A cricket ball is hit `30^(@)` with the horizontal with kinetic energy K. The kinetic energy at the highest point is

Options

A. Zero
B. `K//4`
C. `K//2`
D. `3K//4`

Answer

To solve the problem, we need to determine the kinetic energy of the cricket ball at the highest point of its trajectory after being hit at an angle of \(30^\circ\) with the horizontal.

### Step-by-Step Solution:

1. **Identify the Initial Kinetic Energy**:
The initial kinetic energy \(K\) of the cricket ball when it is hit at an angle of \(30^\circ\) with the horizontal is given.

2. **Determine the Initial Velocity**:
The kinetic energy is related to the velocity by the formula:
\[ K = \frac{1}{2} m u^2 \]
From this, we can express the initial velocity \(u\) as:
\[ u = \sqrt{\frac{2K}{m}} \]

3. **Resolve the Initial Velocity into Components**:
The initial velocity \(u\) can be resolved into horizontal and vertical components:
- Horizontal component \(u_x = u \cos(30^\circ)\)
- Vertical component \(u_y = u \sin(30^\circ)\)

Using the values of cosine and sine for \(30^\circ\):
\[ u_x = u \cdot \frac{\sqrt{3}}{2}, \quad u_y = u \cdot \frac{1}{2} \]

4. **Calculate the Horizontal Component at the Highest Point**:
At the highest point of the trajectory, the vertical component of the velocity becomes zero, and only the horizontal component remains. Thus, the horizontal component \(u_x\) at the highest point is:
\[ u_x = u \cos(30^\circ) = \sqrt{\frac{2K}{m}} \cdot \frac{\sqrt{3}}{2} \]

5. **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:
\[ K' = \frac{1}{2} m (u_x)^2 \]
Substituting \(u_x\):
\[ K' = \frac{1}{2} m \left(\sqrt{\frac{2K}{m}} \cdot \frac{\sqrt{3}}{2}\right)^2 \]
Simplifying this expression:
\[ K' = \frac{1}{2} m \cdot \left(\frac{3}{4} \cdot \frac{2K}{m}\right) = \frac{3K}{4} \]

6. **Final Result**:
Therefore, the kinetic energy at the highest point is:
\[ K' = \frac{3}{4} K \]

### Conclusion:
The kinetic energy of the cricket ball at the highest point is \(\frac{3}{4} K\).