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 = 10 ${\mathrm{ms}}^{-2}$)

Here,

Mass of the gun, M = 100 kg

Mass of the ball, m = 1 kg

Height of the cliff, h = 500 m

g = 10 ${\mathrm{ms}}^{-2}$

Time taken by the ball to reach the ground is

$\mathrm{t}=\sqrt{\frac{2\mathrm{h}}{\mathrm{g}}} = \sqrt{\frac{2\times 500}{10}} = 10 \mathrm{s}$

Horizontal distance covered = ut

400 = u $\times$10 $\Rightarrow u=40 m/s$

According to law of conservation of linear momentum, we get

0 = Mv + mu

$\mathrm{v} = -\frac{\mathrm{mu}}{\mathrm{M}} = -\frac{(1 \mathrm{kg})(40 {\mathrm{ms}}^{-1})}{100 \mathrm{kg}} = -0.4 {\mathrm{ms}}^{-1}$

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