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: