Ball is dropped from a height, $s=90 m$

Initial velocity of the ball, $u=0$ Acceleration, $a=g=9.8 m / s ^{2}$

Final velocity of the ball $= v$

From second equation of motion, time (t) taken by the ball to hit the ground can be obtained as:

$s=u t+(1 / 2) a t^{2}$

$90=0+(1 / 2) \times 9.8 t^{2}$

$t=\sqrt{18.38}=4.29 s$

From first equation of motion, final velocity is given as:

$v=u+a t$

$=0+9.8 \times 4.29=42.04 m / s$

Rebound velocity of the ball, $u_{r}=9 v / 10=9 \times 42.04 / 10=37.84 m / s$

Time $(t)$ taken by the ball to reach maximum height is obtained with the help of first equation of motion as

$v=u_{r}+a t^{\prime}$

$0=37.84+(-9.8) t$

$t'=-37.84 /-9.8=3.86 s$

Total time taken by the ball $=t+t^{\prime}=4.29+3.86=8.15 s$

As the time of ascent is equal to the time of descent, the ball takes 3.86 s to strike back on the floor for the second time.

The velocity with which the ball rebounds from the floor $=9 \times 37.84 / 10=34.05 m / s$

Total time taken by the ball for second rebound $=8.15+3.86=12.01 s$