A body is projected at an angle `60^@` with horizontal with kinetic energy K. When the velocity makes an angle `30^@` with the horizontal, the kinetic energy of the body will be

Options

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

Explanation

To solve the problem, we need to find the kinetic energy of a body projected at an angle of \(60^\circ\) with the horizontal when its velocity makes an angle of \(30^\circ\) with the horizontal. The initial kinetic energy is given as \(K\).

### Step-by-Step Solution:

1. **Initial Kinetic Energy**:
The initial kinetic energy \(K\) is given by the formula:
\[ K = \frac{1}{2} m u^2 \]
where \(u\) is the initial velocity of the body.

2. **Components of Initial Velocity**:
When the body is projected at an angle of \(60^\circ\), the horizontal and vertical components of the initial velocity \(u\) are:
\[ u_x = u \cos(60^\circ) = u \cdot \frac{1}{2} = \frac{u}{2} \]
\[ u_y = u \sin(60^\circ) = u \cdot \frac{\sqrt{3}}{2} = \frac{u \sqrt{3}}{2} \]

3. **Velocity at \(30^\circ\)**:
When the body makes an angle of \(30^\circ\) with the horizontal, we denote the velocity at this point as \(v\). The horizontal and vertical components of this velocity \(v\) are:
\[ v_x = v \cos(30^\circ) = v \cdot \frac{\sqrt{3}}{2} \]
\[ v_y = v \sin(30^\circ) = v \cdot \frac{1}{2} \]

4. **Horizontal Velocity Conservation**:
Since there is no horizontal acceleration, the horizontal component of velocity remains constant:
\[ u_x = v_x \]
Therefore,
\[ \frac{u}{2} = v \cdot \frac{\sqrt{3}}{2} \]
Solving for \(v\):
\[ v = \frac{u}{\sqrt{3}} \]

5. **Calculating Kinetic Energy at \(30^\circ\)**:
The kinetic energy \(K'\) when the velocity makes an angle of \(30^\circ\) is given by:
\[ K' = \frac{1}{2} m v^2 \]
Substituting \(v = \frac{u}{\sqrt{3}}\):
\[ K' = \frac{1}{2} m \left(\frac{u}{\sqrt{3}}\right)^2 = \frac{1}{2} m \cdot \frac{u^2}{3} = \frac{1}{3} \cdot \frac{1}{2} m u^2 \]
Since \(\frac{1}{2} m u^2 = K\), we have:
\[ K' = \frac{K}{3} \]

### Final Answer:
The kinetic energy of the body when the velocity makes an angle of \(30^\circ\) with the horizontal is:
\[ K' = \frac{K}{3} \]