Question

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 1

Ball is dropped from a height, $\text s$ = 90 $\text m$
Initial velocity of the ball, $\text u$ = 0
Acceleration, $\text a$ = $\text g$ = 9.8 $\text m/ \text s^2$
Final velocity of the ball = $\text v$
From second equation of motion, time ($\text t$) taken by the ball to hit the ground can be obtained as: $\text s$ =$\text {ut}+\frac{1}{2}\text{at}^2$

$90$ = $0+\frac{1}{2}\times 9.8\text t^2$

$\text t$ = $\sqrt{18.38}$ = $4.29 \;\text s$

From first equation of motion, final velocity is given as :
$\text v$ = $\text u$ + $\text {at}$

= $0+9.8\times 4.29$ = $42.04 \;\text m/\text s$

Rebound velocity of the ball, $u_r$= $\frac{9}{10}v$ =$\frac{9}{10}\times 42.04$ = $37.84\; \text m /\text s$

Time (t) taken by the ball to reach maximum height is obtained with the help of first equation of motion as:
$\text v$ = $u_r+at$

0 = 37.84 + (– 9.8) $\text t$

$\text t$ = $\frac{-37.84}{-9.8}$ = $3.86 \;\text s$

Total time taken by the ball = $\text t+\text t$= 4.29 + 3.86 = 8.15 $\text s$

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 = $\frac{9}{10}\times37.84$= $34.05 \;\text m/ \text s$

Total time taken by the ball for second rebound = $8.15+3.86$ = $12.01 \text s$
The speed-time graph of the ball is represented in the given figure as: