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 solve the problem, we need to determine the kinetic energy of a cricket ball at its highest point after being hit at an angle of 45 degrees with an initial kinetic energy \( K \).

### Step-by-Step Solution:

1. **Understanding the Initial Conditions:**
- The cricket ball is hit at an angle \( \theta = 45^\circ \) with an initial kinetic energy \( K \).
- The initial kinetic energy can be expressed using the formula:
\[ K = \frac{1}{2} m u^2 \]
where \( m \) is the mass of the ball and \( u \) is its initial velocity.

2. **Finding the Horizontal Component of Velocity:**
- At the highest point of its trajectory, the vertical component of the velocity becomes zero, and only the horizontal component remains.
- The horizontal component of the initial velocity \( u \) can be calculated as:
\[ u_x = u \cos(45^\circ) = u \cdot \frac{1}{\sqrt{2}} = \frac{u}{\sqrt{2}} \]

3. **Calculating the Kinetic Energy at the Highest Point:**
- The kinetic energy at the highest point \( K' \) can be expressed as:
\[ K' = \frac{1}{2} m (u_x)^2 \]
- Substituting \( u_x \) into the kinetic energy formula:
\[ K' = \frac{1}{2} m \left(\frac{u}{\sqrt{2}}\right)^2 \]
- Simplifying this expression:
\[ K' = \frac{1}{2} m \cdot \frac{u^2}{2} = \frac{1}{4} m u^2 \]

4. **Relating \( K' \) to the Initial Kinetic Energy \( K \):**
- We know from the initial kinetic energy expression that:
\[ K = \frac{1}{2} m u^2 \]
- Therefore, we can express \( K' \) in terms of \( K \):
\[ K' = \frac{1}{4} m u^2 = \frac{1}{2} \left(\frac{1}{2} m u^2\right) = \frac{1}{2} K \]

5. **Conclusion:**
- The kinetic energy of the ball at the highest point is:
\[ K' = \frac{K}{2} \]

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