When a particle is projected at an angle to the horizontal, it has range R and time of flight `t_(1)`. If the same projectile is projected with same speed at another angle to have the saem range, time of flight is `t_(2)`. Show that:
`t_(1)t_(2)=(2R//g)`

Options

A. `t_(1) + t_(2) = (2R)/(g)`
B. `t_(1) - t_(2) = (R )/(g)`
C. `t_(1) t_(2) = (2R)/(g)`
D. `t_(1) t_(2) = (R )/(g)`

Answer

To solve the problem, we need to show that the product of the time of flight for two different angles of projection that yield the same range is equal to \( \frac{2R}{g} \).

### Step 1: Understanding the Range Formula
The range \( R \) of a projectile launched with an initial speed \( u \) at an angle \( \theta \) is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
where \( g \) is the acceleration due to gravity.

### Step 2: Setting Up the Equations for Two Angles
Let the first angle of projection be \( \theta \) with time of flight \( t_1 \) and the second angle be \( \alpha \) with time of flight \( t_2 \). Since both projectiles have the same range \( R \), we can write:
\[ \frac{u^2 \sin 2\theta}{g} = \frac{u^2 \sin 2\alpha}{g} \]
This simplifies to:
\[ \sin 2\theta = \sin 2\alpha \]
From this, we can conclude that:
\[ 2\theta = 2\alpha + n\pi \quad \text{(for some integer } n\text{)} \]
This implies:
\[ \theta = \alpha \quad \text{or} \quad \theta = 90^\circ - \alpha \]

### Step 3: Finding Time of Flight for Each Angle
The time of flight \( t \) for a projectile is given by:
\[ t = \frac{2u \sin \theta}{g} \]
Thus, for the first angle \( \theta \):
\[ t_1 = \frac{2u \sin \theta}{g} \]
And for the second angle \( \alpha \):
\[ t_2 = \frac{2u \sin \alpha}{g} \]

### Step 4: Product of Time of Flights
Now, we need to find the product \( t_1 t_2 \):
\[ t_1 t_2 = \left(\frac{2u \sin \theta}{g}\right) \left(\frac{2u \sin \alpha}{g}\right) = \frac{4u^2 \sin \theta \sin \alpha}{g^2} \]

### Step 5: Using the Identity for Sine
Using the identity \( \sin \theta \sin \alpha = \frac{1}{2} [\cos(\theta - \alpha) - \cos(\theta + \alpha)] \), we can express \( \sin \theta \sin \alpha \) in terms of \( \sin 2\theta \):
Since \( \sin 2\theta = 2 \sin \theta \cos \theta \), we can write:
\[ \sin 2\theta = \sin 2\alpha \]
Thus:
\[ t_1 t_2 = \frac{4u^2 \sin \theta \sin (90^\circ - \theta)}{g^2} = \frac{4u^2 \sin \theta \cos \theta}{g^2} \]

### Step 6: Relating to Range
From the range formula, we know:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Thus:
\[ \sin 2\theta = \frac{gR}{u^2} \]
Substituting this into our expression for \( t_1 t_2 \):
\[ t_1 t_2 = \frac{4u^2 \cdot \frac{gR}{u^2}}{g^2} = \frac{4R}{g} \]

### Final Step: Conclusion
We can simplify \( t_1 t_2 \) to:
\[ t_1 t_2 = \frac{2R}{g} \]
This shows that:
\[ t_1 t_2 = \frac{2R}{g} \]
Thus, we have proven the required relation.