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 find the other possible angle of projection for a projectile that gives the same range \( R \) when the initial angle of projection is \( 40^\circ \).

### Step-by-Step Solution:

1. **Understand 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. **Apply the Formula for the Given Angle**:
For the angle \( \theta = 40^\circ \):
\[ R = \frac{U^2 \sin(80^\circ)}{g} \]

3. **Set Up the Equation for the Other Angle**:
Let the other angle of projection be \( \theta_2 \). The range for this angle can also be expressed as:
\[ R = \frac{U^2 \sin(2\theta_2)}{g} \]
Since both angles give the same range, we can equate the two expressions:
\[ \frac{U^2 \sin(2\theta_2)}{g} = \frac{U^2 \sin(80^\circ)}{g} \]

4. **Cancel Common Terms**:
We can cancel \( U^2 \) and \( g \) from both sides of the equation:
\[ \sin(2\theta_2) = \sin(80^\circ) \]

5. **Find Possible Solutions**:
The sine function has two solutions in the range of \( 0^\circ \) to \( 180^\circ \):
- \( 2\theta_2 = 80^\circ \)
- \( 2\theta_2 = 180^\circ - 80^\circ = 100^\circ \)

6. **Solve for \( \theta_2 \)**:
From the first equation:
\[ 2\theta_2 = 80^\circ \implies \theta_2 = 40^\circ \]
From the second equation:
\[ 2\theta_2 = 100^\circ \implies \theta_2 = 50^\circ \]

7. **Identify the Other Angle**:
Since we are looking for the other angle of projection, the answer is:
\[ \theta_2 = 50^\circ \]

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

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