Two balls `A and B` are thrown with speeds `u and u//2`, respectively. Both the balls cover the same horizontal distance before returning to the plane of projection. If the angle of projection of ball `B is 15^@` with the horizontal, then the angle of projection of `A` is.

Options

A. `sin^-1 ((1)/(8))`
B. `(1)/(2) sin^-1 ((1)/(8))`
C. `(1)/(3) sin^-1 ((1)/(8))`
D. `(1)/(4) sin^-1((1)/(8))`

Explanation

To solve the problem, we need to find the angle of projection of ball A, given that both balls A and B cover the same horizontal distance before returning to the plane of projection.

### Step-by-Step Solution:

1. **Identify the Given Information**:
- Speed of ball A = \( u \)
- Speed of ball B = \( \frac{u}{2} \)
- Angle of projection of ball B = \( 15^\circ \)

2. **Use the Range Formula**:
The horizontal range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
where \( u \) is the initial speed, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

3. **Write the Range for Both Balls**:
- For ball A (angle of projection = \( \theta \)):
\[ R_A = \frac{u^2 \sin 2\theta}{g} \]
- For ball B (angle of projection = \( 15^\circ \)):
\[ R_B = \frac{\left(\frac{u}{2}\right)^2 \sin 30^\circ}{g} = \frac{\frac{u^2}{4} \cdot \frac{1}{2}}{g} = \frac{u^2}{8g} \]

4. **Set the Ranges Equal**:
Since both balls cover the same horizontal distance:
\[ R_A = R_B \]
Therefore:
\[ \frac{u^2 \sin 2\theta}{g} = \frac{u^2}{8g} \]

5. **Cancel Common Terms**:
We can cancel \( \frac{u^2}{g} \) from both sides (assuming \( u \neq 0 \)):
\[ \sin 2\theta = \frac{1}{8} \]

6. **Find the Angle \( \theta \)**:
To find \( \theta \), we take the inverse sine:
\[ 2\theta = \sin^{-1}\left(\frac{1}{8}\right) \]
Therefore:
\[ \theta = \frac{1}{2} \sin^{-1}\left(\frac{1}{8}\right) \]

### Final Answer:
The angle of projection of ball A is:
\[ \theta = \frac{1}{2} \sin^{-1}\left(\frac{1}{8}\right) \]