A body has kinetic energy E when projected at angle of projection for maximum range. Its kinetic energy at the highest point of its path will be

Options

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

Explanation

To solve the problem, we need to determine the kinetic energy of a body at the highest point of its projectile motion when it is projected at an angle for maximum range.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- The body is projected at an angle of 45 degrees for maximum range.
- The initial kinetic energy (E) when projected can be expressed as:
\[ E = \frac{1}{2} m u^2 \]
where \( u \) is the initial velocity of projection and \( m \) is the mass of the body.

2. **Finding the Components of Velocity**:
- When the body reaches the highest point of its trajectory, its vertical component of velocity becomes zero, and only the horizontal component remains.
- The horizontal component of the initial velocity \( u \) at 45 degrees is:
\[ V' = u \cos(45^\circ) = u \cdot \frac{1}{\sqrt{2}} = \frac{u}{\sqrt{2}} \]

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

4. **Relating \( E' \) to \( E \)**:
- From the initial kinetic energy expression \( E = \frac{1}{2} m u^2 \), we can relate \( E' \) to \( E \):
\[ E' = \frac{1}{4} m u^2 = \frac{1}{2} \left(\frac{1}{2} m u^2\right) = \frac{E}{2} \]

5. **Conclusion**:
- Therefore, the kinetic energy at the highest point of the projectile's path is:
\[ E' = \frac{E}{2} \]

### Final Answer:
The kinetic energy at the highest point of its path will be \( \frac{E}{2} \).