The range of a projectile is R when the angle of projection is `40^(@)`. For the same velocity of projection and range, the other possible angle of projection is

Options

A. `45^(@)`
B. `50^(@)`
C. `60^(@)`
D. `40^(@)`

Explanation

To solve the problem, we need to determine the other possible angle of projection for a projectile that gives the same range when projected with the same initial velocity.

### Step-by-step Solution:

1. **Understand the Problem**: We are given that the range \( R \) of a projectile is achieved at an angle of projection \( \theta = 40^\circ \). We need to find another angle \( \theta' \) such that the range remains the same with the same initial velocity.

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

3. **Identify the Relationship**: It is known that for a given initial velocity, the range for angles \( \theta \) and \( 90^\circ - \theta \) is the same. Thus, if \( \theta = 40^\circ \), the other angle \( \theta' \) can be found using:
\[ \theta' = 90^\circ - \theta \]

4. **Calculate the Other Angle**: Substitute \( \theta = 40^\circ \) into the equation:
\[ \theta' = 90^\circ - 40^\circ = 50^\circ \]

5. **Conclusion**: Therefore, the other possible angle of projection that gives the same range \( R \) is \( 50^\circ \).

### Final Answer:
The other possible angle of projection is \( 50^\circ \).