Two identical shells are fired from a point on the ground with same muzzle velocity at angle of elevation `alpha=45^(@)` and `beta=tan^(-1)3` towards top of a cliff, `20 m` away from the point of firing. If both the shells reach the top simultaneously, then
time interval between two frings is

Options

A. `sqrt(10)-sqrt(2)s`
B. `sqrt(10)+sqrt(2)s`
C. `sqrt(10)-sqrt(3)s`
D. `sqrt(10)+sqrt(3)s`

Explanation

To solve the problem, we need to analyze the projectile motion of both shells and find the time interval between their firings.

### Given Data:
- Distance to the cliff, \( d = 20 \, \text{m} \)
- Angle of elevation for shell 1, \( \alpha = 45^\circ \)
- Angle of elevation for shell 2, \( \beta = \tan^{-1}(3) \)

### Step 1: Determine the time of flight for shell 1
For shell 1, fired at an angle of \( 45^\circ \):
- The horizontal component of the velocity, \( V_{x1} = V \cos(45^\circ) = \frac{V}{\sqrt{2}} \)
- The vertical component of the velocity, \( V_{y1} = V \sin(45^\circ) = \frac{V}{\sqrt{2}} \)

The time taken to reach the cliff can be calculated using the horizontal motion:
\[ t_1 = \frac{d}{V_{x1}} = \frac{20}{\frac{V}{\sqrt{2}}} = \frac{20 \sqrt{2}}{V} \]

### Step 2: Determine the time of flight for shell 2
For shell 2, fired at an angle of \( \beta = \tan^{-1}(3) \):
- The horizontal component of the velocity, \( V_{x2} = V \cos(\beta) \)
- The vertical component of the velocity, \( V_{y2} = V \sin(\beta) \)

Using the identity \( \tan(\beta) = 3 \), we can find \( \sin(\beta) \) and \( \cos(\beta) \):
\[ \sin(\beta) = \frac{3}{\sqrt{10}}, \quad \cos(\beta) = \frac{1}{\sqrt{10}} \]

The time taken to reach the cliff is:
\[ t_2 = \frac{d}{V_{x2}} = \frac{20}{V \cos(\beta)} = \frac{20}{V \cdot \frac{1}{\sqrt{10}}} = \frac{20 \sqrt{10}}{V} \]

### Step 3: Set the times equal
Since both shells reach the top of the cliff simultaneously, we can set \( t_1 = t_2 \):
\[ \frac{20 \sqrt{2}}{V} = \frac{20 \sqrt{10}}{V} \]

### Step 4: Solve for the time interval
To find the time interval between the firings, we need to find the difference in time taken for each shell to reach the top of the cliff. Since both shells reach the same height at the same time, we can express the time interval \( \Delta t \) as:
\[ \Delta t = t_2 - t_1 \]

From the equations for \( t_1 \) and \( t_2 \):
\[ \Delta t = \frac{20 \sqrt{10}}{V} - \frac{20 \sqrt{2}}{V} = \frac{20}{V} (\sqrt{10} - \sqrt{2}) \]

### Final Step: Conclusion
The time interval between the firings of the two shells is given by:
\[ \Delta t = \frac{20}{V} (\sqrt{10} - \sqrt{2}) \]