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 ms-2)

Here,Mass of the gun, M = 100 kgMass of the ball, m = 1 kgHeight of the cliff, h = 500 mg = 10 ms-2Time taken by the ball to reach the ground ist = 2hg = 2×50010 ms-2 = 10 sHorizontal distance covered = ut400 = u ×10According to law of conservation of linear momentum, we get0 = Mv + muv = -muM = -(1 kg)(40 ms-1)100 kg = -0.4 ms-1-ve sign shows that the direction-ve sign shows that the direction of recoil of