Question

An elevator, in which a man is standing, is moving upward with a constant acceleration of $2m/s^2$ At some instant when speed of elevator is $10m/s$, the man drops a coin from a height of $1.5m$ Find the time taken by the coin to reach the floor.

• A. $\frac{1}{2}\text{sec}$
• B. $\frac{1}{\sqrt{2}}\text{sec}$
• C. $\frac{1}{\sqrt{3}}\text{sec}$
• D. $1\text{sec}$

Answer 1

The correct option is A $\frac{1}{2}\text{sec}$

Step 1: Find the acceleration of coin with respect to elevator $\left({\overrightarrow{a}}_{ce}\right)$
Given:
Acceleration of elevator $\left({\overrightarrow{a}}_{eg}\right)=2m/s^2$
Acceleration of the coin
${\overrightarrow{a}}_{cg}(=g)=10m/s^2$
${\overrightarrow{a}}_{ce}={\overrightarrow{a}}_{cg}+{\overrightarrow{a}}_{eg}$
${\overrightarrow{a}}_{ce}=(-10)-2$
${\overrightarrow{a}}_{ce}=-12m/s^2$

Step 2:Find the velocity of coin with respect to elevator $\left({\overrightarrow{V}}_{ce}\right)$
As we know,
$\text{Velocity of elevator}=\text{Velocity of coin}=10m/s$
${\overrightarrow{V}}_{ce}={\overrightarrow{V}}_{cg}-{\overrightarrow{V}}_{eg}=10-10$
${\overrightarrow{V}}_{ce}=0m/s$

Step 3: Find the time taken by the coin to reach the floor.
As given, $d=-1.5m$
Initial velocity of coin $u=0m/s$
Therefore, from $2^{nd}$ equation of motion
$d_{rel}=ut+\frac{1}{2}{a}_{rel}{t}^2$
Hence, $d_{rel}=\frac{1}{2}{a}_{rel}{t}^2$
By putting the given values, we have,
$-1.5=\frac{1}{2}\times (-12)\times t^2$
$1.5=\frac{1}{2}\times 12\times t^2$
$t=\frac{1}{2}\text{sec}$
Final Answer: (c)