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 $12 s$ .

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Solution

Ball is drooped 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,
$\begin{matrix} S = u t + \frac{1}{2} a t^{2} \Rightarrow 90 = 0 + \frac{1}{2} \times 9.8 t^{2} \Rightarrow t = \sqrt{18.38} = 4.29 s \end{matrix}$
From first equation of motion,
$\begin{matrix} V = u + a t = 0 + 9.8 \times 4.29 = 42.04 m / s \end{matrix}$
Rebound velocity of the ball, $u r = \frac{9 V}{10} = \frac{9 \times 42.04}{10}$
$= 37.84$ m/s
Time taken by the ball to reach maximum height is obtained with the help of first equation of motion as,
$\begin{matrix} V = 37.84 + (- 9.8) t^{'} \Rightarrow t^{'} = \frac{- 37.84}{- 9.8} = 3.86 s \end{matrix}$
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
$= \frac{9 \times 37.84}{10} = 34.05 m / s$
Total time taken by the ball for second rebound $= 8.15 + 3.86 = 12.01$ s

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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.

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 $12 s$ .