Question

A lift in which a man is standing, is moving upwards with a constant speed of $10{\text{m s}}^{-1}$ . The man drops a coin from a height of $4.9\text{m}$. If $g=9.8{\text{m s}}^{-2}$, the coin reaches the floor of the lift after a time

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

Answer 1

The correct option is B $1\text{s}$

Let us solve the problem in terms of relative initial velocity, relative acceleration and relative displacement of the coin with respect to the floor of the lift.
$u_{rel}=10-10=0{\text{ms}}^{-1},a_{rel}=9.8{\text{ms}}^{-2}$
$S_{rel}=4.9\text{m}$
Using second equation of motion, we get
$S_{rel}=u_{rel}t+\frac{1}{2}{a}_{rel}{t}^2$
$4.9=0\times t+\frac{1}{2}\times 9.8\times t^2$
or $4.9t^2=4.9$ or $t=1\text{s}$