enough about temperature to know why it has its own dimension). It took till the early 1900s for anyone to understand temperature. (They were in the same state, we are now, about gravity.)
In the meantime people had invented thermometers, which gave consistent results and which indicated something (whatever it was) that agreed with what you felt as temperature by sticking your finger in it. You could produce a temperature scale (Fahrenheit, Celsius) allowing people to brew beer and control steam in boilers (without them blowing up) and until something better came along, everyone agreed that a thermometer measured temperature.
thermometer:
a device with a property that changes with temperature e.g. a volume of liquid that expands with temperature along a bore of constant cross section; a piece of metal whose resistance changes with temperature; a bimetallic strip (http://en.wikipedia.org/wiki/Bimetallic_strip).
FIXME Various temperature scales are in use
 Centigrade/Celsius: defined by the temperature of liquid and solid water in equilibrium at atmospheric pressure=0°C and the temperature of gaseous and liquid water in equilibrium at atmospheric pressure=100°C
 Kelvin;
I'm not going to tell you what temperature is just yet. First let's look at the things people discovered on the way to figuring out what temperature is.
Materials differ in the the temperature rise produced by a fixed amount of heat. Each type of material has a different heat capacity, called the specific heat capacity, or specific heat (for short). "specific" because it's for a fixed amount of material (usually a gram or a mole). If you have a large blob of the material, the heat capacity is the specific heat * mass.
The symbol for specific heat is C (from capacity? from calorific value?). The Specific heat (http://en.wikipedia.org/wiki/Specific_heat_capacity) of water is 4.2J/g.K.
(I'm just giving you practice in dimensions. There's nothing particular here.) What is the dimension of specific heat in the S.I. system ^{[252]} ?
In the cgs system, the unit of heat is the calorie, being the amount of heat needed to raise 1g of water by 1°C. The calorie is a useful unit, giving water an easy to remember heat capacity of 1.0. It's most unfortunate that nature didn't cooperate giving a heat capacity of 1.0 in the self consistent SI scheme. The unit of energy used for food in the US is the Calorie being the amount of heat that will raise the temperature of 1kg of water by 1°C (i.e. Calorie = 1000 calories). Since most people don't realise that case matters here, for food labelling, the "Calorie" is usually spelt "calorie". If car manufacturers described their cars as having a 3cc engine, when it had a 3l engine, they would be mocked. However when you do the same thing with food, noone notices.
Note  

The rest of the world uses Joules to describe energy, including energy in food. 
Example: a 50kg person ingests 2000 Calories a day. If the energy is just used as heat (which it mostly is), and the person did not loss any heat (to the air), what temperature should their body increase in a day assuming their body is mostly water (which it is) ^{[253]} ? With your basal temperature at 37° your blood would boil in about 1.5days if you were completely insulated.
One of the ways of loosing heat is to evaporate water, either by sweating or by evaporation of water from your lungs. These mechanisms are only effective at elevated temperatures. To turn water into water vapor, the water needs to absorb 540calories/gram (the latent heat of vaporisation of water). How much water do you need to loose/evaporate to maintain body temperature (assuming there are no other mechanisms for loosing heat) ^{[254]} ? This is about as much water as a 50kg person could loose without having problems from dehydration. If it was 37°C and you couldn't offload 2000Calories of heat into the air, you would have to evaporate 3.7kg of water/day to maintain temperature.
Look at the specific heat of a few substances. Notice that water has one of the highest specific heats, measured in J/(gK). If you want to raise the temperature of a pan of water by 100° you need more heat than if you were doing the same thing to a pan of aluminium.
In the 1700s and 1800s industrialists wanted to know the efficiency of their steam engines (and later electric generators and motors). How much work should they expect from the heat produced by burning coal? Until they knew the conversion between heat and work, they couldn't tell.
Earlier work by Count Rumford (1798) boring canons (http://en.wikipedia.org/wiki/An_Experimental_Enquiry_Concerning_the_Source_of_the_Heat_which_is_Excited_by_Friction)
Rumford had observed the frictional heat generated by boring cannon at the arsenal in Munich. Rumford immersed a cannon barrel in water and arranged for a specially blunted boring tool. He showed that the water could be boiled within roughly two and a half hours and that the supply of frictional heat was seemingly inexhaustible. Rumford confirmed that no physical change had taken place in the material of the cannon by comparing the specific heats of the material machined away and that remaining were the same.
Rumford argued that the seemingly indefinite generation of heat was incompatible with the caloric theory. He contended that the only thing communicated to the barrel was motion.
Rumford made no attempt to further quantify the heat generated or to measure the mechanical equivalent of heat.
Note  

The canon boring operation was powered by horses. 
Joule (http://en.wikipedia.org/wiki/James_Prescott_Joule) was an educated brewery manager from a wealthy family. He was regarded by the Royal Society (the premier scientific body in England at the time) as a provincial dillettante. The prevailing theory at the time was that heat was a property of a fluid caloric, that moved from one object to another as they changed temperature. In an exacting set of experiments (financed by and using the physical facilities of the brewery), Joule determined in multiple independant ways (chemical, mechanical, electrical, hydraulic), the conversion factor (the mechanical equivalent of heat), showing that heat was energy (and hence not caloric). His paper in 1843, read before British Association for the Advancement of Science in Cork, was greeted with silence. As it is with scientists, new ideas are not well accepted and it often takes the death of the old generation before new ideas become widespread.
Although the calorie (the amount of work needed to raise 1g of water by 1°) and the Calorie (the amount of work needed to raise 1kg of water by 1°) are no longer used (except in some holdout countries, to measure the amount of energy in food), the conversion factor is 4.1860 J.cal^{1}.
Note  

What is the dimension(s) of this conversion factor, the Joule constant? The factor converts between two different units of work; the joule and the calorie. It is a number; it has no dimensions. Although it is dimensionless, it does have units J.cal^{1}. (Note the similarities with the dimensionless factor, which converts between metres and feet = 3.2809ft/m, or the number which converts dozens to units.) 
As we all know, the main benefit of travel is self education and broadening your viewpoint on the different cultures in the world. If you read the experts on travel (the travel section of magazines and newspapers), you'll find that the best education comes from eating at the destination's most expensive restaurants. Joule, on his honeymoon, didn't have the benefit of the New York Times travel section and
marital enthusiasm notwithstanding, Joule and Thomson (later Lord Kelvin) arranged to attempt an experiment a few days later to measure the temperature difference between the top and bottom of thethough this subsequently proved impractical. The height of the waterfall (http://www.worldwaterfalls.com/waterfall.php?num=202) is 365m top to bottom, with the largest single drop of 270m.
 Cascade de Sallanches waterfall (http://www.flickr.com/photos/sharman/253526381/)
 Cascade de Sallanches waterfall (http://www.flickr.com/photos/louistib/3567924954/)
Looking at the photos of the Cascade de Sallanches waterfall, what sources of errors will confound an attempt to measure the increase in temperature of water falling through the waterfall. (hint: what factors, other than falling through a gravitational field, will affect the temperature of the water) ^{[255]} ?
We recognise Joule's attempt to measure the increase in temperature accompanying the fall of massive objects in gravity, as the same problem as calculating the energy used by a hiking climbing a mountain, a programmer running stairs with a pack on his back, lifting an elevator against gravity, or in the energy released by water falling through gravity in a hydroelectric plant (next section).
We've seen examples of the conversion of potential and kinetic energy into heat for events in our experience. Now let's recreate Joule's experiment at Cascade de Sallanches in your own home.
Prerequisites: piece of pvc pipe with end caps, thermometer, weighing scales and massive object (small bag of rice).
FIXME experimental details
The Barringer Meteor Crater (http://en.wikipedia.org/wiki/Meteor_Crater), was caused by the impact of the Canyon Diablo meteorite about 50kyA. The 50m diameter nickeliron meteorite, weighing 150ktonnes, hit the ground at about 12.8km/sec.
Note  

They even know the direction it was travelling and the angle it struck the ground. Why don't they tell you the time of day it hit too? The explanation for how these numbers were determined should be understandable to many people. However I don't know where to find it. Although I don't know where these numbers come from, they will be the starting point for our calculations. 
Note  

The area at the time was cooler and wetter than now, being grassland and open woodland. How likely was it that the impact was witnessed by humans? Should we look for evidence in cave paintings ^{[256]} ? 
100yrs ago, geologists thought that the crater was a steam vent from a volcano. Barringer thought that the crater was from a meteor and went bankrupt trying to find and excavate it for it's iron content. The origin of the crater was settled about 1960 by Gene Shoemaker who produced enough evidence to convince people of the impact, making the Barringer crater the first identified impact site on earth.
At the time (1960's) the mountains of the Moon were also thought (by many) to be volcanic in origin, despite the characteristic central peak in the craters (see impact craters http://en.wikipedia.org/wiki/Impact_crater). Here's a photo of Tycho (http://www.lpi.usra.edu/education/timeline/gallery/slide_61.html), a relatively young (160Mya) impact site, easily visible from earth with binoculars. Rays of ejecta from Tycho can be followed across most of the visible side of the the Moon. (from Impact Crater Geology http://www.lpi.usra.edu/expmoon/orbiter/orbitercraters.html)
The crater Tycho, 85 kilometers in diameter, is the youngest large impact crater on the Moon's nearside. Ejecta from this crater was spread across much of the nearside of the Moon and is visible in the form of bright rays at full Moon. One such ray crosses the Apollo 17 landing site, 2000 kilometers from Tycho. Arrival of this material from Tycho is believed to have triggered a landslide from the mountains surrounding the Apollo 17 landing site. Laboratory analysis of samples from this landslide suggest that Tycho's age is about 100 million years.
Note  

Tycho (http://www.daviddarling.info/encyclopedia/T/Tycho_crater.html) features in the film and book "2001: A Space Odyssey" as the location of TMA1 (Tycho Magnetic Anomaly 1), which turns out to be one of the enigmatic monoliths placed in the solar system, by an alien civilization. In the story, Dr. Heywood Floyd travels from Clavius Base to see the artifact. 
The Apollo program was a cold war effort to intimidate the Russians. The astronauts were military pilots and science was not on the table. Gene Shoemaker recognised that on arrival at the moon, the astronauts would be seeing rocks and not communists. A photo of astronauts saluting under a US flag would be shallow reward for the efforts to get there. Gene is to be remembered for his successful efforts to use the lunar landings (and the reluctant astronauts) to advance science. (Gene's ashes are on the moon Moon Trivia http://home.hiwaay.net/~krcool/Astro/moon/moonglossary/moontrivia.htm) The Apollo program found that the craters on the moon were impacts and not volcanoes and provided much useful information on the early history of the solar system (on earth much evidence has been erased by erosion). After beating the Russians to the moon, and there being no other reason to be in space, the lunar exploration program was discontinued, revealing the manned space program for the sham it was. The best the US can do at the moment is use the ISS to rediscover the knowledge the Russians gained with Mir (http://en.wikipedia.org/wiki/Mir).
Note  

