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 is (Take $g = 10\,{\text{m}} \cdot {{\text{s}}^{ - 2}}$)
A. $10\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$
B. $14\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$
C. $20\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$
D. $28\,{\text{m}} \cdot {{\text{s}}^{ - 1}}$

Answer 1

Use the formula for kinetic energy and potential energy of the object. Derive the equation for initial energy of the ball at height h, the energy of ball when it strikes the ground and final energy of the ball when it rebounds to the height h. Apply law of conservation of energy to the ball at the ground and when it rebounds and determine the initial velocity.

Formulae used:
The kinetic energy $K$ of an object is
$K = \dfrac{1}{2}m{v^2}$ …… (1)
Here, $m$ is the mass of the object and $v$ is the velocity of the object.
The potential energy of an object is
$U = mgh$ …… (2)
Here, $m$ is the mass of the object, $g$ is acceleration due to gravity and $h$ is the height of the object from the ground.

Complete step by step answer:
We have given that a ball is thrown vertically downwards from the height of $20\,{\text{m}}$ with an initial velocity ${v_0}$.
$h = 20\,{\text{m}}$

Let $m$ be the mass of the ball.
When the particle is at the height $h$, it has kinetic as well as potential energy.
${E_i} = \dfrac{1}{2}mv_0^2 + mgh$
When the ball strikes the ground, its energy $E$ becomes half of its initial energy ${E_i}$.
$E = \dfrac{1}{2}{E_i}$
The ball has only kinetic energy when it strikes the ground which is given by
$E = \dfrac{1}{2}m{v^2}$

When the ball rebounds again from the ground, it reaches the same height as its initial height. The final energy ${E_f}$ of the ball at this height is the potential energy of the ball as the kinetic energy is zero (velocity is zero at maximum height).
${E_f} = mgh$
According to the law of conservation of energy, the total energy $E$ of the ball at the ground is equal to the total final energy ${E_f}$ of the ball at the height h.
$E = {E_f}$
Substitute $\dfrac{1}{2}{E_i}$ for $E$ in the above equation.
$\dfrac{1}{2}{E_i} = {E_f}$
Substitute $\dfrac{1}{2}mv_0^2 + mgh$ for ${E_i}$ and $mgh$ for ${E_f}$ in the above equation.
$\dfrac{1}{2}\left( {\dfrac{1}{2}mv_0^2 + mgh} \right) = mgh$
$\Rightarrow \dfrac{1}{4}v_0^2 + \dfrac{1}{2}gh = gh$
$\Rightarrow v_0^2 = 2gh$
$\Rightarrow {v_0} = \sqrt {2gh}$

Substitute $10\,{\text{m/}}{{\text{s}}^2}$ for $g$ and $20\,{\text{m}}$ for $h$ in the above equation.
$\Rightarrow {v_0} = \sqrt {2\left( {10\,{\text{m/}}{{\text{s}}^2}} \right)\left( {20\,{\text{m}}} \right)}$
$\Rightarrow {v_0} = \sqrt {400}$
$\therefore {v_0} = 20\,{\text{m/s}}$
Therefore, the initial velocity of the ball is $20\,{\text{m/s}}$.

Hence, the correct option is C.

Note: The students may get confused that when initially the ball is at height h, it has initial velocity but when the ball rebounds to the same height its velocity is zero. But the students should keep in mind that we have given that the ball has some initial velocity but the velocity of the ball at maximum height is always zero.