A ball is dropped from a height of 20m and rebounds with a velocity which is 3/4 of the velocity with which it hit the ground. What is the time interval between the first and second bounces?

the acceleration as it's coming down is +10. Let the velocity at which it hit the ground be V. So, the velocity at which it rebounds will be 3/4V. Thus, 4rm v=u+gt. Where u is 0, v=10t, d time at which it lands is t=v/10. 4rm s=(v-u)/2 *t, 20=(v-3/4v)/2 *v/10. Thus, 400=v^2/4. V^2=1600, v=40. Now, d time for hitting d ground is t=40/10, t=4s. 4rm v=u+(-g)t, the time for rebounce t=30-40/-10, t=1s. Thus, d time interval btw the first & second bounce is t1-t2=4-1=3s

Ans. 3 second

Please I don't understand where the 30 in the " the time for rebounds" coming from. Please explain to me...

Please I need to know where that 30 is coming from please.....

To find the time interval between the first and second bounces, we first need to calculate the time it takes for the ball to hit the ground in the first bounce.

Step 1: Find the time it takes for the ball to hit the ground in the first bounce.
We can use the kinematic equation for the vertical motion of an object:
h = ut + (1/2)gt^2,

where:
h = height (20m),
u = initial velocity (unknown),
g = acceleration due to gravity (-9.8 m/s^2),
t = time.

Since the ball is dropped from rest, the initial velocity (u) is 0 m/s.

We can rearrange the equation to solve for time (t):
h = (1/2)gt^2,
20 = (1/2)(-9.8)t^2,
t^2 = (20 * 2) / 9.8,
t^2 = 40 / 9.8,
t^2 ≈ 4.08.

Taking the square root of both sides gives:
t ≈ √4.08,
t ≈ 2.02 seconds.

Step 2: Calculate the time interval between the first and second bounces.
Since the ball rebounds with 3/4 of the velocity it hit the ground, we can assume that the time it takes for the ball to reach its maximum height after the first bounce is equal to the time it takes for the ball to fall from its maximum height to the ground.

Therefore, the time interval between the first and second bounces is twice the time it takes for the ball to hit the ground in the first bounce.

The time interval between the first and second bounces is:
2 * t ≈ 2 * 2.02,
≈ 4.04 seconds.

So, the time interval between the first and second bounces is approximately 4.04 seconds.