Most of the craters on the Moon (and by inference the terrestrial planets) were created during the late heavy bombardment (http://en.wikipedia.org/wiki/Late_Heavy_Bombardment) a short period about 4Ga, when possibly a Mars sized planet was disrupted by Jupiter's gravitational field, and the shards roamed the inner solar system, being swept up by the terrestrial planets. 
Why wasn't the Canyon Diablo meteorite never found? Let's look at what happened on arrival of the meteorite.
How long did it take the meteorite to traverse the atmosphere? Well, how thick is the atmosphere? The atmosphere gets exponentially thinner with altitude and thus is infinite in extent. We could take some arbitary cutoff, like the altitude below which lies 50% (or say 90%) of the mass of the atmosphere. A less arbitary, but mostly convenient, artifice is to calculate the height of an equivalent atmosphere with constant density, temperature and pressure, the same as the real atmosphere has at ground level. The top of the equivalent atmosphere is a sharp cutoff, with no atmosphere above it.
Note  

This height has a name like "height equivalent", but this name is used for many things, so presumably you have to specify it further, but I don't know how. A meteor traversing this height equivalent of atmosphere (all at a pressure of 1Atm) would encounter the same number of air molecules at it would in the real atmosphere. You would use this height if you wanted to calculate the refraction of stars near the horizon, using the refractive index of air at the ground. 
Here's some more info.
What's the height equivalent of the atmosphere (hint 1 ^{[257]} , hint 2 ^{[258]} ) ^{[259]} ? (The height of Mt Everest is 8.8km.)
At the meteorite's impact speed, how long would it have taken to traverse the height equivalent of the atmosphere (assume that deceleration did not change its velocity appreciably; you have to later check this assumption, which we will do below) ^{[260]} ?
What force was acting on the meteorite while it was traversing the atmosphere? We're going to make a quick side trip into aerodynamics and fluid mechanics (thanks to Matt at work for explaining the equation we're about to derive). We need a formula for the force acting on an irregularly shaped object falling through the atmosphere. Where are we going to get it (no, we don't need a book or wikipedia) ^{[261]} ? Make a list of factors that could affect the force acting on an object falling through the atmosphere. Here's my list ^{[262]} . Setup the simultaneous equations to determine the power to which you must raise each factor to get force. Do not solve them just yet. ^{[263]} .
We've got 4 unknowns, w,x,y,z but only 3 equations. We don't have enough information to uniquely solve for the 4 unknowns. We need an extra piece of information. We'll use experience (we could be wrong, but we'll find out when we test the equation by experiment). From eqn (2); y = 1. Next we tackle eqn (1). Obvious values to try are w=0, z=1 or w=1, z=0. We expect from experience that the aerodynamic braking will be dependant on the density of the fluid, so the combination w=1,z=0 is out. It's possible that w=½,z=½ or other combinations of square roots or inverse square roots. However that would result in strange values for x and z; e.g. the force would depend on the √A i.e. the diameter rather than A. Let's assume then that w=0 and see what that gets us. Here's my derivation of the force acting on the meteorite ^{[264]} . The formula we find for the force acting on an object falling through a fluid is
F = k.A.v^2.rho 
This is the answer confirmed by experiment. In fluid dynamics this formula is written F=½.ρ.v^{2}.A.C_{d} where C_{d} is the coefficient of drag (it's our missing dimensionless constant; I don't know where the ½ comes from). Fluid dynamics is a bit of a black art and AFAIK the coefficients of drag can't be calculated simply, even for a sphere. They are measured, or calculated by supercomputers. Have a look at a few values at Drag coefficient (http://en.wikipedia.org/wiki/Drag_coefficient). Notice how streamlined (aerodynamically slippery) objects have lower coefficients of drag.
Let's assume our meteor had C_{d}=1.0. Calculate the aerodynamic force acting on it at impact velocity ^{[265]} . What's the deceleration (calculate it in terms of g if you like) ^{[266]} ? This is not a particularly strong deceleration. Humans can survive this for a short time. The Apollo 11 astronauts, returning from the Moon, endured 10g when they first hit the earth's atmosphere. The deceleration wouldn't cause a rocky meteorite to break up, much less a Fe/Ni meteorite.
What velocity would the meteorite have had when it hit the ground (assume deceleration was constant) ^{[267]} ? What is the change in velocity? It's a.t. A back of the envelope calculation gives a change in velocity of a.t=104m/sec^2*0.65sec=68m/sec, a miniscule change in speed for an object doing 12.8km/sec. (Our initial estimate of the time to traverse the atmosphere assumed that the change in velocity was small. We've confirmed our assumption, thus validating the calculations so far.)
The atmosphere has little effect on the meteorite's velocity. As well only a miniscule fraction of the meteorite's energy is lost in the atmosphere (the K.E. on entering the atmosphere  the K.E. on impact).
The size and mass of the meteorite affect how much it's slowed by the atmosphere. A large enough meteorite, e.g. the Chicxulub Meteor (http://en.wikipedia.org/wiki/Chicxulub_crater), implicated in the extinction of the dinosaurs at the end of the Cretaceous, and estimated to be 10km in diameter, wouldn't even notice the ocean and would punch straight through the ocean floor to the mantle. A 1cm meteorite burns up in the atmosphere.
Even though only a miniscule fraction of the meteorite's energy is dissipated in the atmosphere, calculate the rate at which energy is dissipated (i.e. the power generated) ^{[268]} . Remember this power is being generated in a ball 50m in diameter.
Note  

Peak electrical consumption over the whole US in 2007 was 0.8Tw (peak consumption is in summer). (http://www.eia.doe.gov/cneaf/electricity/epa/epa_sum.html http://www.eia.doe.gov/cneaf/electricity/epa/epa_sum.html). You could calculate the length of time this meteorite could power the US. 
How much energy was produced by the meteor passing through the atmosphere ^{[269]} ? Below, we'll compare this to the energy produced on impact.
The StefanBoltzmann law (http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law) allows you to calculate the temperature of a black body from the radiance (power/area). (I'm not going to derive this. I'm just going to give you the result.)
StefanBoltmann constant: sigma = 5.67*10^8 W.m^2.K^4 surface area of meteor diameter 50m = 7850m^2 power radiated = 200Tw thus 200*10^12w = 5.67 * 10^8 W.m^2.K^4 * 7850m^2 *T^4 T^4 = 200*10^12/(5.67 *10^8 * 7850)K T = 26,000K 
This temperature (assuming a black body, reasonable in this case) allows you to determine the color of the maximum light intensity via Wien's displacement law (http://en.wikipedia.org/wiki/Wien%27s_displacement_law) which says λ_{max}=b/T.
lambda_max = b/T where b = 2.9 *10^6 nmK. 
Wien's displacement law gives rise to the concept of color temperature. As a body gets hotter, the wavelength of the emmitted light gets shorter (it shifts from red at 750nm to blue at 390nm; UV is shorter wavelengths). An incandescent light has a temperature of 3900°K and is a redbrown color with λ_{max}=743nm; the sun has a temperature of 5,500°K and is yellow with λ_{max}=527nm;; the blue sky is blue light scattered from sunlight and if it were to come from a black body, would have a temperature of 9000°K with λ_{max}=322nm. You can tell the temperature (and hence radiance) of an object from its color (e.g. the temperature of red hot metal, the sun, stars).
The eye/brain has a mechanism for color balancing, so that a white surface looks white whether it's illuminated by incandescent light, the sun, fluorescent light or blue sky. Cameras (film, CCD, MOS) have an equivalent mechanism; you tell the camera the color temperature of the light, or later, after taking the photo, you can correct the image. A photo taken in a room with incandescent light looks red when taken by a camera/film which is expecting daylight illumination.
A black body at 26,000°K would have its maximum radiation at
lambda_max = b/T b = 2.9 *10^6 nmK T = 26000K lambda_max = 2.9*10^6nmK/26*10^3K = 111nm 
The meteorite would be emitting mostly UV light.
Here's more useful information.
With the surface of the meteorite at 26,000°, we could expect that Fe would boil off the surface of the meteorite.
Note  

Meteorites pass too quickly through the atmosphere to heat up the insides. Only the outer 1cm shows any sign of heating. 
Assume that instead of heating the atmosphere to 26,000° the energy was used to boil off Fe (the meteorite then would be at 3000° or so and you would see a dull red ball streaking through the sky).
Here's some more info.
How much Fe was boiled off ^{[270]} ? The energy available could have boiled off 21kTonnes of Fe. Here's the estimated loss of mass (I don't know how they know this).
The meteorite must have been loosing mass from other mechanisms as well. It's likely that the meteorite was not a single fused piece of iron. It was more likely a loose aggregate of lumps of iron.
At what rate (mass/sec) was Fe/Ni ablating from the meteorite ^{[271]} ?
This material wasn't just falling off the side of the meteor after being blown off by the passing breeze. The meteorite was travelling at 40times the speed of sound and was preceded by a shock wave at about 10^{5}° ( The Magnetic Effect and Shock Wave of a Meteor http://adsabs.harvard.edu/full/1965SvA.....8..890I). This would have been a very bright 0.65secs.
Calculate the K.E. of the meteorite at impact ^{[272]} . Compare this to the energy that the meteorite lost in passing through the atmosphere  130TJ (0.13 * 10^{15}J). This is only about 1% of the meteorite's K.E. For impacts and explosions, energy released is usually converted into tons of TNT. The energy released by detonating TNT is 2.72MJ/kg (see strength of explosives http://en.wikipedia.org/wiki/Strength_(explosive)). How many tonnes of TNT is the Barringer Meteor Crater worth ^{[273]} ? Websites estimate the energy at 220 MTonnes. One that does a comparable calculation is Tswaing Crater S.A. (http://www.hartrao.ac.za/other/tswaing/tswaing.html)
Note  

