Question

A ball is dropped from a height of $30\text{m}$ in downward direction on the stationary floor. If the value of coefficient of restitution is $0.6$ and ball rebounds back up to the height $\left(h^{\prime }\right)$, find the height $\left(h^{\prime }\right)$ after first collision.
(Take $g=10{\text{m/s}}^2$)

• A. $21.6\text{m}$
• B. $22.8\text{m}$
• C. $10.8\text{m}$
• D. $26.8\text{m}$

Answer 1

The correct option is C $10.8\text{m}$

The speed of ball at which it strike to the floor $\left(u_1\right)$
$u_1=\sqrt{2gh}=\sqrt{2\times 10\times 30}=\sqrt{600}\text{m/s}$
If $u_2\&v_2$ is the initial and final velocity of floor and since floor remains stationary,
$u_2=0\text{m/s}$ and $v_2=0\text{m/s}$
Let $v_1$ is the final velocity of ball after collision.
Coefficient of restitution,
$e=\frac{\text{velocity of separation}}{\text{velocity of approach}}=\frac{v_2-v_1}{u_1-u_2}$
$0.6=\frac{0-(-v_1)}{\sqrt{600}-0}$
[$\because$ take velocity negative in $+y$ direction]
$v_1=0.6\times \sqrt{600}\text{m/s}$
From energy conservation,
$\frac{1}{2}m{v}_1^2=mg{h}^{\prime }$
$h^{\prime }=\frac{v_1^2}{2g}=\frac{(0.6\times \sqrt{600}{)}^2}{2\times 10}=\frac{0.6\times 0.6\times 600}{2\times 10}$
$h^{\prime }=10.8\text{m}$