Question

A ball falls under gravity from a height of 10m with an initial downward velocity “u”. It collides with the ground and loses $50\%$ of its kinetic energy in the collision and then rises back to the same height. Find the initial velocity “u”.
$\begin{aligned} & (A)14m{{s}^{-1}} \\\ & (B)10m{{s}^{-1}} \\\ & (C)11m{{s}^{-1}} \\\ & (D)15m{{s}^{-1}} \\\ \end{aligned}$

Answer 1

The total energy at the height of 10 meters will be the sum of the Kinetic energy of the ball and its potential energy at that height. Since, after collision the ball reaches the same height, this implies the fact that the total gain in energy after bouncing back to the same height should be equal to the energy left within the ball.
Complete step by step solution: Let the height from which the ball falls under gravity be given by $h$ and let the mass of the ball be denoted by $m$ .
Then, we can write the total initial energy at height $h$ as follows:
$\Rightarrow E=\dfrac{1}{2}m{{u}^{2}}+mgh$
Where, $u$ is given to be the speed at the start of descent.
It has been given in the question that the ball loses 50% of its kinetic energy in the collision.
Now, just at the time of time of collision, the total energy of the ball has been converted into kinetic energy. So, by principle of conservation of energy: 50% loss in kinetic energy at the time of collision is equal to 50% loss in total energy.
Therefore, the energy required to move to the same height, that is, $mgh$ should be equal to half the total energy.
$\Rightarrow mgh=\dfrac{1}{2}\left( \dfrac{1}{2}m{{u}^{2}}+mgh \right)$
$\Rightarrow gh=\dfrac{1}{4}{{u}^{2}}+\dfrac{gh}{2}$
$\begin{aligned} & \Rightarrow \dfrac{gh}{2}=\dfrac{1}{4}{{u}^{2}} \\\ & \Rightarrow u=\sqrt{2gh} \\\ \end{aligned}$
Putting the values of:
$\begin{aligned} & \Rightarrow h=10m \\\ & \Rightarrow g=9.8m{{s}^{-2}} \\\ \end{aligned}$
We get:
$\begin{aligned} & \Rightarrow u=\sqrt{2\times 9.8\times 10} \\\ & \Rightarrow u=\sqrt{196} \\\ & \Rightarrow u=14m{{s}^{-1}} \\\ \end{aligned}$
Hence, the initial downward velocity comes out to be $14m{{s}^{-1}}$ .

Hence, option (A) is the correct option.

Note:
We didn’t know the velocity at the time of collision. Though, this could have been calculated but applying conservation of energy saved our time by directly providing us with kinetic energy at the time of collision in terms of total energy at the start of the descent. This not only made our calculations easier but also saved our time. One should be able to figure out the shortest method to solve a problem.