A ball is dropped from a height of  $90 \mathrm{m}$ on a floor. At  each collision with the floor, the ball loses $one-tenth$ of its speed. The total time of motion is:

The time taken by the ball to fall a height $90 \mathrm{m}$ :

$\left(i\right)90=ut+\frac{1}{2}g{t}^2$

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

which gives $t=4.3 \mathrm{s}$

Now velocity just before collision with the floor

$$\begin{gathered} v=u+gt \\ =0+9.8\times 4.3=42 \mathrm{m}/\mathrm{s}The velocity between 0 to 4.3 \mathrm{s} \end{gathered}$$

v=g t=9.8 t 

In this time velocity varies linearly with time from 0 to $42 \mathrm{m}/\mathrm{s}$ during downward motion.

(ii) After first collision with the floor, speed lost by ball

$=\frac{1}{10}\times 42=4.2 \mathrm{m}/\mathrm{s}$

Thus ball rebound with a speed of$v=\frac{9}{10}\times 42=37.8 \mathrm{m}/\mathrm{s}.$

For upward motion, the velocity is given by the equation

$$\begin{gathered} v=u-gt \\ =37.8-9.8\times t \end{gathered}$$

The speed decreases linearly and becomes zero at;

$$\begin{gathered} 0=37.8-9.8t \\ \Rightarrow t=3.9 \mathrm{s} \end{gathered}$$

Thus, the ball reaches the highest point again after time $t=4.3+3.9=8.2 \mathrm{s}$ from the start of the motion.

(iii) At the highest point, the speed of ball becomes zero. It again starts falling. At any time its speed is given by $v=0+9.8 t$.

The speed increases linearly with time from 0 to$37.8 \mathrm{m}/\mathrm{s}$ in the next time interval of $3.9 \mathrm{s}$. The total time of motion from start now becomes $=4.3+3.9+3.9=12.1 \mathrm{s}$.

The motion of the ball is shown by graph as in figure 3.50.