Question

A ball is thrown vertically downwards from a height of $20\text{m}$ with an initial velocity $v_0$. It collides with the ground, loses $50$ percent of its energy in collision and rebounds to the same height. The initial velocity $v_0$ is (Take $g=10{\text{ms}}^{-2})$

• A. $28{\text{ms}}^{-1}$
• B. $14{\text{ms}}^{-1}$
• C. $20{\text{ms}}^{-1}$
• D. $10{\text{ms}}^{-1}$

Answer 1

The correct option is C $20{\text{ms}}^{-1}$

The situation is shown in the figure.

Let $v$ be the velocity of the ball with which it collides with ground. Then according to the law of conservation of energy,
Gain in kinetic energy = loss in potential energy
i.e. $\frac{1}{2}m{v}^2-\frac{1}{2}m{v}_0^2=\text{mgh}$
(where $m$ is the mass of the ball)

$v^2-v_0^2=2gh....\left(i\right)$

Now, when the ball collides with the ground, $50\%$ of its energy is lost and it rebounds to the same height $h$.

$\therefore \frac{50}{100}\left(\frac{1}{2}m{v}^2\right)=\text{mgh}$
$\frac{1}{4}{v}^2=gh$
$v^2=4gh$

Now,
$v_0^2=4gh-2gh=2gh$
$v_0=\sqrt{2gh}$

Here, $g=10{\text{ms}}^{-2}$ and $h=20\text{m}$
$\therefore v_0=\sqrt{2\left(10{\text{ms}}^{-2}\right)\left(20\text{m}\right)}=20{\text{ms}}^{-1}$