### Q.12:- 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.

**Answer:-**

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 = ut + (1/2)at^{2}

90 = 0 + (1/2) × 9.8 t^{2}

t = √*18.38* = 4.29 s

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} = 9v / 10 = 9 × 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} + *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 × 37.84 / 10 = 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: