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 solve the problem of finding the range of a projectile that has the same range for two angles of projection, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a projectile that is launched at two different angles, say \( \theta \) and \( \phi \), but both achieve the same range \( R \). The times of flight for these angles are \( t_1 \) and \( t_2 \).

2. **Formula for Range**:
The range \( R \) of a projectile launched with an initial velocity \( u \) at an angle \( \theta \) is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Similarly, for the angle \( \phi \):
\[ R = \frac{u^2 \sin 2\phi}{g} \]

3. **Setting the Ranges Equal**:
Since both ranges are equal, we can set the two equations for range equal to each other:
\[ \frac{u^2 \sin 2\theta}{g} = \frac{u^2 \sin 2\phi}{g} \]
The \( u^2/g \) terms cancel out, leading to:
\[ \sin 2\theta = \sin 2\phi \]

4. **Finding Relationships Between Angles**:
The equation \( \sin 2\theta = \sin 2\phi \) implies two scenarios:
- \( 2\theta = 2\phi \) (which gives \( \theta = \phi \))
- \( 2\theta = 180^\circ - 2\phi \) (which gives \( \theta = 90^\circ - \phi \))

5. **Time of Flight Equations**:
The time of flight \( t \) for a projectile is given by:
\[ t = \frac{2u \sin \theta}{g} \]
Therefore, we can write:
- For angle \( \theta \):
\[ t_1 = \frac{2u \sin \theta}{g} \]
- For angle \( \phi \):
\[ t_2 = \frac{2u \sin \phi}{g} \]

6. **Using the Relationship Between Angles**:
If we take \( \theta = 90^\circ - \phi \), then:
\[ \sin \theta = \sin(90^\circ - \phi) = \cos \phi \]
Thus, we can rewrite \( t_1 \) as:
\[ t_1 = \frac{2u \cos \phi}{g} \]

7. **Relating \( t_1 \) and \( t_2 \)**:
Now we have:
\[ t_1 = \frac{2u \cos \phi}{g} \quad \text{and} \quad t_2 = \frac{2u \sin \phi}{g} \]
Therefore, we can express \( t_1 \) and \( t_2 \) in terms of \( R \):
\[ t_1 t_2 = \left(\frac{2u \cos \phi}{g}\right) \left(\frac{2u \sin \phi}{g}\right) = \frac{4u^2 \sin \phi \cos \phi}{g^2} = \frac{2u^2 \sin 2\phi}{g^2} \]

8. **Finding the Range**:
Since \( R = \frac{u^2 \sin 2\phi}{g} \), we can express \( t_1 t_2 \) in terms of \( R \):
\[ t_1 t_2 = \frac{4R}{g^2} \]
Rearranging gives:
\[ R = \frac{g}{4} t_1 t_2 \]

### Final Result:
The range \( R \) of the projectile is given by:
\[ R = \frac{g}{4} t_1 t_2 \]