A body is projected horizontally with a velocity of `u ms^(-1)` at an angle `theta` with the horizontal. The kinetic energy at the highest point is `(3)/(4)th` of the kinetic energy. The value of `theta` is :

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `120^(@)`

Answer

To solve the problem, we need to find the angle \( \theta \) at which a body is projected such that the kinetic energy at the highest point of its trajectory is \( \frac{3}{4} \) of the kinetic energy at the lowest point.

### Step-by-Step Solution:

1. **Understanding Kinetic Energy**:
- The kinetic energy (KE) of an object is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]
- At the lowest point (when the body is projected), the total velocity \( u \) can be split into horizontal and vertical components.

2. **Components of Velocity**:
- When projected at an angle \( \theta \):
- Horizontal component: \( u_x = u \cos \theta \)
- Vertical component: \( u_y = u \sin \theta \)

3. **Kinetic Energy at the Lowest Point**:
- The total kinetic energy at the lowest point (ground level) is:
\[ KE_{\text{lowest}} = \frac{1}{2} m (u_x^2 + u_y^2) = \frac{1}{2} m (u^2 \cos^2 \theta + u^2 \sin^2 \theta) = \frac{1}{2} m u^2 \]

4. **Kinetic Energy at the Highest Point**:
- At the highest point of the projectile motion, the vertical component of velocity becomes zero, so:
- Horizontal component remains: \( u_x = u \cos \theta \)
- Thus, the kinetic energy at the highest point is:
\[ KE_{\text{highest}} = \frac{1}{2} m (u \cos \theta)^2 = \frac{1}{2} m u^2 \cos^2 \theta \]

5. **Setting Up the Equation**:
- According to the problem, the kinetic energy at the highest point is \( \frac{3}{4} \) of the kinetic energy at the lowest point:
\[ KE_{\text{highest}} = \frac{3}{4} KE_{\text{lowest}} \]
- Substituting the expressions for kinetic energy:
\[ \frac{1}{2} m u^2 \cos^2 \theta = \frac{3}{4} \left( \frac{1}{2} m u^2 \right) \]

6. **Simplifying the Equation**:
- Canceling \( \frac{1}{2} m u^2 \) from both sides:
\[ \cos^2 \theta = \frac{3}{4} \]

7. **Finding \( \theta \)**:
- Taking the square root:
\[ \cos \theta = \frac{\sqrt{3}}{2} \]
- Therefore, the angle \( \theta \) is:
\[ \theta = 30^\circ \]

### Final Answer:
The value of \( \theta \) is \( 30^\circ \).