A gun of mass M fires a bullet of mass m horizontally. The energy of firing is such that it is sufficient to project the bullet vertically to a height h. The velocity of recoil of the gun is

Let the recoil velocity of the gun be  V

Applying conservation of momentum:          0 = mv − MV

mv = MV        ⟹ v = $\frac{\mathrm{MV}}{\mathrm{m}}$                   ............. (1)

Explosion energy must be equal to the total kinetic energy of the system after the explosion

∴   $\mathrm{E} = \frac{1}{2}{\mathrm{mv}}^2 + \frac{1}{2}{\mathrm{MV}}^2$                

Using (1),        $\mathrm{E} = \frac{1}{2}\mathrm{m}{\left(\frac{\mathrm{MV}}{\mathrm{m}}\right)}^2 + \frac{1}{2}{\mathrm{MV}}^2$

⟹$\mathrm{E} = \frac{\mathrm{M}(\mathrm{m}+\mathrm{M})}{2\mathrm{m}}{\mathrm{V}}^2$                     ...........(2)

Also, the explosion energy is sufficient to project the shell to vertical height h

∴    E = mgh                      ..............(3)

From (2) and (3), we get          mgh = $\frac{\mathrm{M}(\mathrm{m}+\mathrm{M})}{2\mathrm{m}}{\mathrm{V}}^2$    

⟹V=$\sqrt{\frac{2{\mathrm{m}}^2\mathrm{gh}}{\mathrm{M}(\mathrm{m}+\mathrm{M})}}$​