Question

Review Conceptual Example 15 before attempting this problem. Two identical pellet guns are fired simultaneously from the edge of a cliff. These guns impart an initial speed of 42.6 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

Answer:

t=6.122s

Explanation:

The time required for pellet to reach the maximum height, so what describe the motion is by

$v_{f}=v_{o}+a*t$

Given:

$v_{a}=v_{b}=v_{o}=30\frac{m}{s}$

$a=g=-9.8\frac{m}{s^{2} }$

To describe the time of A and B it takes from

$t_{a}=t_{1}+t\\t_{a}-t_{b}=t_{1}+t-t\\t_{a}-t_{b}=t$

So replacing:

$v_{f}=v_{o}-g*t$

$v_{f}=-v_{o}=-30\frac{m}{s}$

$-30=30-9.8*t$

$t=\frac{(-30-30)\frac{m}{s} }{-9.8\frac{m}{s^{2}}}$

$t=6.122s$