A ball dropped from rest on to horizontal ground 20m between rebounds with 3/4 of the velocity with which it hits the ground find time that elapses between the first and second in part of the ball with the ground

So the KE of the rebound is (3/4)^2 of the original energy (height 20), so the ball goes up 9/16 * 20 meters.

time to go up that height = time to fall that height.
time to fall that height:
h= 1/2 g t^2 or t= sqrt2h/g
total time on bounce then is 2t or 2 sqrt(9/16*20/9.8)= 3/2 sqrt(20/9.8) seconds

check my thinking.

To find the time that elapses between the first and second rebound of the ball with the ground, we can use the equations of motion.

Let's break down the information given in the problem:

- Initial velocity of the ball when it hits the ground (v₁) = unknown
- Distance traveled between rebounds (d) = 20 meters
- Velocity of the ball after the second rebound (v₂) = 3/4 of the velocity with which it hits the ground

First, let's find the velocity of the ball after the first rebound.

When a ball rebounds, it bounces back with the same magnitude of velocity but in the opposite direction. Therefore, after the first rebound, the magnitude of the velocity will be the same as the initial velocity but in the opposite direction.

We can use the equation of motion for vertical displacement to find the initial velocity of the ball (v₁):

s = ut + (1/2)gt²

Here, s is the displacement (vertical distance), u is the initial velocity, t is time, and g is the acceleration due to gravity (approximately 9.8 m/s² near the Earth's surface).

Since the ball is dropped from rest, the initial velocity (u) is 0. Additionally, the displacement (s) is the distance between rebounds, which is 20 meters. We can substitute these values into the equation:

20 = 0 + (1/2)(9.8)t²
20 = 4.9t²

Now, solving for t:

t² = 20/4.9
t ≈ √4.08
t ≈ 2.02 seconds

So, the time it takes for the ball to reach the ground after being dropped from rest is approximately 2.02 seconds.

Now, to find the time between the first and second rebound, we know that the velocity after the second rebound (v₂) is 3/4 of the velocity with which it hits the ground. This means v₂ = (3/4)u.

Using the equation of motion for vertical velocity, we can relate the initial velocity (u) and the final velocity (v₂) to find the time it takes to travel between the rebounds:

v₂ = u + gt

Substituting v₂ = (3/4)u and rearranging the equation:

(3/4)u = u + gt
(1/4)u = gt

Since we know u = 0 (the ball is initially at rest), the equation simplifies to:

(1/4)(0) = gt
0 = gt

This implies that the ball takes no time to travel between rebounds since the time (t) cancels out. Therefore, the time that elapses between the first and second rebound of the ball with the ground is effectively zero.