A $100 kg$ gun fires a ball of $1 kg$ horizontally from a cliff of height $500 m$. It falls on the ground at a distance of $400 m$ from the bottom of the cliff. The recoil velocity of the gun is (Take $g=10m{s}^{-2}$)

Here, Mass of the gun, $M=100 kg$ Mass of the ball, $m=1kg$ Height of the cliff, $h=500m$

$g=10 m{s}^{-2}$. Time taken by the ball to reach the ground is $t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 500m}{10 m{s}^{-2}}}=10s$

Horizontal distance covered $=ut \Rightarrow 400=u\times 10$ where $u$ is the velocity of the ball $u=40 m{s}^{-1}$ According to law of conservation of linear momenturn, we get, $0=Mv+mu$

$v=-\frac{mu}{M}=-\frac{\left(1kg\right)\left(40 m{s}^{-1}\right)}{100kg}=-0.4 m{s}^{-1}$

$-$ve sign shows that the direction of recoil of the gun is opposite to that of the ball.