The chemical reaction of explosives is not particularly energetic
compared to combustion.
The useful property of explosives is that they denotate.
A shock wave moving through the explosive sets it off.
The result is that all the material explodes at once,
rather than the material being blown apart by the explosion at the initiating site,
ibefore it can explode.
Here's some
heats of combustion
(http://en.wikipedia.org/wiki/Heat_of_combustion)

Here's some more info
Calculate the temperature of the meteorite immediately after impact, assuming no energy is transferred to the ground (all stays in the meteorite). ^{[274]} . So we now know why Barringer never found the iron from the meteorite  it vaporised on impact.
Assuming the K.E. of impact, instead of heating up the meteorite, was transferred to K.E. of excavated rock. Assume that the rock was all ejected at 45° (i.e. travelling the furthest distance for its initial velocity) (in the actual impact, the rock would get a distribution of energy, some travelling further and some travelling less distances). How far would the ejecta travel? Here's some input data.
First calculate the velocity of the ejected material ^{[275]} .
Note  

Escape velocity (http://en.wikipedia.org/wiki/Escape_velocity) from Earth is 11.2km/sec. The velocity of the ejecta would have a wide distribution. It's reasonable to expect that some of it would be blasted into space, never to return. Escape velocity from the Moon (2.4km/sec) and Mars (5.0km/sec) is lower than Earth's and for a similar impact, more ejecta would escape from these planet's gravitational field. In fact ejecta from other planets e.g. Lunar meteorites (http://en.wikipedia.org/wiki/Lunar_meteorite) and Martian meteorites (http://en.wikipedia.org/wiki/Mars_meteorite) have been found on the Earth's surface. finding a meteorite is hard enough. Realising that it's from the Moon or Mars is beyond my understanding. 
The next part is how far away did the material, ejected at 196m/sec, fall ^{[276]} ?
^{[1] }
F = mg = 2kg * 9.8m/sec2 = 18.6N 
^{[2] }
aerodynamics. The experiment works in a vacuum.
^{[3] }
F = mg = 9.8*1000kg=9.8kN P = F*s/t = F*v = 9.8kN*9.1m/sec = 90kJ/sec = 90kw 
^{[4] }
1.5kV: 90kW requires 60A cables (415V requires 214A).
^{[5] }
F = mg = 70 kg * 9.8 m/sec^2 = 686N W = F*s = 686N * 1000m = 686kJ P = W/t = 686kJ/(3*3600)sec = 63.5W 
^{[6] }
2 * 686kJ 
^{[7] }
Above I said that you need 2000 calories to produce 100W (approx) continuously all day.
to produce 63.5w for 6hrs you'd need (2000 * 6 * 63.5)/(100 * 24) = 317calories 
^{[8] }
Energy used = 2 * 686J = (2 * 686J)/(3600J) kwhr = 0.38kWhr cost @ 6c/kwhr = 0.38kwhr * 6c/kwhr = 2.3c 
^{[9] }
weight of coal burned = 0.38kwhr * 1 tonne / 2460 kwhr = 154g 
^{[10] }
infinite  the pendulum has no force to return it to "vertical" (whatever vertical means in zero gravity).
^{[11] }
t = 1/sqrt(g) = 1/sqrt(2.34) = 0.65sec 
^{[12] }
frequency = 1/sec dimension of frequency = 1/t = t^1 
^{[13] }
the unit of wavelength is a distance. the dimension is l 
^{[14] }
dimension frequency = t^1 dimension wavelength = l We don't know the answer, so let's look around; first divide wavelength by frequency The missing phyical quantity then would have dimensions l.t Do you know anything with dimensions l.t? Nope second try; multiply frequency and wavelength The missing phyical quantity then would have dimensions l.t^1 (l/t) Know anything which has dimensions l/t? Yes  velocity. The missing quantity is the velocity of the wave. Dimensional analysis says v = freq.wavelength 
^{[15] }
lambda = velocity/freq = 1024 ft/sec/(256/sec) = 4' 
^{[16] }
in water lambda = 4.3 * 4' = 17.2' in iron lambda = 5 * 4' = 20' 
^{[17] }
period = f(F,m,l) dimensions of F = m.l.t^2 equating dimensions on both sides m^0.l^0.t^1 = (m.l.t^2)^x.m^y.l^z Dimensions of LHS m^0 l^0 t^1 Dimensions of RHS m^(x+y) l^(x+z) t^2x Equating dimensions on both sides x + y = 0 (1) x +z = 0 (2) 2x = 1 (3) from (3) x = 0.5 (4) using (4) on (1),(2) y = 0.5 z = 0.5 The period of a wave on a string then is t = k.sqrt(m.l/F) 
^{[18] }
v = nu * lambda nu = 1/t v = l /(sqrt(m.l/T) = sqrt(T.l/m) for an infinitely long string, the mass isn't a useful number. Instead use mass/length = M v = k.sqrt(T/M) 
^{[19] }
dimension of energy = F*s = mas = m^1.l^2.t^2 properties of the ball: m,v let K.E. = m^x.v^z = m^x.l^z.t^z since the dimensions of KE = dimensions of E m^x.l^z.t^z = m^1.l^2.t^2 by inspection x = 1 y = 2 Thus K.E. = k*m.v^2 ;k is a dimensionless constant 
^{[20] }
dimensions of area = gt = l.t^2.t = l.t^1 = velocity 
^{[21] }
The velocity is initially ve and increases through the flight. The speed is the same in any frame of reference, so it's the same as above; the speed decreases till the ball stops and then increases again.
^{[22] }
inputs u = 0 v = 97kph = 97000/3600m/s a = ? t = 8.4sec v = u + at 97000/3600 = 0 + a*8.4 a = 3.2m/sec^2 = 3.2/9.8g = 0.33g 
^{[23] }
P = F*v v = 0 no power is being used 
If you hold the car with your foot on the brake, while you have the other foot on the accelerator, no power is being delivered to the road. Sure the water coolant and the transmission fluid are being heated up, but until a force moves through a distance, no power is being delivered.
^{[24] }
P = F*v F = ma = 1.5Tonne*3.2m/sec^2 v = 97kmph P = 1.5*10^3kg * 3.2m/sec^2 * (97*10^3/3600)m/sec = 130kW = 130/0.746HP = 170HP 
^{[25] }
v = u + at 290*10^3/3600m/sec = 0 + a*72 a = 1.1m/sec^2 = 1.1/9.8g = 0.11g F = m*a = 413*10^3kg*1.1m/sec^2 P = F*v = 413*10^3kg*1.1m/sec^2*290*10^3/3600m/sec = 40Mw 
^{[26] }
The ball will return to the ground when the vertical velocity is the negative of the initial vertical velocity. The time taken will be a function of ϑ.
^{[27] }
In the vertical plane v = u + at a = g # the numerical value of g is ve, i.e. 9.8m/s^2 uvert = u.sin(theta) therefore v = u.sin(theta) + gt the ball returns to the ground when v = u.sin(theta) thus u.sin(theta) = u.sin(theta) + gt time of hitting ground t = 2u.sin(theta)/g ;this is +ve, since g is ve 
^{[28] }
In the horizontal plane initial horizontal velocity = u.cos(theta) s = ut # use x for horizontal distance uhorz = u.cos(theta) t = 2u.sin(theta)/g thus x = u.cos(theta).2u.sin(theta)/g = (2.u^2/g).cos(theta).sin(theta) checking dimensions of RHS l/t.l/t.t^2/l = l 
^{[29] }
from trig identities
sin(ϑ).cos(ϑ)=sin(2ϑ)/2
∴ x = (2u^{2}/g).sin(2ϑ)/2
= u^{2}sin(2ϑ)/g
^{[30] }
the maximum value for sin(x) is 1 (occurs for x=90°). Thus the maximum range occurs when 2.ϑ=90° i.e. for ϑ=45°.
To maximise range (in a vacuum), elevate your canon at 45°.
^{[31] }
x = u^{2}/g (note: g is ve)
^{[32] }
s = u^2/g u = 328kph = 328*10^3/3600m/s = 91m/s s = 91^2/9.8 m = 845m 
^{[33] }
from the equation s = u^2/g the distance will be in the inverse ratio of accelaration. Thus the distance is multiplied by 6 distance on moon = 200m*6 = 1200m 
^{[34] }
A good golfer, on the moon, unencumbered by a space suit, using both hands and a driver could shoot the ball 1200m. I have a hard time believing that a one handed shot, in a space suit, with a 6iron, would go 800m.
^{[35] }
i+j = i + j i = i+j  j j = i+j  i 
^{[36] }
2i+2j = 2i + 2j = 2(i + j) i = (2i+2j  2j)/2 2j = 2i+2j  2i 
^{[37] }
(2i+2j)  2(i)  2j = 0 
^{[38] }
Presumably there's a formal proof of this, but I don't know it. Just saying "I can't figure out a way that this can be done, so it must be impossible", doesn't rate in mathematics.
^{[39] }
Two (remember, the number of basis vectors is the definition of the dimension of a space).
^{[40] }
You can't make the 0 vector by any linear combination of u,v.
^{[41] }
w = √2.v  u = i + j  i = j
y   u (0,1)</ (1.0,1.0) ^ /v (0.7,0.7) w  /  / / u (1,0) > x 
^{[42] }
work through the problem backwards. Start with the result you want then see what you need to do to get it from the basis vectors.
^{[43] }
w = v  √2.u = (i+j)/√2  2i/√2 = (i+j)/√2
y   sqrt(2)u </v (0.7,0.7) (0.7,0.7)  /  / / u (1,0) > x 
^{[44] }
w is a vector being resolved with respect to its basis set u,v, thus
w = au +bv
substituting values in Cartesian space for w,u,v
(1/√2,1/√2) = a(1,0) + b(1/√2, 1/√2)
resolving into x,y components
1/√2 = a + b/√2
1/√2) = b/√2
giving a,b = √2,1 and w = √2u,v or with respect to the basis vectors w = (√2,1).
^{[45] }
dimensions of (F = ma) = mlt^{2}
dimensions of (G = Fr^{2}/m^{2}) = (mlt^{2})(l^{2}m^{2}) = m^{1}l^{3}t^{2}
^{[46] }
No. Even if r is a vector, then r^{2} would be a dot product, which is number (or maybe a scalar, I don't know). This is a sloppy equation given to high school students. In college you multiply the RHS by r/r, i.e. the unit vector along the line joining the two masses. I'll use the symbol û for the unit vector along the radius of the orbit (there isn't an html entity r with a circumflex). Thus the vector form of the equation is
F = Gm_{1}m_{2}/r^{2}.û
The high school equation gives the magnitude of the force, but not the direction. For problems with only two masses, the direction is usually conveniently defined (e.g. along the xaxis, or you know that it's between the two masses). In this case, the direction of the force doesn't need to be explicitely defined in the equation.
I think the reason that people are taught bad physics, like equations in which one side is a vector and the other side isn't, is that the education system has decided that it's better to teach bad physics than none at all. Noone looks at the choice of teaching good physics or bad physics. The problem with bad physics is that it fits best with the students who aren't interested in it and who will forget it as soon as they can. The rest of us, some time in the future, will find ourselves, trying to apply our inadequate tools to some new problem. We'll be saying "I haven't a clue how to apply this to a multibody problem", when if we'd been taught to proper tools, we'd be saying "what's the multibody version of this vector operation?" With the latter realisation, you can find the answer. With the former realisation, you can't go anywhere. You're left feeling stupid.
If typing were taught on this principle, we'd all learn two finger huntandpeck. For those of us who go on to type for real, we'd have to unlearn our old style of typing and relearn to type with all fingers.
^{[47] }
a is the acceleration associated with a force. The force exerted on the mass at the end of the string points to the center of the circle. Thus the acceleration points to the center of the circle.
r is a vector pointing along the string holding the mass. It's not obvious which direction is +ve. Let's do some inspired guessing; is the focus of our interest the mass and what's making it move in a circle, or the force at the center of the circle which is doing the accelerating? If you don't already know the answer, you could make a case for either possibility. Let's look at the equation a = rω^{2} for hints. Since you've decided that a and r are in the same (parallel) (or opposite, antiparallel) direction, then ω^{2} must be a scalar. You know that ω is a vector and unless there's some complex math trickery going on (there isn't), then ω^{2}, which you have to assume is actually the dot product, must be +ve. So r is parallel to a. This means that the center of the framework of reference for our accelerated mass it the mass itself, rather than the center which is doing the attracting.
^{[48] }
The vector is along the axis of rotation i.e. orthogonal to the plane of rotation of the mass.
^{[49] }
The siderial period. In the geocentric framework (the one we should use to talk about the moon's orbit around the earth), the sun moves in the sky against the fixed stars. You use the fixed stars, not the sun, as a reference point for determining the orbital period of the moon about the earth.
^{[50] } libration
change in size
Note  

