Two particles are projected with same initial velocity one makes angle `theta` with horizontal while other makes an angle `theta` with vertical. If their common range is R then product of their time of flight is directly proportional to :

Options

A. R
B. `R^2`
C. `1/R`
D. `R^0`

Explanation

To solve the problem step by step, we will analyze the motion of the two particles and derive the required relationship.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have two particles projected with the same initial velocity \( u \). One particle is projected at an angle \( \theta \) with the horizontal, and the other at an angle \( \theta \) with the vertical (which means it makes an angle of \( 90^\circ - \theta \) with the horizontal).

2. **Range of the Projectiles**:
The range \( R \) of a projectile launched at an angle \( \theta \) with the horizontal is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
For the second particle, which is launched at an angle \( 90^\circ - \theta \) with the horizontal, the range is:
\[ R = \frac{u^2 \sin(2(90^\circ - \theta))}{g} = \frac{u^2 \sin(180^\circ - 2\theta)}{g} = \frac{u^2 \sin(2\theta)}{g} \]
Thus, both particles have the same range \( R \).

3. **Time of Flight**:
The time of flight \( t_1 \) for the first particle is given by:
\[ t_1 = \frac{2u \sin(\theta)}{g} \]
The time of flight \( t_2 \) for the second particle is:
\[ t_2 = \frac{2u \sin(90^\circ - \theta)}{g} = \frac{2u \cos(\theta)}{g} \]

4. **Product of Time of Flight**:
We need to find the product \( t_1 \cdot t_2 \):
\[ t_1 \cdot t_2 = \left(\frac{2u \sin(\theta)}{g}\right) \cdot \left(\frac{2u \cos(\theta)}{g}\right) \]
Simplifying this, we get:
\[ t_1 \cdot t_2 = \frac{4u^2 \sin(\theta) \cos(\theta)}{g^2} \]
Using the identity \( \sin(\theta) \cos(\theta) = \frac{1}{2} \sin(2\theta) \), we can rewrite it as:
\[ t_1 \cdot t_2 = \frac{2u^2 \sin(2\theta)}{g^2} \]

5. **Direct Proportionality**:
Since the range \( R \) is given by \( R = \frac{u^2 \sin(2\theta)}{g} \), we can express \( \sin(2\theta) \) in terms of \( R \):
\[ \sin(2\theta) = \frac{gR}{u^2} \]
Substituting this back into the expression for \( t_1 \cdot t_2 \):
\[ t_1 \cdot t_2 = \frac{2u^2 \cdot \frac{gR}{u^2}}{g^2} = \frac{2gR}{g^2} = \frac{2R}{g} \]
Thus, we find that the product of the times of flight \( t_1 \cdot t_2 \) is directly proportional to the range \( R \).

### Conclusion:
The product of the time of flight \( t_1 \cdot t_2 \) is directly proportional to the range \( R \).