The range of a projectile at an angle `theta` is equal to half of the maximum range if thrown at the same speed. The angel of projection `theta` is given by

Options

A. `15^@`
B. `30^@`
C. `60^@`
D. data insufficient

Explanation

To solve the problem, we need to find the angle of projection \( \theta \) such that the range of the projectile at this angle is equal to half of the maximum range when thrown with the same speed \( u \).

### Step-by-step Solution:

1. **Understanding the Range Formula**:
The range \( R \) of a projectile launched at an angle \( \theta \) with initial velocity \( u \) is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
where \( g \) is the acceleration due to gravity.

2. **Finding the Maximum Range**:
The maximum range \( R_{\text{max}} \) occurs when \( \sin 2\theta = 1 \). This happens when \( 2\theta = 90^\circ \) or \( \theta = 45^\circ \). Thus, the maximum range is:
\[ R_{\text{max}} = \frac{u^2}{g} \]

3. **Setting Up the Equation**:
According to the problem, the range at angle \( \theta \) is equal to half of the maximum range:
\[ R = \frac{1}{2} R_{\text{max}} \]
Substituting the expressions for \( R \) and \( R_{\text{max}} \):
\[ \frac{u^2 \sin 2\theta}{g} = \frac{1}{2} \left(\frac{u^2}{g}\right) \]

4. **Simplifying the Equation**:
We can cancel \( \frac{u^2}{g} \) from both sides (assuming \( u \neq 0 \)):
\[ \sin 2\theta = \frac{1}{2} \]

5. **Solving for \( 2\theta \)**:
The equation \( \sin 2\theta = \frac{1}{2} \) gives us:
\[ 2\theta = 30^\circ \quad \text{or} \quad 2\theta = 150^\circ \]
This leads to:
\[ \theta = 15^\circ \quad \text{or} \quad \theta = 75^\circ \]

6. **Choosing the Correct Angle**:
Since the problem typically refers to the smaller angle of projection, we take:
\[ \theta = 15^\circ \]

### Final Answer:
The angle of projection \( \theta \) is \( 15^\circ \).

---