Question

4.
A ball is dropped from height of 45 m and rebounds
with a velocity 3/5th of the velocity with which it hits
the ground. The time interval between first and
second bounces is [Take g = 10 m/s×s]​

Answer 1

Answer:

Let the velocity of the ball with which it hits the ground be denoted as “v” m/s.

Case (i):

Since the ball is dropped, so, the initial velocity of the ball, u = 0 m/s

The height from which the ball is dropped = 45 m

Using the eq. of motion,

v² = u² + 2as

⇒ v² = 0 + [2*10*45]

⇒ v = √900 = 30 m/s

We are given that the velocity of the ball after it rebounds is 3/5th of the velocity with which it hits the ground, therefore,

The velocity after rebound = [3/5] * 30 = 18m/s

Case (ii):

Now, we know that the time interval between first and second bounces is equal to the time taken by ball to reach highest point after first rebound + the taken to reach the ground again for the second rebound.

Using the eq. of motion ,

v = u + at

here a = 10 m/s² will be -ve since the motion is in the upward direction

⇒ 0 = 18 + (-10t)

⇒ t = 1.8 s

∴The total time interval = 2 * t = 2 * 1.8 = 3.6 sec

Thus, the time interval between first and second bounces is 3.6 s.