You can't see the moon's changing velocity, as the camera puts the moon in the middle of the field no matter where it is. 
^{[51] }
we see more of the southern hemisphere and then more of the norther hemisphere.
^{[52] }
This is a number you learn by heart, or remember the derivation. Napoleon (actually the French Academy of Science under his direction) defined the metre (http://en.wikipedia.org/wiki/Metre) such that the length of the Earth's meridian along a quadrant through Paris was 10Mm. The radius then is 40Mm/2π= 6366km (6000km for a backoftheenvelope calculation). Inconveniently for the French Academy of Science, the earth is not a sphere. As well the Academy didn't get the length of the meridian through Paris correct. Based on the provisional metre, the length of the meridian is 10,002 km. This estimate isn't too bad for the era (late 1700's). The radius of the earth varies from 6353 to 6384km (earth radius http://en.wikipedia.org/wiki/Earth_radius)
^{[53] }
He timed balls rolling down an inclined slope and found that they accelarated. His timing device was a klepsydra.
^{[54] }
The gravitational field G(r) at the surface of the earth is g, the accelaration due to gravity. The units of the gravitational field are are those of accelaration. Make sure you can derive this result from the formula for the force of gravity.
^{[55] }
The distance moved in each second is the difference between the total distance moved for each second. I'm using downward as the +ve direction.
t total distance,m in distance,m 1 sec interval 0 0 1 0.5 0.5 2 2.0 1.5 3 4.5 2.5 4 8 3.5 5 12.5 4.5 
^{[56] }
The distance moved per interval = average velocity.
Note  

