It is possible to project a particle with a given speed in two possible ways so that it has the same horizontal range 'R'. The product of time taken by it in the two possible ways is

Options

A. `R//g`
B. 2R/g
C. 3R/g
D. 4R/g

Answer

To solve the problem of finding the product of the time taken by a particle projected at a given speed in two different angles that yield the same horizontal range \( R \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Horizontal Range Formula**:
The horizontal range \( R \) of a projectile is given by the formula:
\[ R = \frac{v_0^2 \sin 2\theta}{g} \]
where \( v_0 \) is the initial speed, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

2. **Set Up the Equations for Two Angles**:
Let’s denote the angles of projection as \( \theta_1 \) and \( \theta_2 \). Since both projections yield the same range \( R \), we can write:
\[ R = \frac{v_0^2 \sin 2\theta_1}{g} = \frac{v_0^2 \sin 2\theta_2}{g} \]
This implies:
\[ \sin 2\theta_1 = \sin 2\theta_2 \]

3. **Relate the Angles**:
From the property of sine, we know:
\[ 2\theta_1 = 180^\circ - 2\theta_2 \]
This leads us to:
\[ \theta_1 = 90^\circ - \theta_2 \]

4. **Time of Flight Formula**:
The time of flight \( t \) for a projectile is given by:
\[ t = \frac{2v_0 \sin \theta}{g} \]
Therefore, the times of flight for the two angles are:
\[ t_1 = \frac{2v_0 \sin \theta_1}{g} \]
\[ t_2 = \frac{2v_0 \sin \theta_2}{g} \]

5. **Calculate the Product of Time of Flight**:
The product of the times of flight is:
\[ t_1 \cdot t_2 = \left(\frac{2v_0 \sin \theta_1}{g}\right) \left(\frac{2v_0 \sin \theta_2}{g}\right) = \frac{4v_0^2 \sin \theta_1 \sin \theta_2}{g^2} \]

6. **Use the Angle Relation**:
Since \( \theta_1 = 90^\circ - \theta_2 \), we have:
\[ \sin \theta_1 = \cos \theta_2 \]
Thus, we can rewrite the product:
\[ t_1 \cdot t_2 = \frac{4v_0^2 \sin \theta_1 \cos \theta_2}{g^2} = \frac{4v_0^2 \sin \theta_2 \cos \theta_2}{g^2} \]
Using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \), we find:
\[ t_1 \cdot t_2 = \frac{2v_0^2 \sin 2\theta_2}{g^2} \]

7. **Relate to Horizontal Range**:
From the range formula, we know:
\[ R = \frac{v_0^2 \sin 2\theta_2}{g} \]
Therefore:
\[ \sin 2\theta_2 = \frac{gR}{v_0^2} \]
Substituting this back into the equation for \( t_1 \cdot t_2 \):
\[ t_1 \cdot t_2 = \frac{2v_0^2 \cdot \frac{gR}{v_0^2}}{g^2} = \frac{2R}{g} \]

8. **Final Answer**:
Thus, the product of the time taken by the particle in the two possible ways is:
\[ t_1 \cdot t_2 = \frac{2R}{g} \]

### Conclusion:
The correct answer is \( \frac{2R}{g} \).