A body is dropped from a certain height above the ground. When it is half way doen, it process.

Options

A. Only Kinetic energy
B. Only Prtential energy
C. Boh Kinetic and Potential energy
D. Neither kinetic nor Potential energy

Answer

To solve the problem of a body dropped from a height and analyzing its energy when it reaches halfway down, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- A body is dropped from a height \( h \).
- At the starting point (Point A), the body has potential energy and no kinetic energy since it is at rest.

2. **Calculate the Initial Energy at Point A**:
- The potential energy (PE) at height \( h \) is given by the formula:
\[ \text{PE}_A = mgh \]
- The kinetic energy (KE) at this point is:
\[ \text{KE}_A = 0 \]
- Therefore, the total mechanical energy at Point A is:
\[ E_A = \text{PE}_A + \text{KE}_A = mgh + 0 = mgh \]

3. **Analyze the Energy at Halfway Point (Point B)**:
- When the body reaches halfway down (Point B), the height is \( \frac{h}{2} \).
- The potential energy at this point is:
\[ \text{PE}_B = mg\left(\frac{h}{2}\right) = \frac{mgh}{2} \]
- The body is now in motion, so it has kinetic energy. By the conservation of mechanical energy, the total energy at Point B must equal the total energy at Point A:
\[ E_B = E_A \]
- Thus, we have:
\[ \text{PE}_B + \text{KE}_B = mgh \]
- Substituting for \(\text{PE}_B\):
\[ \frac{mgh}{2} + \text{KE}_B = mgh \]
- Rearranging gives:
\[ \text{KE}_B = mgh - \frac{mgh}{2} = \frac{mgh}{2} \]

4. **Conclusion**:
- At Point B, the body has both potential energy and kinetic energy:
- Potential Energy: \( \frac{mgh}{2} \)
- Kinetic Energy: \( \frac{mgh}{2} \)
- Therefore, the correct answer is that at halfway down, the body possesses both kinetic and potential energy.