The velocity is linear with time (as it's supposed to be). 
^{[57] }
s = at^2 a = 1m/sec^2 s = 18 t = sqrt(2*18) = +/6 secs 
^{[58] }
v = u + at u = 0 a = 1m/sec^2 t = +/6sec v = +/6m/sec 
^{[59] }
The vertical distance travelled s=0.
^{[60] }
s=ut+½a.t^{2}
s=0
a=g
u_{vert}=u.sin(ϑ)
∴ 0=u.sin(ϑ).t+g.t^{2}
t=0
or
t=2u.sin(ϑ)/g
^{[61] }
v^2 = u^2 + 2as u = 0 v = ? a = g s = 10m v^2 = 2as = 2*9.8*10 (m/sec)^2 v = 14m/s 
^{[62] }
v^2 = u^2 + 2as v = 60m/sec u = 0 a = g s = ? s = 60*60/(2*9.8) = 183m 
^{[63] }
v^2 = u^2 + 2as v = ? u = 14m/sec a = g s = 7.5m v^2 = 14^2 2*9.8*7.5 = 49 v = +/7m/sec 
^{[64] }
v^2 = u^2 + 2as v = ? u = 14m/sec a = g s = 15m v^2 = 14^2 2*9.8*15 = 98 can't take sqrt() of negative number. 
^{[65] }
The ball never gets to 15m above the launch point.
^{[66] }
v^2 = u^2 + 2as v = 0 u = 14m/sec a = g s = ? 0^2 = 14^2 2*9.8*s s = 10m 
^{[67] }
W = Fs = mas W/m = as (1) v^2 = u^2 + 2as (2) substituting (1) in (2) v^2  u^2 = 2W/m W = (1/2)mv^2  (1/2)mu^2 
^{[68] }
Height is acheived by converting kinetic energy (from running in the horizontal plane) into potential energy. Thus the athlete's job is to provide the maximum value of u just as they're about to leave the ground and convert it into velocity in the vertical direction. We'll assume that u, after leaving the ground, is the same on the earth and the moon.
^{[69] }
v^{2} = u^{2} + 2as.
Adopting the Cartesian conventions (up is +ve)
a=g # what sign is g?
At the top of the athlete's trajectory
v = 0
Thus
s=(u^{2})/2g=constant/g # note ve sign
At 1/6th gravity, the records would be 6 times as high (15m for high jump, 36m for pole vault).
^{[70] }
High jump record = 2.5m = 2.0m raising centre of gravity + 0.5m agility
On the moon, the high jump record = 2.0m*6 + 0.5m = 12.5m
^{[71] }
The pole vaulter's muscles give the athlete's body a certain amount of velocity and hence kinetic energy on leaving the ground. This energy is converted to potential energy as the athlete crosses the bar and this energy is converted back to kinetic energy when the athlete hits the ground again. The amount of energy is the same whether the athlete is on the moon or on the earth. You would use the same amount of cushioning on the moon.
^{[72] }
v=u+at
a=g
v=0
t=u/g
The time the athlete is in the air would be 6 times higher on the moon.
^{[73] }
I'm assuming that walking is a balancing act. If the gravity is 2.5 times higher, then to counteract any departure from equilibrium (i.e. stay upright), the life form has to react 2.5 times faster than we do, to any small changes in balance.
^{[74] }
v^{2}=u^{2}+2as
a=g
u_{vert}=u.sin(ϑ)
v=0
s=h
∴ h=(u.sin(ϑ))^{2}/2g
^{[75] } FIXME  make a note about a previous answer being 1/sqrt(2), and this one being ½
h_{45°}/h_{90°}=(sin(45°)/sin(90°))^{2}=½
^{[76] }
1::1. When launched straight up (ϑ=90°) the ball reaches v=0 at the top of the flight (i.e. K.E.=0 and all energy is P.E.). At the top of the arc for ϑ=45deg;, the projectile reaches half the height of that reached when launched straight up. This means for the same amount of initial K.E., half of it has been converted to P.E.
^{[77] }
K.E.=½mv^{2}. The height reached, h=(u.sin(ϑ))^{2}/2g is a function of the square of the initial velocity and hence is a function of the K.E. (h=K.E.*sin^{2}(ϑ)/mg).
^{[78] }
Height reached is determined by (potential) energy. For a particular geometry of launch (i.e. for any fixed ϑ), you'd expect that the height reached would be a function of the initial (kinetic) energy of the ball.
^{[79] }
K.E.=½mv^{2}. The distance travelled s=u^{2}sin(2.ϑ)/g. is a function of the square of the initial velocity and hence is a function of the K.E. (s=2*K.E.*sin(2ϑ)/mg).
^{[80] }
The distance travelled is a function of the initial (horizontal) velocity and the time of flight. The time of flight is a function of the initial (vertical) velocity. The distance travelled then is a function of the vertical velocity multiplied by the horizontal velocity, hence is dependant on the square of velocity, or energy. I don't know how you could see this without the equations.
^{[81] } Substitute u = v  at (from the first equation) in either of the other two equations. The result is
s = vt  ½at^{2} # doesn't involve u
^{[82] }
equation: s = vt  ½at^{2}
initial values:
v = (541*10^{3}/3600) = 150 m/sec
s = 400m
t = 4.428sec
answer:
a = 25m/sec^{2} = 2.6g
^{[83] }
1. from v = u + at
initial values
v = 150 m/sec
a = 25m/sec^{2}
t = 4.428sec
answer:
u = v  at = 150  110
= 40m/sec
2. from s = ut + ½at^{2}
initial values:
s = 400m
t = 4.428sec
a = 25m/sec^{2}
answer:
u = 400  ½25*4.4^{2}
= 36m/sec
^{[84] }
equation: v = u + at
input values: u = 0, a = 25m/sec^{2}, t = 4.4sec
answer: v = 110m/sec
^{[85] }
equation: v = u + at
input values: u = 0, t = 4.4sec
answer: a = 34m/sec^{2} = 3.5g
^{[86] }
equation: v^{2} = u^{2} + 2as
input values: u = 0, a = 34m/sec^{2} = 3.5g, t = 4.4sec
answer: s = 330m
^{[87] }
from v = u + at, a = (v  u)/t
substituting in s = ut + at^{2}
^{[88] }
0. If the particle moved, another force, with a component in the allowed direction of movement, must have set the particle in motion.
^{[89] }
ϑ=0 so cos(ϑ) is +ve, so F⋅s is +ve.
^{[90] }
P.E. = F⋅s = ma⋅s = mg⋅h
^{[91] }
slope = mg #not mg (g is ve)
^{[92] }
v^{2} = u^{2} + 2a⋅s
s = h
a = g # g is ve (it's in the opposite direction to h).
∴ v^{2} = u^{2} + 2g⋅h
K.E. = ½mv^{2} = ½m(u^{2} + 2g⋅h) = ½mu^{2} + mg⋅h
^{[93] }
slope = mg #note: not mg
^{[94] }
E_{tot} = K.E. + P.E. = ½mu^{2}
^{[95] }
P.E = K.E + ½mu^{2},
slope = 1
intersecting the P.E. axis at (K.E.,P.E)=(0,½mu^{2})
intersecting the K.E. axis at (K.E.,P.E)=(½mu^{2},0)
^{[96] }
Conservation of energy. Any change in the energy along one of the axes (P.E. or K.E.) requires an equal but opposite change in the other.
^{[97] }
1/2.mu^2 o h=1/2.u^2/g  o  o PE  o  o  o h=0  KE 1/2.mu^2 
The ball starts at bottom right, moves up the line to top left (at the top of the flight) and then returns to the bottom right (at the end of the flight).
^{[98] }
The ball would head off in the SE direction, with P.E. decreasing and K.E. increasing (without limit).
^{[99] }
P.E. changes linearly with h, (and by conservation of E, so will K.E.), so the ball will move with uniform velocity, as function of height, from the bottom right to the top left and then return with uniform velocity back to the bottom right.
^{[100] }
(ΔKE,ΔPE) = (mg⋅h, mg⋅h)
Note: with g ve, ΔKE is initially ve, and ΔPE is initially +ve.
^{[101] }
You could do any of the following (since the line has slope=1, they're all equivalent). Use the position of the ball at h=0 as the origin.
^{[102] }
abscissa of ball in (K.E.,P.E.) system = ½mu^{2} + mg⋅h
starting abscissa of ball in (K.E.,P.E.) system = ½mu^{2} (this is E_{tot} = K.E._{t=0})
projection on ΔKEaxis = abscissa  starting abscissa = mg⋅h
^{[103] }
ordinate of ball in (K.E.,P.E) coordinates = P.E. = mg⋅h
starting ordinate in (K.E., P.E) coordinates = 0
projection on ΔPEaxis = ordinate  starting ordinate = mg⋅h
^{[104] }
Note: the values of ΔKE and ΔPE are the ve of each other. By Pythagoras then, the distance is √2 times the distances just found. distance then = √[2*(mg⋅h)^{2}]
^{[105] }
equation: s = ut + ½at^{2}
initial value: a = g
initial value: s = h
P.E. = mas = mg⋅h = mg(ut + ½gt^{2})
^{[106] }
P.E. = mg⋅ut  ½m(gt)^{2}
d(P.E.)/dt = mg(u + gt)
We set d(P.E.)/dt = 0 and find t. To be rigourous, we should then differentiate again to see whether the stationary point is a maximum or minimum, but we already know that it's a maximum.
t = u/g
^{[107] }
P.E. = mg(ut + ½gt^{2})
for t = u/g
P.E. = ½mu^{2}
^{[108] }
equation: v = u + at
initial value: a = g
K.E. = ½mv^{2} = ½m(u + at)^{2} = ½mu^{2} + mgut + ½m(gt)^{2}
^{[109] }
K.E. = ½mu^{2} + mgut + ½m(gt)^{2}
d(K.E.)/dt = mgu + mgt
∴ t = u/g
K.E._{min} = 0
^{[110] }
E_{tot} = K.E. + P.E. = ½mu^{2}
^{[111] }
The ball would head off in the SE direction, with P.E. decreasing and K.E. increasing (without limit). We're using the same diagram as for E=f(h), so the ball will move the same way as we calculated for E=f(h).
^{[112] }
When P.E. is a maximum, the P.E. changes slowly with time (one of the defining features of a stationary point). We can expect the locus of the ball on the (K.E.,P.E.) graph to move slowly near the top left of the graph (K.E.,P.E.)=(0,½mu^{2}).
^{[113] }
(ΔKE,ΔPE) = (½mu^{2} + mgut + ½m(gt)^{2} , mg(ut + ½gt^{2}))
^{[114] }
abscissa of ball in (K.E.,P.E. system) = ½mu^{2} + mg(ut + ½gt^{2})
starting abscissa in (K.E.,P.E. system) = ½mu^{2} (this is E_{tot} = K.E._{t=0})
projection on ΔKEaxis = abscissa  starting abscissa = mg(ut + ½gt^{2})
^{[115] }
ordinate of ball in (K.E.,P.E) coordinates = P.E. = mg(ut + ½gt^{2})
starting ordinate in (K.E.,P.E) coordinates = 0
projection on ΔPEaxis = ordinate  starting ordinate = mg(ut + ½gt^{2})
^{[116] }
Note: the values of ΔKE and ΔPE are the ve of each other. By Pythagoras then, the distance is √2 times the distances just found. distance then = √[2*(mg(ut + ½gt^{2}))^{2}]
^{[117] }
P.E. = mg⋅h
v^{2} = u^{2} + 2a⋅s
a = g, s = h
h = (v^{2}  u^{2})/2g
∴ P.E. = ½m(u^{2}  v^{2})
^{[118] }
v = u or v = u.
^{[119] }
u is a constant. P.E. is maximum when v^{2} is a minimum (not when v is a minimum).
^{[120] }
a parabola
^{[121] }
Yes. You know this two ways
^{[122] }
roots = [b ± √(b^{2}  4ac)]/2a
if b = 0, then both roots are symmetrical around the yaxis roots = ±√(c/a)
^{[123] }
Concave downwards. P.E. is a maximum.
^{[124] }
concave upwards
^{[125] }
¼mv^{2}. You can see this by eyeballing the graphs looking for P.E._{max}.
^{[126] }
In the flight (graphs with s or t on the abscissa) v is ve at the finish (end) of the flight i.e. on the RHS of the graph, not the LHS as here.
^{[127] }
E_{tot} = K.E.+ P.E. = ½mu^{2}
^{[128] }
1/2mu^2 o v=0  o  o PE  o  o  o v=u,u  KE 1/2mu^2 
As before in the sections on E=f(h),f(t), the ball starts at bottom right, moves up the line to top left (at the top of the flight) and then returns to the bottom right (at the end of the flight). If there were no ground, the ball would continue to the bottom right.
^{[129] }
K.E.=f(v^{2}). When v≅0, K.E. will change slowly with v. I would expect that the ball would move slowly along the K.E.,P.E. line near the top left. At large v (+ve or ve), K.E. and P.E. change as v^{2} so the locus of the ball would move quickly near the origin.
^{[130] }
(ΔKE,ΔPE) = (½m(v^{2}  u^{2}) , ½m(u^{2}  v^{2}))
^{[131] }
projection on ΔKEaxis = ½m(v^{2}  u^{2})
^{[132] }
projection on ΔPEaxis = ½m(u^{2}  v^{2})
^{[133] }
Note: the values of ΔKE and ΔPE are the ve of each other. By Pythagoras then, the distance is √2 times the distances just found. distance then = √[2*(½m(u^{2}  v^{2}))^{2}]
^{[134] }
P.E. = mgh
equation: s = ut + ½at^{2}
values:
s = h
a = g
u_{vert} = u.sin(ϑ)
∴ P.E. = mg(ut.sin(ϑ) + ½gt^{2})
^{[135] }
parabola. We already know that the height of the ball in the 2D problem is a parabola. As well, P.E. is proportional to height (i.e. the original parabola multiplied by a constant, in this case a constant with dimensions), so the function P.E. = f(t) is also a parabola.
^{[136] }
P.E. = mg(ut.sin(ϑ) + ½gt^{2})
if P.E. = 0, then
t = 0 or
(u.sin(ϑ) + ½gt) = 0
i.e. t = 2u.sin(ϑ)/g
^{[137] }
P.E. = mg(ut.sin(ϑ) + ½gt^{2})
substituting t_{top} = u.sin(ϑ)/g
gives P.E._{top} = ½mu^{2}sin^{2}(ϑ) (= ½m(u_{vert})^{2})
^{[138] }
P.E. = mg(ut.sin(ϑ) + ½gt^{2})
substitute t = T  u.sin(ϑ)/g.
giving P.E. = ½m(g^{2}T^{2}  u^{2}sin^{2}(ϑ))
∴ P.E.  ½mu^{2}sin^{2}(ϑ) = ½mg^{2}T^{2}
^{[139] }
concave down. P.E. goes in the same direction as height (P.E. becomes more +ve as height becomes more +ve), so the P.E. parabola is concave downwards (as is the path of the parabola describing the x,y coordinates).
^{[140] } T = 0. The point of translating the xaxis was so that T would be 0 at the vertex of the parabola. The parabola is symmetrical about T = 0 (we get the same value for the LHS for T = x, T = x) confirming our initial assumption that this value of T is the center of the flight.
^{[141] }
P.E. = ½m(g^{2}T^{2}  u^{2}sin^{2}(ϑ))
if P.E. = 0 then
g^{2}T^{2}  u^{2}sin^{2}(ϑ) = 0
∴ T = ±u.sin(ϑ)/g
Note  

there are two values for T, symmetrical about the line T = 0. 
To find t for P.E. = 0 (if we didn't already know it), we substitute our axis translation formula t = T  u.sin(ϑ)/g, to give us t for the start and the end of the flight.
^{[142] }
J.E. = ½mg^{2}T^{2}
^{[143] }
Vertical component of K.E. as a function of time
K.E.=½mv^{2}
equation for velocity: v=u+at
input values:
u_{vert}=u.sin(ϑ)
a=g
∴ K.E._{vert}=½m(u.sin(ϑ)+gt)^{2}
Note  

Do a dimensional check. (You don't have to go right down to m,l,t; just show that all terms look like mv^{2}.) RHS = m(u+gt)^{2} = m(v+v)^{2} 
^{[144] }
Horizontal component K.E. as a function of time
K.E.=½mv^{2}
equation for velocity: v=u_{horz}=u.cos(ϑ)
Note  

the horizontal velocity is constant (it's not a function of t), so the horizontal K.E. will also be constant. 
∴ K.E._{horz}=½m.(u*cos(ϑ))^{2}
Note  

