A gun of mass `M`. fires a shell of mass `m` horizontally and the energy of explosion is such as would be sufficient to project the shell vertically to a height `'h'` . The recoil velocity of the gun is

Options

A. `((2m^(2)gh)/(M(m+M)))^(1/2)`
B. `((2m^(2)gh)/(M(m-M)))^(1/2)`
C. `((2m^(2)gh)/(2M(m-M)))^(1/2)`
D. `((2m^(2)gh)/(2M(m+M)))^(1/2)`

Explanation

To solve the problem, we need to find the recoil velocity of the gun when it fires a shell vertically to a height \( h \). We can use the principle of conservation of momentum and the relationship between potential energy and kinetic energy.

### Step-by-Step Solution:

1. **Calculate the Potential Energy of the Shell:**
The potential energy (PE) gained by the shell when it reaches a height \( h \) is given by:
\[ PE = mgh \]
where \( m \) is the mass of the shell, \( g \) is the acceleration due to gravity, and \( h \) is the height.

2. **Relate Potential Energy to Kinetic Energy:**
At the moment of firing, all the potential energy gained by the shell is equal to the kinetic energy (KE) imparted to it. The kinetic energy of the shell when it is fired is given by:
\[ KE = \frac{1}{2} mv^2 \]
where \( v \) is the velocity of the shell just after it is fired. Setting the potential energy equal to the kinetic energy gives:
\[ mgh = \frac{1}{2} mv^2 \]

3. **Solve for the Velocity of the Shell:**
We can simplify the equation by canceling \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gh = \frac{1}{2} v^2 \]
Rearranging gives:
\[ v^2 = 2gh \]
Taking the square root:
\[ v = \sqrt{2gh} \]

4. **Apply Conservation of Momentum:**
Before firing, the total momentum of the system (gun + shell) is zero. After firing, the momentum of the shell and the gun must still sum to zero. Let \( V \) be the recoil velocity of the gun. The conservation of momentum gives us:
\[ mv + MV = 0 \]
Rearranging gives:
\[ MV = -mv \]
Therefore, the recoil velocity of the gun is:
\[ V = -\frac{mv}{M} \]

5. **Substitute the Velocity of the Shell:**
Now, substituting \( v = \sqrt{2gh} \) into the equation for \( V \):
\[ V = -\frac{m \sqrt{2gh}}{M} \]

6. **Final Expression for Recoil Velocity:**
Thus, the recoil velocity of the gun is:
\[ V = -\frac{m \sqrt{2gh}}{M} \]
The negative sign indicates that the gun recoils in the opposite direction to the motion of the shell.

### Summary:
The recoil velocity of the gun when it fires a shell vertically to a height \( h \) is:
\[ V = -\frac{m \sqrt{2gh}}{M} \]