Two projectiles are thrown same speed in such a way that their ranges are equal. If the time of flights for the two projectiles are `t_(1)` and `t_(2)` the value of `'t_(1)t_(2)'` in terms of range `'R'` and `'g'` is

Options

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

Answer

To solve the problem, we need to find the product of the time of flights \( t_1 \) and \( t_2 \) of two projectiles thrown with the same speed such that their ranges are equal. Let's denote the initial speed of the projectiles as \( u \), the range as \( R \), and the acceleration due to gravity as \( g \).

### Step-by-Step Solution:

1. **Understanding the Time of Flight**:
The time of flight \( t \) for a projectile launched at an angle \( \theta \) with an initial speed \( u \) is given by:
\[ t = \frac{2u \sin \theta}{g} \]
Therefore, for the first projectile:
\[ t_1 = \frac{2u \sin \theta_1}{g} \]
And for the second projectile:
\[ t_2 = \frac{2u \sin \theta_2}{g} \]

2. **Finding the Product of Time of Flights**:
We can find the product \( t_1 t_2 \):
\[ t_1 t_2 = \left(\frac{2u \sin \theta_1}{g}\right) \left(\frac{2u \sin \theta_2}{g}\right) = \frac{4u^2 \sin \theta_1 \sin \theta_2}{g^2} \]

3. **Using the Range Formula**:
The range \( R \) of a projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
For the two projectiles, since their ranges are equal, we can write:
\[ R = \frac{u^2 \sin 2\theta_1}{g} = \frac{u^2 \sin 2\theta_2}{g} \]

4. **Relating Angles**:
Since the ranges are equal, the angles \( \theta_1 \) and \( \theta_2 \) must be complementary:
\[ \theta_2 = 90^\circ - \theta_1 \]
This means:
\[ \sin \theta_2 = \cos \theta_1 \]

5. **Substituting the Sine Values**:
Now substituting \( \sin \theta_2 \) into the product:
\[ t_1 t_2 = \frac{4u^2 \sin \theta_1 \cos \theta_1}{g^2} \]
We can use the identity \( \sin \theta_1 \cos \theta_1 = \frac{1}{2} \sin 2\theta_1 \):
\[ t_1 t_2 = \frac{4u^2}{g^2} \cdot \frac{1}{2} \sin 2\theta_1 = \frac{2u^2 \sin 2\theta_1}{g^2} \]

6. **Expressing in Terms of Range**:
From the range equation, we know:
\[ R = \frac{u^2 \sin 2\theta_1}{g} \]
Thus, we can express \( \sin 2\theta_1 \) as:
\[ \sin 2\theta_1 = \frac{gR}{u^2} \]
Substituting this back into our equation for \( t_1 t_2 \):
\[ t_1 t_2 = \frac{2u^2}{g^2} \cdot \frac{gR}{u^2} = \frac{2R}{g} \]

### Final Answer:
The value of \( t_1 t_2 \) in terms of range \( R \) and \( g \) is:
\[ t_1 t_2 = \frac{2R}{g} \]