1. Initial total mechanical energy

A rubber ball is dropped from height $h$.

So initially, $E_i = mgh$

2. Energy after rebound

It rebounds to height $\dfrac{h}{2}$, so just after collision its mechanical energy is enough to raise it to $\dfrac{h}{2}$.

Thus, $E_f = mg\left(\frac{h}{2}\right)=\frac{mgh}{2}$

3. Loss of energy during collision

$\Delta E = E_i - E_f = mgh - \frac{mgh}{2} = \frac{mgh}{2}$

Hence percentage loss of total energy is $\frac{\Delta E}{E_i}\times 100 = \frac{\frac{mgh}{2}}{mgh}\times 100 = 50\%$

4. Velocity just before striking the ground

Using conservation of mechanical energy during the fall: $mgh = \frac{1}{2}mv^2$

Cancelling $m$, $gh = \frac{v^2}{2}$ $v^2 = 2gh$ $v = \sqrt{2gh}$

5. Match with options

The required pair is:

• Percentage loss of energy = $50\%$
• Velocity before striking ground = $\sqrt{2gh}$

So the correct option is A.