A particle is projected at ` 60(@)` to the horizontal with a kinetic energy `K` . The kinetic energy at the highest point is

Options

A. K
B. zero
C. `K//4`
D. `K//2`

Explanation

To solve the problem, we need to find the kinetic energy of a particle at its highest point when it is projected at an angle of 60 degrees to the horizontal with an initial kinetic energy \( K \).

### Step-by-Step Solution:

1. **Identify the Initial Kinetic Energy**:
The initial kinetic energy \( K \) of the particle when it is projected can be expressed as:
\[ K = \frac{1}{2} m v^2 \]
where \( m \) is the mass of the particle and \( v \) is its initial velocity.

2. **Resolve the Initial Velocity into Components**:
When the particle is projected at an angle \( \theta = 60^\circ \), we can resolve the initial velocity \( v \) into horizontal and vertical components:
- Horizontal component: \( v_x = v \cos(60^\circ) \)
- Vertical component: \( v_y = v \sin(60^\circ) \)

We know that:
\[ \cos(60^\circ) = \frac{1}{2} \quad \text{and} \quad \sin(60^\circ) = \frac{\sqrt{3}}{2} \]
Thus, we have:
\[ v_x = v \cdot \frac{1}{2} = \frac{v}{2} \]
\[ v_y = v \cdot \frac{\sqrt{3}}{2} \]

3. **Determine the Kinetic Energy at the Highest Point**:
At the highest point of the projectile's motion, the vertical component of the velocity becomes zero (i.e., \( v_y = 0 \)). Therefore, the kinetic energy at the highest point depends only on the horizontal component of the velocity:
\[ K_{\text{highest}} = \frac{1}{2} m v_x^2 \]
Substituting \( v_x \):
\[ K_{\text{highest}} = \frac{1}{2} m \left(\frac{v}{2}\right)^2 = \frac{1}{2} m \cdot \frac{v^2}{4} = \frac{1}{8} m v^2 \]

4. **Relate the Kinetic Energy at the Highest Point to the Initial Kinetic Energy**:
We know from Step 1 that \( K = \frac{1}{2} m v^2 \). Therefore, we can express \( K_{\text{highest}} \) in terms of \( K \):
\[ K_{\text{highest}} = \frac{1}{8} m v^2 = \frac{1}{8} \cdot \left( \frac{1}{2} m v^2 \cdot 4 \right) = \frac{K}{4} \]

5. **Conclusion**:
The kinetic energy at the highest point is:
\[ K_{\text{highest}} = \frac{K}{4} \]

### Final Answer:
The kinetic energy at the highest point is \( \frac{K}{4} \).