A man of mass 50 kg is standing in a gravity-free space at a height of 10 m above the floor. he throws a stone of mass 0.5 kg downwards with a speed 2 m/s. when the stone reaches the floor, the distance of the man above the floor will be?
From the law of conservation of momentum and we obtain the �recoil speed� v of the man from the equation, 50� v = 0.5�2, from which v = 0.02 ms�1.
The time taken by the stone to reach the floor is 10/2 = 5 s. During this time the man will move up through a distance 0.02�5 = 0.1 m so that when the stone reaches the floor, the man will be at a distance 10 + 0.1 = 10.1 m above the floor