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(D) The time taken for an object to fall a distance $h$ under constant acceleration $a$ is given by $t = \sqrt{\frac{2h}{a}}$.When the elevator is at

Vedclass · Accepted Answer

$t_{1} = t_{2}$

Vedclass · Answer

(D) The time taken for an object to fall a distance $h$ under constant acceleration $a$ is given by $t = \sqrt{\frac{2h}{a}}$.When the elevator is at rest,its acceleration is $0$,so the effective acceleration of the coin relative to the floor is $g$.When the elevator is moving uniformly (constant velocity),its acceleration is also $0$. Therefore,the effective acceleration of the coin relative to the floor remains $g$.Since the effective acceleration is the same in both cases $(a_{real} = g)$,the time taken for the coin to reach the floor is the same.Thus,$t_{1} = t_{2}$.

$A$ person standing on the floor of an elevator drops a coin. The coin reaches the floor in time $t_{1}$ if the elevator is at rest and in time $t_{2}$ if the elevator is moving uniformly. Then

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