For a given velocity, a projectile has the same range R for two angles of projection. If `t_(1)` and `t_(2)` are the time of flight in the two cases, then `t_(1)*t_(2)` is equal to

Options

A. `(2R)/(g)`
B. `(R)/(g)`
C. `(4R)/(g)`
D. `(R)/(2g)`

Explanation

To solve the problem, we need to find the product of the time of flights \( t_1 \) and \( t_2 \) for two angles of projection that give the same range \( R \).

### Step-by-Step Solution:

1. **Understanding the 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, \( \theta \) is the angle of projection, and \( g \) 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 can be expressed as:
\[ \theta_1 = \theta \quad \text{and} \quad \theta_2 = 90^\circ - \theta \]

3. **Calculating Time of Flight**:
The time of flight \( t \) for a projectile is given by:
\[ t = \frac{2u \sin \theta}{g} \]
Therefore, for the first angle \( \theta_1 \):
\[ t_1 = \frac{2u \sin \theta_1}{g} = \frac{2u \sin \theta}{g} \]
For the second angle \( \theta_2 \):
\[ t_2 = \frac{2u \sin \theta_2}{g} = \frac{2u \sin(90^\circ - \theta)}{g} = \frac{2u \cos \theta}{g} \]

4. **Finding the Product \( t_1 \cdot t_2 \)**:
Now, we can find the product of the two times of flight:
\[ t_1 \cdot t_2 = \left(\frac{2u \sin \theta}{g}\right) \cdot \left(\frac{2u \cos \theta}{g}\right) \]
Simplifying this gives:
\[ t_1 \cdot t_2 = \frac{4u^2 \sin \theta \cos \theta}{g^2} \]

5. **Using the Double Angle Identity**:
We know from trigonometric identities that:
\[ 2 \sin \theta \cos \theta = \sin 2\theta \]
Therefore, we can rewrite \( t_1 \cdot t_2 \) as:
\[ t_1 \cdot t_2 = \frac{2u^2 \sin 2\theta}{g^2} \]

6. **Substituting the Range**:
From the range formula, we have:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Thus, substituting \( R \) into our expression for \( t_1 \cdot t_2 \):
\[ t_1 \cdot t_2 = \frac{2R}{g} \]

### Final Result:
The product of the time of flights \( t_1 \cdot t_2 \) is given by:
\[ t_1 \cdot t_2 = \frac{2R}{g} \]