Question

A ball is dropped from a height of $10m$. If the energy of the ball reduces by $40\%$ after striking the ground, how high can the ball bounce back? ($g=10m{s}^{-2}$) .

Answer 1

Step 1: Given Data

Initial height from where the ball is dropped $h=10m$

Acceleration due to gravity $g=10m{s}^{-2}$

The energy of the ball reduced $=40\%$

Let $m$ be the mass of the ball.

Let the height of the ball after rebound be $h\text{'}$.

Step 2: Calculate the total potential energy

We know that the initial potential energy of the ball is

$PE=mgh$

$=m\times 10\times 10$

$=100m$

Step 3: Calculate the reduced potential energy

Since $100m$ is the total energy, then the reduced energy can be given as

$$\begin{gathered} =100m-\left(40\%of100m\right) \\ =100m-\left(\frac{40}{100}\times 100m\right) \\ =100m-40m \\ =60m \end{gathered}$$

Step 4: Calculate the bounced height

This is the energy of the ball when it bounces back.

$\therefore 60m=mgh\text{'}$

$\Rightarrow h\text{'}=\frac{60}{g}$

$\Rightarrow h\text{'}=\frac{60}{10}=6m$

Hence, the ball can bounce back $6m$ high.