A projectile can have the same range R for two angles of projection. If `t_(1)` and `t_(2)` be the times of flight in the two cases:-

Options

A. `t_(1)t_(2)propR^(2)`
B. `t_(1)t_(2)propR`
C. `t_(1)t_(2)prop(1)/(R)`
D. `t_(1)t_(2)prop(1)/(R_(2))`

Explanation

To solve the problem regarding the relationship between the times of flight \( t_1 \) and \( t_2 \) for a projectile launched at two angles that give the same range \( R \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Angles of Projection**:
- For a projectile to have the same range \( R \) for two different angles, if one angle is \( \theta \), the other angle must be \( 90^\circ - \theta \).

2. **Formula for Time of Flight**:
- The time of flight \( t \) for a projectile launched at an angle \( \theta \) with initial velocity \( u \) is given by:
\[ t = \frac{2u \sin \theta}{g} \]
- Therefore, for the angle \( \theta \), the time of flight \( t_1 \) is:
\[ t_1 = \frac{2u \sin \theta}{g} \]
- For the angle \( 90^\circ - \theta \), the time of flight \( t_2 \) is:
\[ t_2 = \frac{2u \sin(90^\circ - \theta)}{g} = \frac{2u \cos \theta}{g} \]

3. **Finding the Relationship Between \( t_1 \) and \( t_2 \)**:
- Now we have:
\[ t_1 = \frac{2u \sin \theta}{g} \]
\[ t_2 = \frac{2u \cos \theta}{g} \]
- We can express the ratio of \( t_1 \) and \( t_2 \):
\[ \frac{t_1}{t_2} = \frac{\frac{2u \sin \theta}{g}}{\frac{2u \cos \theta}{g}} = \frac{\sin \theta}{\cos \theta} = \tan \theta \]
- Thus, we can conclude that:
\[ t_1 = t_2 \tan \theta \]

4. **Conclusion**:
- The relationship between the times of flight \( t_1 \) and \( t_2 \) for the two angles of projection that yield the same range \( R \) is:
\[ t_1 = t_2 \tan \theta \]