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

Answer

To solve the problem, we need to find the product of the time of flights \( t_1 \) and \( t_2 \) for two projectiles thrown with the same speed such that their ranges are equal. Let's go through the solution step by step.

### Step 1: Understanding the Time of Flight Formula
The time of flight \( T \) for a projectile is given by the formula:
\[ T = \frac{2u \sin \theta}{g} \]
where:
- \( u \) is the initial speed,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity.

### Step 2: Time of Flight for Each Projectile
For the first projectile with angle \( \theta_1 \):
\[ t_1 = \frac{2u \sin \theta_1}{g} \]

For the second projectile with angle \( \theta_2 \):
\[ t_2 = \frac{2u \sin \theta_2}{g} \]

### Step 3: Product of Time of Flights
Now, we can find the product of \( t_1 \) and \( t_2 \):
\[ t_1 t_2 = \left(\frac{2u \sin \theta_1}{g}\right) \left(\frac{2u \sin \theta_2}{g}\right) = \frac{4u^2 \sin \theta_1 \sin \theta_2}{g^2} \]

### Step 4: Relationship Between Angles
Since the ranges of the two projectiles are equal, we can use the property of complementary angles in projectile motion. If \( \theta_1 + \theta_2 = 90^\circ \), then:
\[ \sin \theta_2 = \cos \theta_1 \]

### Step 5: Substitute into the Product Equation
Substituting \( \sin \theta_2 \) into the product equation:
\[ t_1 t_2 = \frac{4u^2 \sin \theta_1 \cos \theta_1}{g^2} \]

### Step 6: Use the Double Angle Identity
We can use the double angle identity for sine:
\[ \sin 2\theta = 2 \sin \theta \cos \theta \]
Thus,
\[ \sin \theta_1 \cos \theta_1 = \frac{1}{2} \sin 2\theta_1 \]
So we can rewrite the product as:
\[ t_1 t_2 = \frac{4u^2}{g^2} \cdot \frac{1}{2} \sin 2\theta_1 = \frac{2u^2 \sin 2\theta_1}{g^2} \]

### Step 7: Relate to Range
The range \( R \) of a projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Thus, we can express \( \sin 2\theta_1 \) in terms of \( R \):
\[ \sin 2\theta_1 = \frac{Rg}{u^2} \]

### Step 8: Substitute Back
Substituting this back into our equation for \( t_1 t_2 \):
\[ t_1 t_2 = \frac{2u^2}{g^2} \cdot \frac{Rg}{u^2} = \frac{2R}{g} \]

### Final Answer
Thus, the product of the time of flights \( t_1 t_2 \) in terms of range \( R \) and gravitational acceleration \( g \) is:
\[ \boxed{\frac{2R}{g}} \]