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$\left( 1 + \frac { \mathrm { m } } { \mathrm { M } } \right)$
• B. h $\left( 2 - \frac { m } { M } \right)$
• C. 2h
• D. A function of m, M, h and u

Answer 1

Answer: (A)

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

Answer 2

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