# Physics

Two identical pellet guns are fired simultaneously from the edge of a cliff. These guns impart an initial speed of 36.3 m/s to each pellet. Gun A is fired straight upward, with the pellet going up and then falling back down, eventually hitting the ground beneath the cliff. Gun B is fired straight downward. In the absence of air resistance, how long after pellet B hits the ground does pellet A hit the ground?

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1. A: d=Va*t1-4.9t1^2
B: d=-Va*t2-4.9t2^2
set them equal
-Va*t2-4.9t2^2=Va*t1-4.9t1^2
-Va(t1+t2)=4.9(t2^2-t1^2)

-Va(t1+t2)=4.9 (t2-t1)(t2+t1)
-Va=4.9(t2-t1)
or t1-t2=Va/4.9

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2. the time for the pellets to hit the ground is

A: h + 36.3t - 4.9t^2
B: h - 36.3t - 4.9t^2

I expect you will need to know h (the height of the cliff) to solve for the two values of t and take their difference.

A: t = (-36.3 ±√(1317.69+19.6h))/9.8
B: t = (36.3 ±√(1317.69+19.6h))/9.8

The difference is .20√(1317.69+19.6h) - 7.41

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posted by Steve
3. Hmmm. bobpursley is usually right on these matters. I'll have to think on it a bit to see why the initial height makes no difference.

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posted by Steve
4. because when the one that was fired upward gets down to cliff level it is traveling downward at the same initial speed down as the first one was.

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posted by Damon

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