Question

A man of 50 kg mass is standing in a gravity free space at a height of 10m above the floor. He throws a stone of mass 0.5 kg downwards with a speed of 2ms^-1. When the stone hits the floor the distance of the man above the floor will be

(1) 20 m

(2) 9.9 m

(3) 10.1 m

(4) 10 m

Answer 1

(3) 10.1 m

When the man throws the ball down, he will get a velocity v upwards, such that

mv = MV (Law of conservation of linear momentum)

or 0.2 × 2 = 50 × V or v = 1/50 ms^-1

Time taken by the ball to reach floor is

t = 10/2 = 5 s (The ball will not accelerate in the absence of gravity)

∴ Distance moved up by the man in this t = 5s is

d = V x t = 1/50 x 5 = 0.1 m ; upwards

Hence the distance of the man from the floor, when the stone hits the floor is

H = h + d = 10 + 0.1 = 10.1m