A 100 kg gun fires a ball of 1 kg horizontally from a cliff of height 500m. 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/$s^2$ )

0.2 m/s 0.4 m/s 0.6 m/s 0.8 m/s

0.4 m/s

Given  Mass of the gun, M = 100 kg Mass of the ball, m= 1 kg Height of the chiff, h= 500 m Time taken by the ball to reach the ground is  $t=\$\backslash sqrt\{\backslash frac\{2h\}\{g\}\}\$\; =\; \$\backslash sqrt\{\backslash frac\{2*500\}\{10\}\}\$\; =\; 10\; sec$ $Horizontal\; velocity\; of\; the\; bullet\; u\; =\; \$\backslash frac\{400\}\{10\}\$\; =\; 40\; m/s$   According to law of conservation of linear momentum, we get, 0=Mv+mu ⇒ v=-mu/M=-40/100 =−0.4m/s -ve sign shows that the direction of recoil of the gun is opposite to that of the ball.