Can you see by inspection that the RHS has the correct dimensions? 
^{[145] } Total K.E. at a function of time
K.E._{total} = K.E._{horizontal} + K.E._{vertical}
K.E. component values:
K.E._{vert} = ½m(u.sin(ϑ) + gt)^{2}
K.E._{horz} = ½m.(u*cos(ϑ))^{2}
K.E._{total} = ½m(u.sin(ϑ)+gt)^{2} + ½m.(u*cos(ϑ))^{2}
After some manipulation (and using a familiar trig identity)
= ½m(u^{2} + 2u.sin(ϑ).gt + (gt)^{2})
Note  

Some of the terms from K.E._{vert} and K.E._{horz} added (via a trig identity) to give a term in u^{2}, i.e. in terms of the initial velocity, rather than its components. 
Note  

confirm that all terms have the same dimensions as m.v^{2}. 
^{[146] }
The roots of the parabola show where it crosses the xaxis. If it has no real roots, then it doesn't cross the xaxis. Let's find the roots of the quadratic equation for K.E. We only need the √(b^{2}4ac) part, since this will tell us if we have real roots.
^{[147] }
Here's our formula for K.E. K.E. = ½m(u^{2} + 2u.sin(ϑ).gt + (gt)^{2}) .
term inside √ for quadratic formula
= 2ug.sin(ϑ))^{2}  4u^{2}g^{2}
= 4u^{2}g^{2}(sin^{2}(ϑ)  1)
^{[148] }
The energy will be constant. A plot of E=f(t) will be a horizontal line. It's relatively simple to prove (left as an exercise for the student) that the two parabolas have the same (but opposite) curvature and their vertices at the same t.
^{[149] }
P.E. + K.E. = ½mu^{2}
^{[150] }
K.E._{tot,ϑ}  K.E._{tot,ϑ=90°}
= (½m(u^{2} + 2u.sin(ϑ).gt + (gt)^{2}))  (½m(u^{2} + 2u.gt + (gt)^{2}))
= mg(1sin(ϑ))ut #remember g is ve
^{[151] }
P.E._{ϑ} = mg(ut.sin(ϑ) + ½gt^{2})
P.E._{ϑ}  P.E._{ϑ90°}
=  mg(ut.sin(ϑ) + ½gt^{2}) + mg(ut + ½gt^{2})
= mg(1sin(ϑ)ut #g is ve
.^{[152] }
There is no horizontal component for P.E.
equation: P.E. = mgh
^{[153] }
equation: K.E. = ½mv2
equation: v^{2} = u^{2} + 2a⋅s
initial value: a = g
initial value: s = h
initial value: u_{vert} = u.sin(ϑ)
∴ K.E._{vert} = ½m(u^{2}sin^{2}(ϑ) + 2g⋅h)
^{[154] }
The horizontal component of velocity is constant, and hence the horizontal component of K.E. is constant.
equation: K.E._{horz} = ½mu^{2}cos^{2}(ϑ)
^{[155] }
equation: K.E._{tot} = K.E._{horz} + K.E._{vert} = ½mu^{2}cos^{2}(ϑ) + ½m(u^{2}sin^{2}(ϑ) + 2g⋅h) = ½mu^{2} + mg⋅h
^{[156] }
E_{tot} = P.E. + K.E._{vert} + K.E._{horz} = ½mu^{2}
^{[157] }
P.E. = mg⋅h
v^{2}_{vert} = u^{2}_{vert} + 2a⋅s
s = h
a = g
u_{vert} = u.sin(ϑ) #note: u_{vert} is a constant, while v varies.
h = (v^{2}_{vert}  u^{2}_{vert})/2g
∴ P.E. = + ½mv^{2}_{vert}  ½m(u.sin(ϑ))^{2}
^{[158] } by Pythagoras
v^{2} = v^{2}_{vert} + v^{2}_{horz}
and v_{horz} = u.cos(ϑ)
thus v^{2}_{vert} = v^{2}  (u.cos(ϑ))^{2}
∴ P.E. = ½mv^{2}  ½[(u.cos(ϑ))^{2} +(u.sin(ϑ))^{2}] = ½m(v^{2}  u^{2})
^{[159] }
K.E._{horz} = ½mu_{horz}^{2} = ½m(u.cos(ϑ))^{2}
^{[160] }
K.E._{vert} = ½mv_{vert}^{2}
^{[161] }
with help from Pythagoras (v^{2}_{vert} = v^{2}  (u.cos(ϑ))^{2})
K.E._{tot} = ½m((u.cos(ϑ))^{2} + v_{vert}^{2}) = ½mv^{2}
^{[162] }
E._{tot} = P.E. + K.E._{horz} + K.E._{vert} = ½mu^{2}
^{[163] }
The coordinates of the vector v as a function of time are
v = (v_{horz}, v_{vert})
v_{horz} = u.cos(ϑ)
This is a constant (i.e. d(v_{horz})/dt = 0 ), so the locus of the line has constant v_{horz}
v_{vert} = u.sin(ϑ) + gt
d(v_{vert})/dt = g
i.e. the locus has v_{vert} varying uniformly with time (at constant acceleration of g).
The velocity (as a cartesian vector) of the locus as a function of time is dv/dt = (0,g)
The locus is a straight line moving uniformly as a function of time (it's moving downwards since g is ve). The dots would be equally spaced if drawn at constant time intervals.
^{[164] }
The cartesian coordinates are (r,angle). Since we're already using ϑ for an angle symbol, use Φ (phi) for the angle. The cartesian coordinates then are (r,Φ).
r = √(v_{horz}^{2} + v_{vert}^{2}) = √( (u.cos(ϑ))^{2} + (u.sin(ϑ) + gt)^{2} ) = √( u^{2} + 2ugt.sin(ϑ) + (gt)^{2}) )
Φ = tan^{1}(v_{vert}/v_{horz}) = tan^{1}((u.sin(ϑ) + gt)/u.cos(ϑ)) = tan^{1}(tan(ϑ) + gt/u.cos(ϑ))
^{[165] }
t = 0
(r,Φ) = (u, 45°)
^{[166] }
^{[167] }
^{[168] }
^{[169] }
h = x.tan(ϑ) + ½g(x/u.cos(ϑ))^{2}
Substituting x_{start} = 0, gives (as expected) h = 0 .
^{[170] }
h = x.tan(ϑ) + ½g(x/u.cos(ϑ))^{2}
Substituting x_{top} = u^{2}sin(ϑ)cos(ϑ)/g .
gives (as expected) h = (u.sin(ϑ))^{2}/g + ½(u.sin(ϑ))^{2}/g = ½(u.sin(ϑ))^{2}/g
^{[171] }
h = x.tan(ϑ) + ½g(x/u.cos(ϑ))^{2}
Substituting x_{end} = 2u^{2}sin(ϑ)cos(ϑ)/g .
gives (as expected) h = (2u.sin(ϑ))^{2}/g + (2u.sin(ϑ))^{2}/g = 0
^{[172] }
h = 0.∞ + ½g(0/u.0)
^{[173] }
P.E. = mg⋅h = mg(x.tan(ϑ) + ½g(x/u.cos(ϑ))^{2})
^{[174] }
K.E._{tot} = ½m(u^{2} + 2gx.tan(ϑ) +(gx/u.cos(ϑ))^{2})
^{[175] }
x_{start}=0; K.E._{start} = ½mu^{2}
^{[176] }
(x_{top} = u^{2}sin(ϑ)cos(ϑ)/g)
K.E._{top} = ½m(u^{2}  2u^{2}sin^{2}(ϑ) + u^{2}sin^{2}(ϑ)) = ½mu^{2}(1  sin^{2}(ϑ)) = ½m(u.cos(ϑ))^{2} = K.E._{horz}
^{[177] }
(x_{end} = 2u^{2}sin(ϑ)cos(ϑ)/g)
K.E._{end} = ½m(u^{2}  4u^{2}.sin^{2} + 4u^{2}.sin^{2}) = ½mu^{2}
^{[178] }
E_{tot} = ½mu^{2}
^{[179] }
The projectile is moving in a conservative field; it has the same total energy anywhere in the path. The P.E. at the beginning and end of the flight is the same, so the K.E. at impact is the same as at the beginning of the flight. It's as if the golf club whacked the ground (or your head) instead of the ball. The ball just acted as the agent to deliver the energy.
^{[180] }
The height is the same so the P.E. is the same (in a gravitational field, P.E. is only a function of height).
^{[181] }
v = u + at
initial conditions
u_{vert} = u.sin(ϑ)
a = g # g is ve
The flight ends when u_{vert} = u.sin(ϑ)
the time of flight is t = 2u.sin(ϑ)/g
This is maximum for ϑ=90°, when t = 2u/g
^{[182] }
the time of flight: t = 2u.sin(ϑ)/g
max time of flight: (ϑ=90°) t = 2u/g
Time of flight for launch angle ϑ, compared to time for longest flight: = sin(ϑ).
launch angle for max throw: ϑ = 45°
Time of flight for maximum throw compared to time of longest flight: sin(45°) = 0.707.
^{[183] }
x x exeeeeeexe  x x KE=x        PE=o   o o oo o o x > 
^{[184] }
equation: s = ut + ½at^{2}
Since we're only interested in the height, we don't need to know anything about the horizontal plane.
initial values
h = s
a = g
u_{vert} = u.sin(ϑ)
The slope of the h=f(t) graph is
dh/dt = u.sin(ϑ) + gt
^{[185] }
equation: s = ut + ½at^{2}
initial conditions:
a = g
u_{vert} = u.sin(ϑ)
s = h
∴ h = ut.sin(ϑ) + ½gt^{2}
^{[186] }
dh/dt = u.sin(ϑ) + gt
^{[187] }
The differential of position with respect to time is velocity. The dimensions of all terms will be velocity.
^{[188] }
dh/dt = u.sin(ϑ) + gt
for t = 0
dh/dt = u.sin(ϑ)
^{[189] }
The parabola is horizontal when the derivate is 0 and the vertical velocity is 0. These two statements are the same thing.
^{[190] }
0 = u.sin(ϑ) + gt
t = u.sin(ϑ)/g
^{[191] }
P.E. = mg(ut.sin(ϑ) + ½gt^{2})
K.E. = ½m(u^{2}+2u.sin(ϑ).gt+(gt)^{2})
P.E./K.E. = mg(ut.sin(ϑ) + ½gt^{2}) / ½m(u^{2}+2u.sin(ϑ).gt+(gt)^{2})
= (2u.sin(ϑ).gt + (gt)^{2}) / (u^{2}2u.sin(ϑ).gt+(gt)^{2})
^{[192] }
u=i+0*j
v=0*i+j
DP=u•v=1*0+0*1=0
^{[193] }
u=i/√2+j/√2
v=i/√2+j/√2
DP=u•v=(1/√2)^{2}+(1/√2)^{2}=1
^{[194] }
u=i/√2+j/√2
^{[195] }
v=i/2+√3j/2
^{[196] }
DP = u1*v1+u2*v2 = 1/2√2 + √3/2√2
x60:~# echo "(1+sqrt(3))/(2*sqrt(2))"  bc l .96592582628906828674 x60:~# echo "c(3.14159*15.0/180.0)"  bc l .96592588352234270171 
the angle between the two vectors is 15°.
^{[197] }
The Lorentz force is at right angles to the velocity, so there in no change in the magnitude of the velocity. However the direction of the velocity vector will change; the electron will swing to the right. Here is the electron a bit later (ignore the limitations of ascii art; assume the sloping lines are all at 45°)
. . . . . . . . . . . . . . . . . . . . . . F. . . . . . ./ . . . . .  e . . . . . v\ . . . . . . . . . . . . . . . . . . . 
Note: the electron is still moving with velocity v and still being acted on by the same force, but now the electron is moving at 135° to the xaxis.
Here's the electron after another interval of the same length.
. . . . . . . . . . . . . . . . . . . ^ . . . . . . .v e >F . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 
The force is still acting at right angles to the velocity, so v doesn't change, and the electron keeps swinging to the right.
The electron moves at constant speed, with a constant force at right angles to its speed. The electron follows a circle at constant angular velocity. In the magnetic field we have here, the path of the electron is a semicircle. It will emerge from the magnetic field with the magnitude of the velocity unchanged, but now it will be going right.
. . . . . . . . >v . . . . . . . . e . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 
^{[198] }
cut the rope and insert a spring weighing scale (find a picture with google if you need to). What will the scale register?
^{[199] }
You calibrate the scale. The original definition of mass had 1kg=(mass of 1000cc of water). You calibrate the spring scale with buckets of water.
^{[200] }
normal
^{[201] }
cross product
^{[202] }
vector
^{[203] }
Note  

