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. (a) Zero
B. (b) ` (E )/(2) `
C. (c)` (E )/(sqrt 2 ) `
D. (d)` E `

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 \(45^\circ\) with an initial kinetic energy \(K\).

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- The ball is hit at an angle of \(45^\circ\) to the horizontal.
- The initial kinetic energy of the ball is given as \(K\).

2. **Kinetic Energy Formula**:
- The kinetic energy \(K\) of an object is given by the formula:
\[ K = \frac{1}{2} mv^2 \]
where \(m\) is the mass of the object and \(v\) is its velocity.

3. **Components of Velocity**:
- When the ball is hit at an angle of \(45^\circ\), its initial velocity can be resolved into two components:
- Horizontal component: \(v_x = v \cos(45^\circ) = v \cdot \frac{1}{\sqrt{2}}\)
- Vertical component: \(v_y = v \sin(45^\circ) = v \cdot \frac{1}{\sqrt{2}}\)

4. **Velocity at the Highest Point**:
- At the highest point of its trajectory, the vertical component of the velocity becomes zero (\(v_y = 0\)), but the horizontal component remains:
\[ v_x = v \cdot \frac{1}{\sqrt{2}} \]

5. **Calculating Kinetic Energy at the Highest Point**:
- The kinetic energy at the highest point is determined solely by the horizontal component of the velocity:
\[ K_{h} = \frac{1}{2} m (v_x)^2 = \frac{1}{2} m \left(v \cdot \frac{1}{\sqrt{2}}\right)^2 \]
- Simplifying this:
\[ K_{h} = \frac{1}{2} m \left(v^2 \cdot \frac{1}{2}\right) = \frac{1}{4} mv^2 \]

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

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

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