A projectile has same range for two angules of projection. If times of flight in two cases are ` t_(1) and t_(2)` then the range of the projectilie is

Options

A. `1/2 g t_(1)t_(2)`
B. `1/4g t_(1)t_(2)`
C. `g t_(1) t_(2)`
D. `1/8 g t_(1)t_(2)`

Answer

To find the range of a projectile that has the same range for two angles of projection, we can derive the relationship using the given times of flight \( t_1 \) and \( t_2 \). Here’s a step-by-step solution:

### Step 1: Understand the relationship between angles and time of flight
When a projectile is launched at an angle \( \theta \), the complementary angle \( 90^\circ - \theta \) will give the same range. Therefore, we can denote the two angles as \( \theta \) and \( 90^\circ - \theta \).

### Step 2: Write the formula for time of flight
The time of flight \( t_1 \) for the angle \( \theta \) is given by:
\[ t_1 = \frac{2v \sin \theta}{g} \]
For the angle \( 90^\circ - \theta \), the time of flight \( t_2 \) becomes:
\[ t_2 = \frac{2v \sin(90^\circ - \theta)}{g} = \frac{2v \cos \theta}{g} \]

### Step 3: Relate the times of flight to the range
The range \( R \) of the projectile can be expressed using the formula:
\[ R = \frac{v^2 \sin 2\theta}{g} \]
Using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \), we can rewrite the range as:
\[ R = \frac{v^2 (2 \sin \theta \cos \theta)}{g} \]

### Step 4: Express the range in terms of \( t_1 \) and \( t_2 \)
From the expressions for \( t_1 \) and \( t_2 \), we can multiply them:
\[ t_1 t_2 = \left(\frac{2v \sin \theta}{g}\right) \left(\frac{2v \cos \theta}{g}\right) = \frac{4v^2 \sin \theta \cos \theta}{g^2} \]
This can be rearranged to express \( v^2 \sin \theta \cos \theta \):
\[ v^2 \sin \theta \cos \theta = \frac{g^2 t_1 t_2}{4} \]

### Step 5: Substitute back to find the range
Substituting this back into the range formula:
\[ R = \frac{v^2 (2 \sin \theta \cos \theta)}{g} = \frac{2 \cdot \frac{g^2 t_1 t_2}{4}}{g} = \frac{g t_1 t_2}{4} \]

### Final Result
Thus, the range \( R \) of the projectile in terms of \( t_1 \) and \( t_2 \) is:
\[ R = \frac{g t_1 t_2}{4} \]