Question

An elevator, in which a man is standing, is moving upwards with a speed of $10m/s.$ If the man drops a coin from a height of $2.45m$ from the floor of the elevator, it reaches the floor of the elevator after time $(Takeg=9.8m/s^2)$

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

Answer 1

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

The elevator moves with a uniform speed of 10 m/s.

When the elevator is considered as frame of reference:

Initial velocity of the coin, u = 0 m/s

Height, h = 2.45 m

Using second equation of motion,$s=ut+\frac{1}{2}g{t}^2\Rightarrow h=0\times t+\frac{1}{2}g{t}^2\Rightarrow 2.45=\frac{1}{2}\times 98t^2\Rightarrow t^2=\frac{4.9}{9.8}\Rightarrow t^2=\frac{1}{2}\Rightarrow t=\frac{1}{\sqrt{2}}s$

Hence, the correct answer is option (b).