You don't know this yet, but all the rotational vector cross products point along the axis of rotation; e.g. torque, moment, angular momentum. 
The axis of rotation is the only thing in 3D that has the same geometric relationship to the object/force, no matter where the force/object is on the circumference of the circle.
^{[204] }
M = F*d*sin(30°) + F*d*sin(30°) = F*d/2 + F*d/2 = F*d
^{[205] }
M = 2*F*sin(30°)*d/2 = F*d/2
^{[206] }
Here's one version. The vector diagram for the forces at the left hand end of the bar make up an equilateral triangle. The translational force then is F pushing to the right. We've already calculated the moment.
>F ^ F\ d/2 d/2 O \ F v 
Here's another version. The translational force is also F pushing to the right. The moment is the same as above (the moment is the same whether the force is 30° to the left or to the right of the normal to the bar).
^ F / d/2 d/2 O / F v >F 
^{[207] }
The steering wheel shaft exerts an equal and opposite force on the steering wheel. The steering wheel doesn't accelerate vertically, but does rotate.
  / \ F   O   d \ ^ / v    F 
Here's the diagram with the force from the right hand moved to the steering wheel shaft and the moment replaced by a couple applied at the circumference of the steering wheel. The force(s) at the circumference of the steering wheel are F/2.
F   ^ / v \ F/2    O   d F/2 \ ^ / v    F 
^{[208] }
M = Fd
M = 160ftlbs
d = 2'
F = 160/2 = 80lbs
^{[209] }
W = Fs = (T/r)*2πr = 2πT
This is the correct answer (you can measure the heat output to the brake).
Note  

The work extracted from a rotating shaft is independant of the distance from the center of the shaft that you extract the force. At a larger distance, the force is less, but the distance the force moves is proportionally more. Thus you can extract the force from a ring gear around the outside of the flywheel, from a clutch plate covering the surface of the flywheel , or connect directly to another shaft. 
^{[210] }
W = F•s = (T/r) • 2πr = 2πT
Note  

What is T/r when both T,r are orthogonal vectors? Is there a cross divide like there's a cross product? You would have to hope, from the definition of torque, that T/r is the vector F. 
Note  

Since W is a scalar, (T/r)•2πr is going to be a scalar. If we just cancel out the two r factors, we're left with T (a vector and not a scalar) as a result. Clearly T*(r/r) = T but (T/r)•r ≠ T*(r/r) Here we see the consequence of not being told about vectors when presented with them. I don't know the answer here, except to say that if you unvectorize/scalarize T, you get the numerical right answer. (It's no wonder Space Shuttles blow up.) 
^{[211] }
P = W/t
f = 1/t
W = 2πT
P = 2πTf
^{[212] }
P = 2πTf
P = 241kW, f = 6400/60 = 107 sec^{1}
T = P/2πf = 241*10^{3}/(2π*107) = 360Nm
Inspection of the graph gives about 375Nm, which is close enough.
^{[213] }
The torque is a line of slope 1. The speed is a line of slope +1. The power out is the product of these two.
^{[214] }
This is a pair of first class levers, with the joint being the fulcrum.
^{[215] }
A cantilever is supported at one end. This type of design is required when the structure can't be supported from above or below. The design problem of a cantilever is that the forces (here F_{1}, F_{2}) become large as the cantilever becomes longer.
mg  v   ^ v  F1 F2      
It's not particularly useful to regard either of the support points as a fulcrum. You have to design both of them to handle the forces involved.
^{[216] }
A shovel is the same as a cantilever, with your hands providing F1,F2. To make the forces smaller, you keep one hand on the handle and move the other hand on the shaft as close to the load as possible (compatible with being able to maneuver the load). There's no way of assigning a fulcrum to either of the places where your hands grip the shovel.
^{[217] }
This is the same problem as the shovel and the cantilever.
^{[218] }
m = 1
^{[219] }
W = mgs
^{[220] }
Note  

