A cannon of mass `5m` (including a shell of mass m) is at rest on a smooth horizontal ground, fires the shell with its barrel at an angle `theta` with the horizontal at a velocity u relative to cannon. Find the horizontal distance of the point where shell strikes the ground from the initial position of the cannon :-

Options

A. `(2v_(1)^(2)v_(2))/(g)`
B. `(2v_(1)v_(2)^(2))/(g)`
C. `(2v_(1)v_(2))/(g)`
D. `(2v_(1)^(2)v_(2)^(2))/(g)`

Explanation

To solve the problem step by step, we will analyze the situation involving the cannon and the shell it fires.

### Step 1: Identify the components of the shell's velocity
The shell is fired at an angle \( \theta \) with a velocity \( u \) relative to the cannon. We can break this velocity into horizontal and vertical components:
- Horizontal component: \( u_x = u \cos \theta \)
- Vertical component: \( u_y = u \sin \theta \)

### Step 2: Determine the velocity of the cannon after firing
Since the cannon has a mass of \( 5m \) (including the shell of mass \( m \)), when the shell is fired, the cannon will recoil. Let \( V_0 \) be the velocity of the cannon after firing. By conservation of momentum:
\[ \text{Initial momentum} = \text{Final momentum} \]
Initially, both the cannon and the shell are at rest, so the initial momentum is \( 0 \). After firing:
\[ m (u \cos \theta - V_0) + 4m (-V_0) = 0 \]
This simplifies to:
\[ m (u \cos \theta - V_0) - 4m V_0 = 0 \]
\[ u \cos \theta - 5V_0 = 0 \]
Thus, we find:
\[ V_0 = \frac{u \cos \theta}{5} \]

### Step 3: Calculate the time of flight of the shell
The time of flight \( T \) for the shell can be calculated using the vertical motion. The vertical motion is influenced only by gravity. The time of flight until the shell hits the ground can be calculated using:
\[ T = \frac{2u \sin \theta}{g} \]

### Step 4: Calculate the horizontal distance traveled by the shell
The horizontal distance \( D \) traveled by the shell can be calculated by considering the horizontal velocity of the shell with respect to the ground. The horizontal velocity of the shell with respect to the ground is:
\[ u_x' = u \cos \theta - V_0 = u \cos \theta - \frac{u \cos \theta}{5} = \frac{4u \cos \theta}{5} \]
Now, using the time of flight \( T \):
\[ D = u_x' \cdot T = \left(\frac{4u \cos \theta}{5}\right) \cdot \left(\frac{2u \sin \theta}{g}\right) \]
This simplifies to:
\[ D = \frac{8u^2 \sin \theta \cos \theta}{5g} \]
Using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \), we can rewrite this as:
\[ D = \frac{4u^2 \sin 2\theta}{5g} \]

### Final Answer
The horizontal distance from the initial position of the cannon to the point where the shell strikes the ground is:
\[ D = \frac{4u^2 \sin 2\theta}{5g} \]