A projectile can have same range `R` for two angles of projection. It `t_1 and t_2` are the times of flight in the two cases, then what is the product of two times of flight ?

Options

A. `t_1 t_2 prop R^2`
B. `t_1 t_2 prop R`
C. `t_1 t_2 prop (1)/( R)`
D. `t_1 t_2 prop (1)/(R^2)`

Explanation

To solve the problem, we need to find the product of the times of flight \( t_1 \) and \( t_2 \) for a projectile launched at two different angles that result in the same range \( R \).

### Step-by-step Solution:

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

2. **Identifying the Angles**:
For a given range \( R \), there are two angles of projection that yield the same range. These angles are complementary, meaning if one angle is \( \theta \), the other angle is \( 90^\circ - \theta \).

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

4. **Finding the Product of Times of Flight**:
Now, we need to find the product \( t_1 \times t_2 \):
\[ t_1 \times t_2 = \left(\frac{2u \sin(\theta)}{g}\right) \times \left(\frac{2u \cos(\theta)}{g}\right) \]
Simplifying this, we get:
\[ t_1 \times t_2 = \frac{4u^2 \sin(\theta) \cos(\theta)}{g^2} \]

5. **Using the Identity**:
We know that \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \), thus:
\[ t_1 \times t_2 = \frac{2u^2 \sin(2\theta)}{g^2} \]

6. **Relating to the Range**:
From the range formula, we have:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
Therefore, we can express \( \sin(2\theta) \) in terms of \( R \):
\[ \sin(2\theta) = \frac{gR}{u^2} \]

7. **Final Expression**:
Substituting this back into our expression for \( t_1 \times t_2 \):
\[ t_1 \times t_2 = \frac{2u^2 \left(\frac{gR}{u^2}\right)}{g^2} = \frac{2R}{g} \]

### Conclusion:
The product of the two times of flight \( t_1 \) and \( t_2 \) is:
\[ t_1 \times t_2 = \frac{2R}{g} \]