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

Given:

Height from which the ball is dropped $s=90m$

Initial velocity of the ball, $u=0$

Let us consider that the final velocity of the ball is $v$

The time taken by the ball to reach the ground is given by:

$s=ut+\frac{1}{2}a{t}^2$

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

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

The final velocity of the ball as it reaches the ground is given by:

$v=u+at$

$\Rightarrow v=0+9.8\times 4.29=42.04m/s$

The ball losses it's one-tenth of the velocity at collision.

So, the rebound velocity of the ball,

$u_r=\frac{9v}{10}$

$=\frac{9\times 42.04}{10}=37.84m/s$

Now, the ball rebounds and reaches a new maximum height from the ground. The time taken by the ball is given by:

$v=u_r+at\text{'}$

$0=37.84+(–9.8)t\text{'}$

$t\text{'}=-\frac{37.84}{-9.8}=3.86s$

Total time the ball takes to reach the maximum height is :

$T=t+t\text{'}$

$=4.29+3.86=8.15s$

Now the ball travels back to the ground in the same time as it takes to reach the maximum height i.e. $3.86s$

The final velocity of the ball with which it reaches the ground will be same as the speed with which it goes up.

So, the velocity of the ball after rebound:

$v^{\prime }=\frac{9\times 37.84}{10}=34.05m/s$

Total time taken by the ball to rebound for the second time:

$T_t=8.15+3.86=12.01s$