Question

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

Answer 1

H = 1/2 g t^2 calculate time to reach ground

t = (2 H / g)^1/2 = (2 * 20 / 9.8)^1/2 = 2.02 sec

V1 = a t = 9.8 m/s^2 * 2.02 sec = 19.8 m/s original speed

V2 = 19.8 * .75 = 14.85 m/s speed of rebound

t2 = V2 / g = 14.85 / 9.8 = 1.51 sec time to get back to top

T = 2 * t2 = 3.02 sec time in air after 1st bounce