A stone is thrown at an angle of `45^(@)` to the horizontal with kinetic energy K. The kinetic energy at the highest point is -
(a) `(K)/(2)` (b) `(K)/(sqrt2)` (c) K (d) zero

Options

A. `(K)/(2)`
B. `(K)/(sqrt2)`
C. K
D. zero

Answer

To solve the problem, we need to determine the kinetic energy of a stone thrown at an angle of 45 degrees to the horizontal at its highest point. Let's break this down step by step.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- The stone is thrown with an initial kinetic energy \( K \).
- The angle of projection is \( 45^\circ \).

2. **Components of Initial Velocity**:
- Let the initial velocity of the stone be \( u \).
- The horizontal component of the initial velocity \( u_x \) is given by:
\[ u_x = u \cos(45^\circ) = \frac{u}{\sqrt{2}} \]
- The vertical component of the initial velocity \( u_y \) is:
\[ u_y = u \sin(45^\circ) = \frac{u}{\sqrt{2}} \]

3. **Velocity at the Highest Point**:
- At the highest point of the projectile's path, the vertical component of the velocity becomes zero (\( v_y = 0 \)).
- Therefore, the velocity at the highest point is only the horizontal component:
\[ v = u_x = \frac{u}{\sqrt{2}} \]

4. **Calculating Kinetic Energy at the Highest Point**:
- The kinetic energy \( K' \) at the highest point can be calculated using the formula:
\[ K' = \frac{1}{2} m v^2 \]
- Substituting \( v = \frac{u}{\sqrt{2}} \):
\[ K' = \frac{1}{2} m \left(\frac{u}{\sqrt{2}}\right)^2 = \frac{1}{2} m \cdot \frac{u^2}{2} = \frac{1}{4} m u^2 \]

5. **Relating Initial Kinetic Energy to Kinetic Energy at Highest Point**:
- The initial kinetic energy \( K \) is given by:
\[ K = \frac{1}{2} m u^2 \]
- Now we can express \( K' \) in terms of \( K \):
\[ K' = \frac{1}{4} m u^2 = \frac{1}{2} \left(\frac{1}{2} m u^2\right) = \frac{K}{2} \]

6. **Conclusion**:
- The kinetic energy at the highest point is:
\[ K' = \frac{K}{2} \]
- Therefore, the correct answer is (a) \( \frac{K}{2} \).