Question

An elevator in which a man is standing is moving upward with a constant speed of 10 m/sec. If the man drops a coin from a height of\[2.45\,m\], it reaches the floor of elevator after a time
A. \[\sqrt 2 \,s\]
B. \[\dfrac{1}{{\sqrt 2 }}s\]
C. \[2\,s\]
D. \[\dfrac{1}{2}s\]

Answer 1

Hint: We have given that the elevator is moving upward with constant velocity. Therefore, the acceleration of the elevator is zero. Determine the relative velocity of the coin when it is dropped from the elevator. Then you can use the kinematic equation to determine the time taken by the coin to reach the ground.

Formula used:
\[s = ut + \dfrac{1}{2}a{t^2}\]
Here, s is the distance, u is the initial velocity, a is the acceleration and t is the time.

Complete step by step answer:
We have given that the elevator is moving upward with constant speed. Therefore, we can see the acceleration of the elevator is zero.
We can see the given question is based on relative velocity of the coin because the coin already has constant initial velocity in the upward direction. At the moment when the coin is dropped from the elevator, the acceleration of the coin is,
\[{\vec a_c} = {\vec a_{cg}} - {\vec a_{ce}}\]
Here, \[{\vec a_{cg}}\] is the acceleration of the coin with respect to ground and \[{\vec a_{ce}}\] is the acceleration of the coin with respect to the elevator.
We know that the elevator is moving with constant velocity, the acceleration of the coin with respect to the elevator is zero.
Therefore, the acceleration of the coin is,
\[{\vec a_c} = {\vec a_{cg}} = - g\]
Here, g is the acceleration due to gravity.
Here, we have assumed that the downward direction to be negative.

When the coin is dropped from the elevator, the velocity of the coin with respect to ground is the same as the velocity of coin with respect to the elevator. Therefore, we can see the relative velocity of the coin
when it is dropped is zero.
Now, we can use kinematic equation to determine the time taken by the coin to reach the ground as follows,
\[s = ut + \dfrac{1}{2}a{t^2}\]
We have to use the above equation in the vertical direction for the coin moving downward. Therefore,
 \[ - s = - \dfrac{1}{2}g{t^2}\]
\[ \Rightarrow t = \sqrt {\dfrac{{2s}}{g}} \]
We have to substitute 2.45 m for s and \[9.8\,m/{s^2}\] for g in the above equation.
 \[t = \sqrt {\dfrac{{2\left( {2.45} \right)}}{{9.8}}} \]
\[ \therefore t = \dfrac{1}{{\sqrt 2 }}s\].

So, the correct answer is “Option B”.

Note:
While solving such questions, students should never forget to assign the direction of movement of the body. For ease of solving the equations correctly, one can assume the upward movement of the body to be positive and negative for the downward movement.
In this question, if the elevator does not have constant velocity, the acceleration of the coin would have been different.