Question

A man standing in a lift falling under gravity releases a ball from his hand. As seen by him, the ball:
A. Falls down
B. Remains stationary
C. Goes up
D. Executes SHM

Answer 1

Hint: Concept of relative motion is to be used.
 For lift falling under gravity, the man in lift feels weightless.

Complete step by step answer:
Now, we know that for the downward motion in a lift which is moving with acceleration “a” in downward direction, the apparent weight of man decreases. As the acceleration of lift increases in downward direction.
For a freely falling lift, its accelerations will be g.
The apparent weight, $R = m\left( {g - a} \right)$
$R = m\left( {g - g} \right) = 0$
i.e. the man feels weightless so, for freely falling, the acceleration of the body does not depend on the man.
Hence, the man will have acceleration \[ = \] acceleration due to gravity
Similarly, the ball will have acceleration $ = $ acceleration due to gravity.
So, we can say that both the man and ball will fall with the same acceleration so, both will have the same velocity at every point.
Hence, their relative velocity will be zero. This means that the ball will appear stationary to the man standing in a lift falling under gravity if he releases a ball from his hand.

Additional Information:
Relative velocity of object B w.r.t. to object A is $ = {v_B} - {v_A}$
$ = $ velocity of B – Velocity of A
If both are same as here, ${v_A} = {v_B}$
Then, the relative velocity $ = 0$
If ${v_A} > {v_B}$, then A crosses B
If ${v_B} > {v_A}$, then B crosses A.

Note:
As both are falling freely, the velocity of both will be the same.
Hence the relative velocity will be zero.