Two projectiles are fired at different angles with the same magnitude of velocity such that they have the same range. At what angles they might have been projected ?

Options

A. `35^(@) and 75^(@)`
B. `25 ^(@) and 65^(@)`
C. `10^(@)and 59^(@)`
D. None of these

Answer

To solve the problem of determining the angles at which two projectiles can be fired with the same initial velocity to achieve the same range, we can follow these steps:

### 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, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of projection.

2. **Setting Up the Problem**:
Let the angles of projection for the two projectiles be \( \theta_1 \) and \( \theta_2 \). Since both projectiles have the same initial velocity and range, we can equate their range formulas:
\[ \frac{u^2 \sin(2\theta_1)}{g} = \frac{u^2 \sin(2\theta_2)}{g} \]
This simplifies to:
\[ \sin(2\theta_1) = \sin(2\theta_2) \]

3. **Using the Sine Function Properties**:
The sine function has the property that:
\[ \sin(x) = \sin(180^\circ - x) \]
Therefore, we can write:
\[ 2\theta_1 = 2\theta_2 \quad \text{or} \quad 2\theta_1 = 180^\circ - 2\theta_2 \]
From the first equation, we get:
\[ \theta_1 = \theta_2 \]
This is not useful since we need different angles.

4. **Finding the Relationship Between Angles**:
From the second equation, we can derive:
\[ 2\theta_1 + 2\theta_2 = 180^\circ \]
Dividing by 2 gives:
\[ \theta_1 + \theta_2 = 90^\circ \]
This means that the two angles of projection are complementary.

5. **Conclusion**:
Therefore, if two projectiles are fired at different angles with the same initial velocity and achieve the same range, the angles must satisfy:
\[ \theta_1 + \theta_2 = 90^\circ \]