Question

A ball initially at rest, is dropped from building of height $10\text{m}$ to ground as shown in the figure below. If the coefficient of restitution is $e=0.7$ and after collision, the ball rebounds back up to height $h$, find the value of $h$. (Take $g=9.8{\text{m/s}}^2)$

• A. $10\text{m}$
• B. $4.9\text{m}$
• C. $9.8\text{m}$
• D. $\text{None of these}$

Answer 1

The correct option is B $4.9\text{m}$

Taking downward direction as $+ve$
Just before collision with ground
Velocity of the ball $\left(\overrightarrow{u_1}\right)=\sqrt{2gh}=\sqrt{2\times 9.8\times 10}=14\text{m/s}$

After collision,
Let velocity of the ball $\left(\overrightarrow{v_1}\right)=-v\text{m/s}$
As ground is stationary
$\therefore$Velocity of the ground before collision $u_2$ and after collision $v_2$ will be $0\text{m/s}$.

From, coefficient of restiution $\left(e\right)=\frac{\overrightarrow{v_2}-\overrightarrow{v_1}}{\overrightarrow{u_1}-\overrightarrow{u_2}}$
$0.7=\frac{0-(-v)}{14-0}$
$v=14\times 0.7=9.8\text{m/s}$

After rebound, at maximum height $h$, speed of the ball will become zero.

$\therefore$ According to law of conservation of energy,
$\frac{1}{2}m{v}^2=mgh$
$h=\frac{v^2}{2g}=\frac{(9.8{)}^2}{2\times 9.8}$
$=4.9\text{m}$