Question

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?

Answer 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

Answer 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

Answer 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.

Answer 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.