A ball is drop from a height of 20m and rebounce with a velocity which is 3/4 of this velocity with which it hit the ground. What is the time interval between the first and second bounces? (g=9.8m/s2)

Answer 1

one needs to calculate the velocity of the bounce:
1/2 m v^2=mgh
v= sqrt 2gh
vrebound=3/4 sqrt 40g

Now, figure the time in air for that ball:
hf=hi+vi*t-4.8t^2
0=0+3/4 sqrt(40g) t-4.8t^2

t=3sqrt(40g)/(4*4.8) seconds

Answer 2

16.8mm

Answer 3

16.8seconds