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

t=6.122s

Step-by-step 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(m)/(s)`

`a=g=-9.8(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(m)/(s)`

`-30=30-9.8*t`

`t=((-30-30)(m)/(s) )/(-9.8(m)/(s^(2)))`

`t=6.122s`