Q. A lift in which a man is standing, is moving upward with a speed of 10$ms^{- 1}.$ The man drops a coin from a height of 4.9m and if $g = 9.8\ ms^{- 2},\ $then the coin reaches the floor of the lift after a time
A) $\sqrt{2}$s
B) 1 s
C) $\frac{1}{2\ }\ $s
D) $\frac{1}{\sqrt{3}}\ $s
Explanation
(b) We will solve the problem is terms of relative initial velocity, relative acceleration and relative displacement of the coin with respect to the floor of the lift.
$u = 10 - 10 = 0ms^{- 1},a = 9.8ms^{- 2}$, $s\ $= 4.9m, $t$=?
$4.9 = 0 \times t + \frac{1}{2} \times 9.8 \times t^{2}$
or $4.9t^{2} = 4.9\ or\ \ t = 1$s
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.