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 will the ball bounce back after striking the ground for the first time (in meters)?
Take $g=10m/s^2$​

• A. 6
• B. 4
• C. 3
• D. 9

Answer 1

The correct option is A 6

Given:
Initial height of the ball, $h=10m$
Acceleration due to gravity, $g=10m/s^2$

Let the mass of the object be $m$.

Initially, the ball has only potential energy stored in it. So, total energy of the ball will be:
$E=U$
Using formula of potential energy:
$U=mgh$

$⟹E=mgh$

After hitting to the ground, 40% energy of the ball reduces. In other words, 60% of energy is retained. Let the energy left be $E_f$.
$\frac{E_f}{E}=\frac{60}{100}$
$E_f=0.6E$

After bouncing, again at the highest point, the entire energy of the ball is in the form of potential energy. Let the final height of the ball be $h_f$. Using this and conservation of energy,
$E_f=mg{h}_f$
$0.6E=mg{h}_f$
$0.6mgh=mg{h}_f$
$h_f=0.6h$
$h_f=0.6\times 10=6m$