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
If just before striking the top of cliff, the two shells get stuck together, considering elastic collision of combined body with the top of cliff, then maximum height reached by the combined body is

Options

A. a. `20m`
B. b. `10m`
C. c. `24m`
D. `12m`

Explanation

To solve the problem step by step, we will follow the trajectory equations for both shells and analyze their motion before they collide and stick together at the top of the cliff.

### Step 1: Determine the equations of motion for both shells

The trajectory of a projectile can be described by the equation:
\[ y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta} \]

Where:
- \( y \) is the height,
- \( x \) is the horizontal distance,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity,
- \( u \) is the initial velocity.

### Step 2: Calculate the height of the first shell fired at \( \alpha = 45^\circ \)

For the first shell:
- \( \theta_1 = 45^\circ \)
- \( \tan(45^\circ) = 1 \)
- \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \)

Substituting into the trajectory equation:
\[ h_1 = 20 \tan(45^\circ) - \frac{g \cdot 20^2}{2 u^2 \cos^2(45^\circ)} \]
\[ h_1 = 20 - \frac{g \cdot 400}{2 u^2 \cdot \frac{1}{2}} \]
\[ h_1 = 20 - \frac{g \cdot 400}{u^2} \]

### Step 3: Calculate the height of the second shell fired at \( \beta = \tan^{-1}(3) \)

For the second shell:
- \( \tan(\beta) = 3 \)
- The opposite side is 3 and adjacent side is 1, so the hypotenuse is \( \sqrt{10} \).
- \( \sin(\beta) = \frac{3}{\sqrt{10}}, \cos(\beta) = \frac{1}{\sqrt{10}} \)

Substituting into the trajectory equation:
\[ h_2 = 20 \tan(\beta) - \frac{g \cdot 20^2}{2 u^2 \cos^2(\beta)} \]
\[ h_2 = 20 \cdot 3 - \frac{g \cdot 400}{2 u^2 \cdot \frac{1}{10}} \]
\[ h_2 = 60 - \frac{5g \cdot 400}{u^2} \]

### Step 4: Set the heights equal since both shells reach the top simultaneously

Setting \( h_1 = h_2 \):
\[ 20 - \frac{g \cdot 400}{u^2} = 60 - \frac{5g \cdot 400}{u^2} \]

### Step 5: Solve for \( u^2 \)

Rearranging gives:
\[ 60 - 20 = \frac{5g \cdot 400}{u^2} - \frac{g \cdot 400}{u^2} \]
\[ 40 = \frac{4g \cdot 400}{u^2} \]
\[ u^2 = \frac{4g \cdot 400}{40} = 40g \]
Assuming \( g = 10 \, \text{m/s}^2 \):
\[ u^2 = 400 \]
\[ u = 20 \, \text{m/s} \]

### Step 6: Calculate the maximum height reached by each shell

Substituting \( u \) back into the height equations:
For the first shell:
\[ h_1 = 20 - \frac{10 \cdot 400}{400} = 20 - 10 = 10 \, \text{m} \]

For the second shell:
\[ h_2 = 60 - \frac{5 \cdot 10 \cdot 400}{400} = 60 - 50 = 10 \, \text{m} \]

### Step 7: Calculate the final velocities just before collision

Using the formula for the final vertical velocity:
\[ v_{y1} = u \sin(45^\circ) - g t_1 \]
\[ v_{y1} = 20 \cdot \frac{1}{\sqrt{2}} - 10 \cdot t_1 \]

For the second shell:
\[ v_{y2} = u \sin(\beta) - g t_2 \]
\[ v_{y2} = 20 \cdot \frac{3}{\sqrt{10}} - 10 \cdot t_2 \]

### Step 8: Apply conservation of momentum

When the shells stick together, using conservation of momentum:
\[ m v_{y1} + m v_{y2} = (m + m) v_{combined} \]
\[ v_{combined} = \frac{v_{y1} + v_{y2}}{2} \]

### Step 9: Calculate the maximum height after the collision

Using the rebound velocity:
\[ h_{max} = \frac{v_{combined}^2}{2g} \]

### Step 10: Add the height of the cliff to find the total height

Total height reached by the combined body:
\[ H = h + h_{max} = 10 + h_{max} \]

### Conclusion

The maximum height reached by the combined body after the collision with the cliff is \( 12 \, \text{m} \).

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