A particle is projected at ` 60 ^@ ` to the horizontal with kinetic energy K. Its 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 analyze the motion of the particle projected at an angle of \(60^\circ\) to the horizontal and determine its kinetic energy at the highest point of its trajectory.

### Step-by-Step Solution:

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

2. **Determine the Components of Initial Velocity**:
When the particle is projected at an angle of \(60^\circ\), its initial velocity can be resolved into horizontal and vertical components:
- Horizontal component:
\[ u_x = u \cos(60^\circ) = u \cdot \frac{1}{2} = \frac{u}{2} \]
- Vertical component:
\[ u_y = u \sin(60^\circ) = u \cdot \frac{\sqrt{3}}{2} \]

3. **Analyze the Motion at the Highest Point**:
At the highest point of the projectile's trajectory, the vertical component of the velocity becomes zero (\(u_y = 0\)). Therefore, the only component of velocity at this point is the horizontal component:
\[ v_x = u_x = \frac{u}{2} \]

4. **Calculate the Kinetic Energy at the Highest Point**:
The kinetic energy at the highest point \(K'\) can be calculated using the horizontal component of the velocity:
\[ K' = \frac{1}{2} m v_x^2 = \frac{1}{2} m \left(\frac{u}{2}\right)^2 = \frac{1}{2} m \cdot \frac{u^2}{4} = \frac{1}{8} m u^2 \]

5. **Relate \(K'\) to the Initial Kinetic Energy \(K\)**:
Since we know that \(K = \frac{1}{2} m u^2\), we can express \(K'\) in terms of \(K\):
\[ K' = \frac{1}{8} m u^2 = \frac{1}{4} \left(\frac{1}{2} m u^2\right) = \frac{K}{4} \]

### Final Answer:
Thus, the kinetic energy at the highest point is:
\[ K' = \frac{K}{4} \]