A ball is dropped from a height of 90 m on a floor

Solution:
Height from which the ball is dropped is H = 90m.
Initial velocity of the ball is u = 0.
Time taken by the ball to reach the ground can be found by using equation of motion for constant acceleration,
s = ut + 12 at2
Putting the values and solving we get,
⇒ t = 4.29 sec
Final velocity of the ball when it reaches the ground,
v = u + at
⇒ v = 42.04 m/s
According to the question, at each collision ball loses one tenth of its speed. Rebound velocity becomes,
u1 = 9v/10 = 37.84 m/s
Time taken by the ball when it attains the maximum height after the ball rebounds is,
0 = u1 + (-9.8)t1
 ⇒ t1 = 3.86 sec
Same amount of time  is taken when the ball reaches the ground which is t1 = 3.86 sec
Velocity of the ball when it reaches the ground will be,
⇒ v1 = 37.84 m/s
So total time from the beginning will be,
T = 4.29 + 3.86+3.86 = 12.01 sec
The graph between velocity and time will look like,