A projectile is thrown at an angle of `40^@` with the horizontal and its range is `R_1` Another projectile is thrown at an angle `40^@` with the vertical and its range is `R_2` What is the relation between `R_1 and R_2` ? 

Options

A. `R_1 =R_2`
B. `R_1 =2R_2`
C. `R_1=R_2/2`
D. `R_1 = (4R_2)/5`

Explanation

To find the relationship between the ranges \( R_1 \) and \( R_2 \) of two projectiles thrown at angles of \( 40^\circ \) with the horizontal and vertical respectively, we can use the formula for the range of a projectile.

### Step-by-Step Solution:

1. **Understanding the Angles**:
- The first projectile is thrown at an angle \( \theta_1 = 40^\circ \) with the horizontal.
- The second projectile is thrown at an angle \( \theta_2 = 40^\circ \) with the vertical. This means the angle with the horizontal is \( 90^\circ - 40^\circ = 50^\circ \).

2. **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.

3. **Calculating \( R_1 \)**:
For the first projectile (angle \( 40^\circ \)):
\[ R_1 = \frac{u^2 \sin(2 \times 40^\circ)}{g} = \frac{u^2 \sin(80^\circ)}{g} \]

4. **Calculating \( R_2 \)**:
For the second projectile (angle \( 50^\circ \)):
\[ R_2 = \frac{u^2 \sin(2 \times 50^\circ)}{g} = \frac{u^2 \sin(100^\circ)}{g} \]

5. **Using the Sine Property**:
We know that \( \sin(100^\circ) = \sin(80^\circ) \) because \( \sin(180^\circ - x) = \sin(x) \).
Therefore, we can say:
\[ R_2 = \frac{u^2 \sin(80^\circ)}{g} \]

6. **Comparing \( R_1 \) and \( R_2 \)**:
Since both \( R_1 \) and \( R_2 \) have the same expression:
\[ R_1 = R_2 \]

### Conclusion:
The relationship between the ranges is:
\[ R_1 = R_2 \]