Question

A ball is dropped from a height of 20m and rebounds with a velocity which is ¾ of the velocity with which it hit the ground. What is the time interval between the first and second bounces?

Answer 1

$v = \sqrt{2\times g \times h}$

velocity after collision, $v = \frac{3}{4} \times \sqrt {2 \times 10 \times 20} = 15m/s$

$t = \frac{2u}{g} = \frac{2\times 15}{10}=3seconds$ (time interval)