A gun of mass M (including the carriage) fires a shot of mass m. The gun along with the carriage is kept on a smooth horizontal surface. The muzzle speed of the bullet $v_r$ is constant. Find  The elevation of the gun with horizontal at which maximum range of bullet with respect to the ground is obtained.

Muzzle velocity$v_r$is given to be constant.

From conservation of linear momentum in horizontal direction, we have

$$\begin{gathered} Mv=m\left(v_r\cos \theta -v\right) \\ \text{or } v=\frac{m{v}_r\cos \theta }{M+m} \end{gathered}$$

Further, range of bullet on horizontal ground

$R=\frac{2v_r,\sin \theta }{g}\left(v_r\cos \theta -v\right)$

$$\begin{gathered} R=\frac{2v_r \sin \theta }{g}\left(v_r\cos \theta -\frac{m{v}_r\cos \theta }{M+m}\right) \\ =\frac{2M{v}_r^2\sin \theta \cos \theta }{(M+m)g} \\ \text{or }R=\left(\frac{M}{M+m}\right)\frac{v_r^2\sin 2\theta }{g} \end{gathered}$$

 From above Eq.  we see that maximum range is at

$\theta =45°$