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

Vf ² = Vi ² + 2ad

Vf ² = 0 + 2(9.8)(20)
Vf ² = 392
Vf ~ 19.80 m/s when it lands.

Initial speed coming back up:
(3/4)(19.80) = 14.85 m/s

Equation of motion for second bounce:
d = Vit + (1/2)at ²
d = 14.85t + (1/2)(-9.8)t ²
Substitute d = 0 (to represent the landing height) and solve for t.