mass M is being lifted up. It isn't doing any work; it's having work done on it. 
You can find the distance mass M is elevated by trigonometry, but by inspection the distance is sl_{M}/l_{m} or s/4 in the example above.
This ratio is called the mechanical advantage. The force you need to move the mass is reduced by the mechanical advantage. The distance you move that force is increased by the mechanical advantage.
The amount of work done by M then is
W = Mgsl_{M}/l_{m}
since M*l_{M} = m*l_{m}
then the amount of work done by M then is
W = mgs
^{[221] }
total work = 0
^{[222] }
The fulcrum must be pushing back on the lever with an equal and opposite force. From this information (that the lever is stationary), we can't tell why it is stationary, only that it is so. To find out why we have to investigate why objects are rigid (they don't collapse when you push on them  they push back).
^{[223] }
The mass M has a zero moment arm about the point under it, so it contributes no moment.
The fulcrum at distance l_{M} pushes anticlockwise with a force (M+m)g. Its moment about the point under mass M is
moment_{fulcrum} = l_{M}*(M+m)g
The mass m at distance (l_{M}+l_{m}) pushes clockwise with force mg. Its moment about the point under mass M is
moment_{m} = (l_{M}+l_{m})*mg
The total moment about the point under mass M then is
moment_{total} = l_{M}*(M+m)g + (l_{M}+l_{m})*mg = g(l_{M}*M l_{M}m + l_{M}*m + l_{m}*m) = g(l_{M}*M + l_{m}*m)
From the lever being in equilibrium about the fulcrum we know that (l_{M}*M + l_{m}*m)=0, so the moment about the point under the mass M is 0.
^{[224] }
The lever is pushing back with force=Mg. Again our observation doesn't tell us why. All we know is that the mass isn't moving, so the resultant force must be 0.
^{[225] }
F_{1},F_{2} get very large (infinity large).
^{[226] }
F_{2}  F_{1} = mg
^{[227] }
Eqn (2)  Eqn (1) gives
F_{2}  F_{1} = mg/s  mg(1s)/s = mg
^{[228] }
If you push on a bar/rod, hinged at the other end, and it doesn't rotate about the hinge, then you have to push along the bar. The resultant has to be along the bar.
^{[229] }
a_{h}/a_{v}=tan(ϑ)
ϑ = 90°, thus tan(ϑ) = ∞
∴ a_{v} = 0
The hinge point A is not taking any of the weight of M. The arm AM pulls out on the wall, and the wall pulls back on AM.
^{[230] }
compression. You should be able to tell this without any math, because BM is supporting the flowers. From the diagrams and our math(s), both hinges are pushing on BM.
^{[231] }
theta multiple 0 1 30 1.15 45 1.41 60 2.0 75 3.86 
^{[232] }
a_{h} exerts a force on M producing a moment about B.
^{[233] }
tension
^{[234] }
multiple theta sloping horizontal arm arm 0 1 0 30 1.15 0.577 45 1.41 1 60 2.0 1.73 75 3.86 3.73 
^{[235] }
We know that the window box is not accelerating horizontally (it's stationary).
^{[236] }
With smaller ϑ the compression increases.
^{[237] }
R_{AC} = ½Mg
Each element takes half the weight of the train.
^{[238] }
R_{AC} = cosec(45°)/cosec(90°) = √2
^{[239] }
Each half of the bridge has a resultant horizontal force of zero. Why this is true is not immediately obvious from looking at the bridge.
It's reasonable that both halves would push (or pull) on each other, and with the bridge being laterally (left/right) symmetrical, the resultant of the forces from each of the two halves is zero. The two halves exert moments on each other, but why do they exert no horizontal force?
Here's half the bridge (supporting half the weight) showing the horizontal forces on it. (Equivalently we could have shown the horizontal forces that the half bridge exerts on its surroundings.) The top of the bridge is being pushed outwards at the top and pulled inwards at the bottom (both by the other half of the bridge). Again, the resultant force is 0.
B o o <bh .  .  .  o theta  oo o bh> A D Mg/2 v 
If each half pushed (or pulled) on the other half, what change would we have to make to our diagram? If the left half above was being pushed by the right half (here the force is in some arbitary location; we can find the exact location later if needed), by force F then our diagram would become
B o o <bh .  .  <F .  o theta  oo o bh> A D Mg/2 v 
(Note we don't know if F is +ve or ve yet.) Now we have a resultant force F on the left half of the bridge, but we know it isn't accelerating (it's stationary), so there must be a corresponding cancelling (or balancing force) Ffrom somewhere, and since we've accounted for all the forces from the right side of the bridge (F remember), then the balancing force must come from the left of the left hand half of the bridge, which is the pylon support. Here's our diagram now.
B o o <bh .  .  <F .  o theta  F> oo o bh> A D Mg/2 v 
Since we've accounted for all moments on the bridge, then the two new forces F must exert no moment, so the right hand F must be colinear with the left F and hence act at D.
B o o <bh .  .  .  o theta  F> oo o bh> <F A D Mg/2 v 
The diagram above indicates that the pylon pushes inward with a force F. We know this force is 0, as we can (in principle) drop a preassembled king post bridge in place, resting on the ground. The pylons don't compress or tension the elements AD,DC.
There is no resultant force on each half of the bridge because there's no horizontal force from the pylons at each end.
major blunder: we calculated the vertical force on the pylons (½Mg) but we didn't allow for a horizontal force. We implicitely assumed the horizontal force was 0. This is a reasonable assumption, but it has to be made explicite. (Forgetting reasonable assumptions is why Space Shuttles blow up.) If we'd explicitely assumed that the horizontal force on our pylons was 0 and later our pylons collapsed inward or outward, we could quickly find which assumption was wrong. With the formulation I presented as the problem here, it took a while to figure out what was missing. If a force is 0 in a problem, you have to say so.
As our statics analysis shows, the pushing at the top is counterbalanced by an equal force pulling the two halves of the bridge together at the bottom (at D). As the equivalence of a wrench with a translational force+moment shows, if these two forces are on the bridge are equal, then there is no resultant force, just a moment.
This isn't the end of the story. Why is the (horizontal) force at B and D equal? The force at B acts on the top of element AB; the force at D acts on the bottom of element AB (in the opposite direction). Element AB doesn't accelerate horizontally, so these two forces are equal.
Conclusion: There is no horizontal resultant force on the two halves of the bridge because
The explanations above require an understanding of statics and the chain of inference is relatively long. Even if there are no mistakes, it's hard to run through the derivation in your head while looking at the bridge. For a simple physical object like a truss bridge, you should be able to look at it and see whether the sections exert horizontal forces on each other. Someone one said that if you can't explain it to your Grandma, you don't understand it. Possibly they were joking; Richard Feynmann, on being awarded the Nobel Prize, was asked by reporters to explain his work. His reply was that if he could explain it to reporters, he wouldn't have got the Nobel Prize for it. Possibly there are Grandma grade explanations for everything and we just haven't found them yet. I came up with this Grandma grade explanation. Cut the bridge into two halves. Support the half of the bridge at B by a cable. Now the moment normally provided by the two horizontal forces at B,D is provided by the cable, so we remove the two b_{h} forces. Here's our half bridge now.
^ F=Mg/2 o o B .  .  .  o theta  oo o A D Mg/2 v 
The cable can't/doesn't provide a horizontal force (do you know why?) and the bridge half is not accelerating horizontally (it's stationary). Thus there must be no horizontal force on the bridge half.
This section required a lot more thinking than I expected.
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The vertical members and the road bed are under tension (you could replace them with cables). The top horizontal bars and the sloping elements are under compression.
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The sloping members and the road bed are under tension (you could replace them with cables). The top horizontal bars and the vertical elements are under compression.
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Elements AD1, D2C have no forces on them. You can remove these elements without affecting the remaining elements of the bridge.
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In the whole bridge, the two halves of the bridge are prevented from rotating into each other by two nodes
In the sectioned bridge, the node D cannot exert any moment on the bridge. The moment then must be handled at node B. Presumably the horizontal force at node B will be double that in the whole bridge.
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m.l^{3}
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You can calculate the number from scratch or you can use the inverse ratio of the height compared to the Nurek Dam.
cups of water used at the Hoover dam = 142*300/233 = 183 
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cups of water used at the Grand Coulee = 142*300/168 = 254 
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cups of water used at the Eno River Mill = 142*300/10 = 4260 
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height to boil cup of coffee = (height of Nurek dam)*142 = 42km 
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v^2 = u^2 + 2as initial conditions u = 0 after falling 42km v = sqrt (2 * 9.8m/sec^2 * 42,000m) = 907m/sec 
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P = F*v 12MW = F * 500*10^3/3600 m/sec F = 12 * 10^6 * (3600/500*10^3) N = 86 kN 
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F = mg 86 kN = m * 9.8m/sec^2 m = 8.8Mg = 8.8 tonnes 
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work = F.s = m.a.s dimensions of work = m^1.l^1.t^2.l = m^1.l^2.l^2 specific heat has units J/g.K dimensions of specific heat = dimensions of work /m.T = l^2.t^2.T^1 
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W = C.m.dT where C = heat capacity = 1 calorie/g.K or 1 Calorie/kg.T m = mass (g or kg) = 50kg W = heat (in calories/Calories) = 2000Calories dT = change in temperature dT = 2000Calories/(50kg * 1Calorie/kg.T) = 40 degC 
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H = latent_heat_of_vaporisation * mass latent heat of vaporisation of water = 540 cal/g = 540Calories/kg H = 2000 Calories mass of water evaporated = 2000/540 = 3.7kg 
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No chance it was observed by humans. Humans crossed over Beringia into the Americas from Siberia about 13kya. migration to the new world (http://en.wikipedia.org/wiki/Models_of_migration_to_the_New_World). Settlement of South America appears to have occured independantly of North America (possibly by coastal migration) with the Mound people of Pennsylvannia having come from S.A. over the Isthmus of Panama. Dates of 33kya at Monte Verde in Chile are not widely accepted.
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what mass resting on 1m^2 would be required to produce this force?
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What height of a column of air 1m^2 on base, would have a mass of 10.2Mg?
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pressure at ground level = 100N/m^2 what mass resting on 1m^2 would be required to produce this force? F = mg 100kN = m*9.8m/sec^2 m = 10.2Mg The air density at ground level = 1.225kg/m^3 What height of a column of air 1m^2 on base, would have a mass of 10.2Mg? height equivalent = 10.2/1.225 = 8.3km 
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time = distance/velocity = 8.3/12.8 sec = 0.65 sec 
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dimensions
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frontal area of the object A, velocity v, density of air ρ. How about we throw in mass just for fun (we'll need mass later to get the deceleration, but let's include mass anyhow).
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F = f(m,A,v, rho) to find: dimensions of F F = ma dimensions of ma = m.l.t^2 dimensions of f(m,A,v,rho) = m^w * l^2x * l^2y/t^2y * m^z/l^3z equating dimensions w + z = 1 (1) 2x + 2y  3z = 1 (2) 2y = 2 (3) 
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w + z = 1 (1) 2x + 2y  3z = 1 (2) 2y = 2 (3) from 3 y = 1 (4) from discussion w = 0 (5) from (1) and (5) z = 1 (6) from (2),(4),(6) x = 1 
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F=(1/2).rho.v^2.A.Cd rho = 1.225 kg/m3 v = 12.8km/sec diam = 50m A = 78m^2 Cd = 1 F = 15.6GN 
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F = ma F = 15.6GN m = 150kTonnes a = F/m = 15.6*10^9N/150*10^6kg = 104m/sec^2 = (104/9.8)g = 10.6g 
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We can do this two ways. Both will give the same answer.
Method 1. v^2 = u^2 + 2as v = 12.8km/sec u = ? a = 104m/sec^2 s = 8.3km u^2 = (12.8*10^3)^2 + 2*104*8.8*10^3 u = 12.87km/sec Method 2. v = u + at t = 0.65sec u = v  at = 12.8^10^3m/sec  104m/sec^2 * 0.65sec = 12867.60m/sec 
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P = F.v F = 15.6GN v = 12.8km/sec P = 15.6*10^9N * 12.8*10^3m/sec = 200Tw 
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W = P.t P = 200Tw t = 0.65sec W = 130Tj 
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heat used = latent heat vaporization * mass 130Tj = (350kj/mol / 56g/mol) * mass mass = 130 * 10^12j * 0.056kg/mol /350*10^3j/mol = 21kTonnes 
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weight loss during passage through atmosphere = 300150 = 150ktonnes time of passage = 0.65sec rate of ablation = 150/0.64 = 230ktonnes/sec 
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K.E.=(1/2)m.v^2 m = 150kTonnes v = 12.8km/sec K.E. = 0.5 * 150*10^6kg * (12.8*10^3m/sec)^2 = 12.3 * 10^15J 
^{[273] }
energy of meteor impact = 12.3*10^15J energy of TNT = 2.27MJ/kg tonnes of TNT equivalent = 12.3*10^15J/2.27*10^6J/kg = 5.4MTonnes 
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However the size (or the mass) don't effect on the temperature rise on conversion of the KE to heat. A larger meteorite will produce more heat on impact, but then has a correspondingly larger mass to absorb the heat.
K.E./g at impact KE/kg = 1/2v^2 = 82*10^6 m^2.sec^2 = 82*10^6 J/kg Specific heat/g = 0.45J/K Specific heat/kg = 449J/K each kg of meteorite has 82MJ of kinetic energy. each kg requires 449J to raise the temperature 1deg. The temperature rise then is 82M/449=182kdeg 
^{[275] }
for rocks mass of rocks = density * volume = 2.6 * Tonnes/m^3 * 1200^2m^2 * 170m = 636MTonnes = 636*10^3kg for impactor m = 150kTonnes v = 12.8km/sec K.E.(impactor) = K.E.(rocks) (1/2)150kTonnes * (12.8km/sec)^2 = (1/2)*636MTonnes *v(rocks)^2 v(rocks) = 12.8m/sec * sqrt(150kTonnes/636MTonnes) = 196m/sec 
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distance travelled for material ejected at 45deg at speed u s = 2u^2/a u = 196m/sec a = g s = 2*196^2/9.8 = 7.8km 
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