Question

An elevator in which a man is standing is moving upwards with a constant speed of 10 m/sec. If the man drops a coin a from a height $2.45m$, it reaches the floor of the elevator after a time

(g= $9.8m/s^2$)

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

Answer 1

The correct option is B $\frac{1}{\sqrt{2}}s$

Lift frame :

Initial speed of the coin $u=0m/s$

Acceleration $a=9.8m/s^2$

Initial height of coin from the floor of elevator $h=2.45m$

Time taken by coin to hit the floor $T=\sqrt{\frac{2H}{g}}$

$⟹T=\sqrt{\frac{2\times 2.45}{9.8}}=\frac{1}{\sqrt{2}}s$

Correct answer is option B.