cq_1_211

Your 'cq_1_21.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

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The acceleration acts through equal and opposite displacements, so the term 2 a `ds of the fourth equation is the same in both cases, except for the sign. Going up, a and `ds are in opposite directions to 2 a `ds is negative, whereas coming down a and `ds are in the same direction so 2 a `ds is positive. Thus

2 a `ds_up = -2 a `ds_down,

with 2 a `ds_down being positive.

The fourth equation says that

vf^2 = v0^2 + 2 a `ds.

Going up we know that vf = 0, so v0_up^2 = - 2 a `ds_up.

Going down we know that v0 = 0 so that vf_down^2 = 2 a `ds_down.

Since 2 a `ds_up = -2 a `ds_down, we conclude that v0_up^2 = vf_down^2 so that | v0_up | = | vf_down |; i.e., the speeds are equal.

It's actually even simpler that that. The displacement `ds from start to end is zero, and the acceleration is constant throughout, so that for the entire up-down motion phase 2 a `ds = 0. Since vf^2 = v0^2 + 2 a `ds, we have vf^2 = v0^2 + 0, or vf^2 = v0^2 and | vf | = | v0 |.

This can also be reasoned from the point of view of energy. The net force up is the same as the net force down, and the displacement up is equal and opposite to the displacement down. Thus the work done by the net force on the way up is equal and opposite to the work done by the net force coming down, so that the work done by the net force for the entire motion is zero. Thus KE change is zero, and 1/2 m v^2, and therefore | v |, must be the same at the beginning as at the end.

Finally we can reason in terms of potential and kinetic energies. The change in gravitational PE going up is equal and opposite to the change going down, since the vertical displacements are equal and opposite. So the change in gravitational PE is zero. Only the gravitational force acts on the object, so nonconservative forces are not present and `dKE = - `dPE. Since `dPE = 0, `dKE = 0 and 1/2 m v^2, and therefore | v |, is the same at the beginning as at the end. [This argument can be further simplified; we don't have to worry about separating the up and down motions. All we need to know is that the altitude at the end is the same as at the beginning, so that `dPE from beginning to end is zero. The same conclusion about velocities follows immediately.]

Air resistance will enhance the slowing effect of gravity on the rising ball, which will as a result not rise as far. As a result of the decreased maximum height, the falling ball won't drop as far, resulting in a lesser final speed.

Air resistance will then oppose the speeding up of the falling ball, so that the final velocity will be even less.

For a dense ball tossed gently upward, the effect of air resistance will be small, perhaps negligible with respect to the accuracy of our instruments. If the ball is less dense, and/or speeds are greater, air resistance will have a greater effect.

In terms of energy conservation, assuming still air (i.e., no wind, no rising and falling of air), air resistance acts always in the direction opposite motion and therefore does negative work on the object. PE doesn't change between the beginning and the end of the interval, so the result is a lesser KE at the end than at the beginning.

If we want to think in terms of the fourth equation of motion, the direction of the air resistance tells us that the acceleration of the ball has greater magnitude on the way up than on the way down. Thus the 2 a `ds term has greater magnitude on the way up than on the way down; since 2 a `ds is negative on the way up and positive on the way down, the value of 2 a `ds for the entire motion is negative. Since vf^2 - v0^2 = 2 a `ds, it follows that vf^2 - v0^2 is negative, i.e., v0^2 is greater than vf^2 so that | v0 | > | vf |. Final speed is less than initial.

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