Ball is dropped from a height, s = 90 m
Initial velocity of the ball, u = 0
Acceleration, a = g = 9.8 m/s2
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, 
Time (t) taken by the ball to reach maximum height is obtained with the help of first equation of motion as: v = ur + at′
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
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:
