Two projectiles A and B are thrown with velocities v and `v/2` respectively. They have the same range. If B is thrown at an angle of `15^@` to the horizontal, A must have been thrown at an angle

Options

A. `sin^(-1) (1/16)`
B. `sin^(-1)(1/4)`
C. `2 sin^(-1)(1/4)`
D. `1/2 sin^(-1)(1/8)`

Explanation

To solve the problem, we need to find the angle at which projectile A is thrown, given that projectile B is thrown at an angle of 15° and both projectiles have the same range.

### Step-by-Step Solution:

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

2. **Write the Range for Projectile A**:
For projectile A, let the initial velocity be \( v \) and the angle of projection be \( \theta_A \). The range \( R_A \) can be expressed as:
\[ R_A = \frac{v^2 \sin 2\theta_A}{g} \quad \text{(Equation 1)} \]

3. **Write the Range for Projectile B**:
For projectile B, the initial velocity is \( \frac{v}{2} \) and the angle of projection is \( 15° \). The range \( R_B \) can be expressed as:
\[ R_B = \frac{\left(\frac{v}{2}\right)^2 \sin 2(15°)}{g} = \frac{\frac{v^2}{4} \sin 30°}{g} \quad \text{(Equation 2)} \]
Since \( \sin 30° = \frac{1}{2} \), we can simplify this to:
\[ R_B = \frac{v^2}{4} \cdot \frac{1}{2g} = \frac{v^2}{8g} \]

4. **Setting the Ranges Equal**:
Since both projectiles have the same range, we can set \( R_A = R_B \):
\[ \frac{v^2 \sin 2\theta_A}{g} = \frac{v^2}{8g} \]

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

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

7. **Conclusion**:
The angle at which projectile A must have been thrown is:
\[ \theta_A = \frac{1}{2} \sin^{-1}\left(\frac{1}{8}\right) \]