To solve this problem, we need to consider the principles of energy conservation and the behavior of the object during the rebound. Let's break it down step-by-step:

1. When the object falls from a height of $10 \mathrm{~m}$, it converts its potential energy to kinetic energy at the point of impact. The potential energy (PE) just before hitting the ground can be calculated using the formula:

$PE = mgh$

where:

• $m$ is the mass of the object
• $g$ is the acceleration due to gravity ($9.8 \mathrm{~m/s^2}$)
• $h$ is the height ($10 \mathrm{~m}$)

The kinetic energy (KE) of the object just before impact is equal to the potential energy it had at the height of $10 \mathrm{~m}$, because of the conservation of energy principle.

2. Upon hitting the ground, the object loses $50\%$ of its kinetic energy. Therefore, the kinetic energy after the impact is:

$$KE_{after\ impact} = \frac{1}{2} KE_{before\ impact}$$

3. The object will now rebound to a height where its kinetic energy is converted back to potential energy. Let this height be $h_{rebound}$. The potential energy at this rebound height is:

$$PE_{rebound} = mgh_{rebound}$$

Since the kinetic energy after the impact is $\frac{1}{2} mgh$, we set this equal to the potential energy at the rebound height:

$$\frac{1}{2} mgh = mgh_{rebound}$$

By canceling out the mass $m$ and the gravitational constant $g$, we get:

$$\frac{1}{2} h = h_{rebound}$$

Substituting $h = 10 \mathrm{~m}$, we find:

$$\frac{1}{2} \times 10 \mathrm{~m} = h_{rebound}$$

$$h_{rebound} = 5 \mathrm{~m}$$

Therefore, the height up to which the object can rebound from the ground is $5 \mathrm{~m}$. The correct answer is:

Option D $5 \mathrm{~m}$