A body is dropped from a certain height from the ground. When it is halfway down, it possesses,

Options

A. only K.E.
B. both K.E. and P.E.
C. only P.E.
D. zero energy

Answer

To solve the problem, we need to analyze the situation of a body dropped from a certain height and determine the energies it possesses when it is halfway down.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- When the body is dropped from a height \( h \), its initial velocity \( u = 0 \).
- At this point, the kinetic energy (KE) is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
Since \( v = 0 \), the initial kinetic energy is:
\[ KE = 0 \]
- The potential energy (PE) at height \( h \) is given by:
\[ PE = mgh \]

2. **At Halfway Down**:
- When the body is halfway down, it is at height \( \frac{h}{2} \).
- The potential energy at this height is:
\[ PE = mg\left(\frac{h}{2}\right) = \frac{mgh}{2} \]

3. **Calculating the Speed at Halfway Point**:
- To find the speed \( v \) of the body when it is halfway down, we can use the equation of motion:
\[ v^2 = u^2 + 2as \]
Here, \( u = 0 \), \( a = g \) (acceleration due to gravity), and \( s = \frac{h}{2} \):
\[ v^2 = 0 + 2g\left(\frac{h}{2}\right) = gh \]
Thus, the speed \( v \) at halfway down is:
\[ v = \sqrt{gh} \]

4. **Calculating Kinetic Energy at Halfway Point**:
- Now, we can calculate the kinetic energy at this point:
\[ KE = \frac{1}{2} mv^2 = \frac{1}{2} m(gh) = \frac{mgh}{2} \]

5. **Total Energy Conservation**:
- The total mechanical energy at the start (when dropped) is:
\[ E_{total} = PE + KE = mgh + 0 = mgh \]
- At halfway down, the total energy is still conserved:
\[ E_{total} = PE + KE = \frac{mgh}{2} + \frac{mgh}{2} = mgh \]

### Conclusion:
At the halfway point, the body possesses both kinetic energy and potential energy, and they are equal in magnitude:
- Kinetic Energy \( KE = \frac{mgh}{2} \)
- Potential Energy \( PE = \frac{mgh}{2} \)

Thus, the correct answer is that the body possesses **both kinetic energy and potential energy**.