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

Answer 1

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.