A cricket ball is hit for a six the bat at an angle of `45^(@)` to the horizontal with kinetic energy K. At the highest point, the kinetic energy of the ball is

Options

A. zero
B. K
C. `K//2`
D. `K//sqrt(2)`

Explanation

To find the kinetic energy of the cricket ball at the highest point after being hit at an angle of \(45^\circ\) with an initial kinetic energy \(K\), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- The cricket ball is hit at an angle of \(45^\circ\) with an initial kinetic energy \(K\).
- The initial velocity \(U\) can be derived from the kinetic energy formula:
\[ K = \frac{1}{2} m U^2 \]

2. **Breaking Down the Initial Velocity**:
- The initial velocity \(U\) can be split into horizontal and vertical components.
- Both components will be equal since the angle is \(45^\circ\):
\[ U_x = U \cos(45^\circ) = \frac{U}{\sqrt{2}} \]
\[ U_y = U \sin(45^\circ) = \frac{U}{\sqrt{2}} \]

3. **Analyzing Motion at the Highest Point**:
- At the highest point of the projectile's trajectory, the vertical component of the velocity becomes zero (\(U_y = 0\)).
- The horizontal component (\(U_x\)) remains unchanged:
\[ U_x = \frac{U}{\sqrt{2}} \]

4. **Calculating Kinetic Energy at the Highest Point**:
- The kinetic energy at the highest point \(K_h\) can be calculated using the horizontal component of the velocity:
\[ K_h = \frac{1}{2} m (U_x)^2 = \frac{1}{2} m \left(\frac{U}{\sqrt{2}}\right)^2 \]
- Simplifying this gives:
\[ K_h = \frac{1}{2} m \left(\frac{U^2}{2}\right) = \frac{1}{4} m U^2 \]

5. **Relating \(K_h\) to \(K\)**:
- From the initial kinetic energy equation, we know:
\[ K = \frac{1}{2} m U^2 \]
- Therefore, we can express \(K_h\) in terms of \(K\):
\[ K_h = \frac{1}{4} m U^2 = \frac{1}{4} \cdot 2K = \frac{K}{2} \]

6. **Final Result**:
- The kinetic energy of the ball at the highest point is:
\[ K_h = \frac{K}{2} \]

### Conclusion:
The kinetic energy of the cricket ball at the highest point is \(\frac{K}{2}\).