In gravity-free space, a man of mass M standing at a height h above the floor throws a ball of mass m straight down with a speed u . When the ball reaches the floor, the distance of the man above the floor will be

Question

In gravity-free space, a man of mass M standing at a height h above the floor throws a ball of mass m straight down with a speed u . When the ball reaches the floor, the distance of the man above the floor will be (A) h(1 + (m/M)) (B) h (2 - (m/M)) (C) 2h (D) a function of m , M , h and u

Tardigrade Admin · Accepted Answer

Due to conservation of momentum in the vertical direction, the man will begin to move up with speed (m u/M) . The ball will reach the floor in time (h/u) . In this time the man will move up by ((m u/M)) ((h/u)) . Total distance of man from floor =h + (m h/M) .

Q. In gravity-free space, a man of mass $M$ standing at a height $h$ above the floor throws a ball of mass $m$ straight down with a speed $u$ . When the ball reaches the floor, the distance of the man above the floor will be

$h (1 + \frac{m}{M})$

$h (2 - \frac{m}{M})$

$2 h$

a function of $m$ , $M$ , $h$ and $u$

Due to conservation of momentum in the vertical direction, the man will begin to move up with speed $\frac{m u}{M}$ . The ball will reach the floor in time $\frac{h}{u}$ . In this time the man will move up by $(\frac{m u}{M}) (\frac{h}{u})$ . Total distance of man from floor $= h + \frac{mh}{M}$ .

Q. In gravity-free space, a man of mass M standing at a height h above the floor throws a ball of mass m straight down with a speed u . When the ball reaches the floor, the distance of the man above the floor will be

h(1+Mm​)

h(2−Mm​)

2h

a function of m , M , h and u