Question

An object falls from a height of 10 m above the ground. After striking the ground it loses $50\%$ of its kinetic energy. The height upto which the object can rebounce from the ground is:

• A. 7.5 m
• B. 10 m
• C. 2.5 m
• D. 5 m

Answer 1

The Correct Option is D

To determine the height to which the object can rebound, we use conservation of energy principles.

1. Initial potential energy at height $h = 10 \, \text{m}$: 

$$ PE = mgh $$

where:

• $m$ is the mass of the object
• $g$ is the acceleration due to gravity $\approx 9.8 \, \text{m/s}^2$
• $h$ is the height, $10 \, \text{m}$

2. Just before hitting the ground:

All potential energy converts to kinetic energy:

$$ KE = \frac{1}{2}mv^2 = mgh $$

3. After rebound:

The object retains only 50% of its kinetic energy due to inelastic collision with the ground.

$$ KE_{\text{after}} = 0.5 \times \left(\frac{1}{2}mv^2\right) = \frac{1}{2} \times mgh $$

4. At peak of rebound:

This remaining KE converts back into potential energy:

$$ \frac{1}{2} \times mgh = mg \cdot h_{\text{rebound}} $$

Cancel $m$ and $g$ from both sides:

$$ \frac{h}{2} = h_{\text{rebound}} $$

5. Substitute $h = 10 \, \text{m}$:

$$ h_{\text{rebound}} = \frac{10}{2} = 5 \, \text{m} $$

The height to which the object rebounds is 5 m

This step-by-step solution illustrates how energy conservation applies to find rebound height and shows the energy loss due to inelastic collision with the ground.