Question

A tennis ball is hit straight up at 27.0 m/s from the edge of a sheer cliff. Sometime later, the ball passes the original height from which it was hit.

1)How fast is the ball moving at that time? (Express your answer to three significant figures.)

2)If the cliff is 36.0 m high, how long will it take the ball to reach the ground level? (Express your answer to three significant figures.)

3)What total distance did the ball travel? (Express your answer to three significant figures.)

Answer 1

The situation described in the question is depicted in the figure below:

Given

the initial velocity of the ball u = 27 m/s

the height of the cliff h = 36 m

1) When the ball passes its original position during its downward motion, then the net displacement is 0.

Using the equation of motion,

So, when the ball passes its original position during its downward motion its speed is 27.0 m/s

2) When the ball reaches the ground the net displacement is -36 m.

Using the equation of motion,

Hence, the time taken for the ball to reach the ground is 6.61 sec.

3) When the ball reaches its maximum height, it momentarily comes to rest.

So,

The total distance travelled by the ball is given by

i.e.

Therefore, the total distance travelled by the ball is 110.4 m.