Question

Q. A man of mass 60 kg is at rest in gravity free space at height 20m above the horizontal surface. If he throws a stone of mass 1kg in downward direction. Find height of man above the surface when stone will collide.​

Answer 1

Answer:

20m

Explanation:

as it is a gravity free space so the ston will not fall.

and

the stone was thrown, but he man remains in the same psition

Answer 2

Now the man is standing at a gravity free space, 20 m above the surface, from where he threw a stone downwards.

As he throws stone downwards, the man tends to move upward as a reaction to his throw according to Newton's Third Law.

Thus the man will move some distance upward as a result of his throwing.

So he has initial velocity for this motion. Let this initial velocity be $\sf{v_m,}$ which acts upward.

Let initial velocity of stone be $\sf{v_s,}$ which acts downward. So both are opposite to each other.

Initially the system of man and stone has no net linear momentum.

By conservation of linear momentum,

$\sf{\longrightarrow 60\,v_m-v_s=0}$

$\sf{\longrightarrow\dfrac{v_m}{v_s}=\dfrac{1}{60}\quad\quad\dots(1)}$

The time taken by the stone to reach the surface will be,

$\sf{\longrightarrow t=\dfrac{20}{v_s}}$

During this time, the distance moved by the man upward will be,

$\sf{\longrightarrow x=v_mt}$

$\sf{\longrightarrow x=\dfrac{20\,v_m}{v_s}}$

From (1),

$\sf{\longrightarrow x=\dfrac{20}{60}}$

$\sf{\longrightarrow x=\dfrac{1}{3}\ m}$

$\sf{\longrightarrow x=0.333\dots\ m}$

Hence, height of man above the surface when stone reaches the ground will be,

$\sf{\longrightarrow H=20+x}$

$\sf{\longrightarrow\underline{\underline{H=20.333\dots\ m}}}$