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 step by step, we will use the kinematic equations for projectile motion.

### Step 1: Understand the problem
We have two balls, A and B, thrown with speeds \( u \) and \( \frac{u}{2} \) respectively. Both cover the same horizontal distance (range) before returning to the plane of projection. The angle of projection for ball B is given as \( 15^\circ \). We need to find the angle of projection for ball A.

### Step 2: Write the formula for the range of a projectile
The range \( R \) of a projectile launched with speed \( v \) at an angle \( \theta \) is given by the formula:
\[ R = \frac{v^2 \sin(2\theta)}{g} \]
where \( g \) is the acceleration due to gravity.

### Step 3: Write the range equations for both balls
For ball A:
\[ R_A = \frac{u^2 \sin(2\theta_A)}{g} \]
For ball B:
\[ R_B = \frac{\left(\frac{u}{2}\right)^2 \sin(2 \cdot 15^\circ)}{g} = \frac{\frac{u^2}{4} \sin(30^\circ)}{g} \]

### Step 4: Set the ranges equal to each other
Since both balls cover the same horizontal distance, we can equate their ranges:
\[ \frac{u^2 \sin(2\theta_A)}{g} = \frac{\frac{u^2}{4} \sin(30^\circ)}{g} \]

### Step 5: Simplify the equation
We can cancel \( \frac{u^2}{g} \) from both sides (assuming \( u \neq 0 \)):
\[ \sin(2\theta_A) = \frac{1}{4} \sin(30^\circ) \]

### Step 6: Calculate \( \sin(30^\circ) \)
We know that:
\[ \sin(30^\circ) = \frac{1}{2} \]
Substituting this value into the equation gives:
\[ \sin(2\theta_A) = \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8} \]

### Step 7: Solve for \( 2\theta_A \)
To find \( \theta_A \), we take the inverse sine:
\[ 2\theta_A = \sin^{-1}\left(\frac{1}{8}\right) \]

### Step 8: Solve for \( \theta_A \)
Now, divide by 2 to find \( \theta_A \):
\[ \theta_A = \frac{1}{2} \sin^{-1}\left(\frac{1}{8}\right) \]

### Conclusion
Thus, the angle of projection of ball A is:
\[ \theta_A = \frac{1}{2} \sin^{-1}\left(\frac{1}{8}\right) \]