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

A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed - time graph of its motion between t = 0 to 12 s.

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 + 1/2 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,

⇒ v= 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,