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:

From first equation of motion, final velocity is given as: v = u + at = 0 + 9.8 × 4.29 = 42.04 m/s

Rebound velocity of the ball, u_{r} = (9/10)v = 9/10 x 42.04 = 37.84m/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} + at′

0 = 37.84 + (-9.8) t'

t' = -37.84/-9.8 = 3.86 s

Total time taken by the ball = t + t′ = 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 = 9/10 x 37.84 = 34.05 m/s

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

The speed-time graph of the ball is represented in the given figure as: