As given that, mass of the gun $\left(m_1\right)=100 \mathrm{~kg}$ Mass of the ball $\left(m_b\right)=1 \mathrm{~kg}$
Height of the cliff $(s)=500 \mathrm{~m}, g=10 \mathrm{~ms}^{-2}$
Horizontal distance travelled by the ball $(x)$ $=400 \mathrm{~m}$


$$
\begin{aligned}
&s=u t+\frac{1}{2} a t^2 \\
&(\because u=0, a=g)
\end{aligned}
$$
From $s=\frac{1}{2} g t^2$
( $\because$ Initial velocity in downward direction is zero)
$$
500=\frac{1}{2} \times 10 t^2 \Rightarrow t=\sqrt{100}=10 \mathrm{sec}
$$
$$
x=v_b t \Rightarrow v_b=\frac{x}{t}=\frac{400}{10}=40 \mathrm{~m} / \mathrm{sec}
$$
If $v$ is recoil velocity of gun, then by the law conservation of linear momentum
$$
\begin{aligned}
&m_b u_b+M_G u_g=m_b v_b+M_G v_G \\
&m_b \times 0+M_G \times 0=1 \times 40+100 v_G \\
&100 v_G=-40 \Rightarrow v_G=\frac{-40}{100}=\frac{1}{5} \mathrm{~ms}^{-1}
\end{aligned}
$$
Opposite to the speed of ball.