Question

a man standing in a lift falling under gravity releases a ball from hi s hand as seen by him the ball
a)falls up b)remains stationery c)goes up

Answer 1

If he releases the ball and moves his hand away, the ball will appear to float in mid-air.

Unless further acted upon (e.g., by his nudging the ball), the ball will stay there, and, recalling Galileo’s experiment,

it will not rise, because it cannot fall more slowly than him, and

it will not fall, because it cannot fall more quickly than him.

Answer 2

A man standing in a lift falling under gravity releases a ball from his hand. As seen by him, the ball remains stationary. Therefore, the correct option is b) remains stationery.

Explanation:

• As we know that the man is standing in a lift which is falling under gravity, therefore, the man feels weightless in the lift.
• The apparent weight of the man decreases for the downward motion in the lift.
• Suppose the acceleration of the lift is $a$.
• The acceleration of the lift increases in the downward direction in free falling.
• For free-falling, the acceleration will be $g$.
• Therefore, the apparent weight of the man is given as:

        $\[\begin{gathered} R = m(g - a) \hfill \\ R = m(g - g) = 0 \hfill \\ \end{gathered} \]$

• Thus, the acceleration of the man will be equal to the acceleration due to gravity.
• The acceleration of the ball will also be equal to the acceleration due to gravity.
• Hence, both the man and the ball will have the same acceleration and thus, the same velocity.
• This means that their relative velocity is zero.
• Hence, the ball will appear stationary to the man.

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