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

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$\frac{1}{\sqrt{3}} s$

$\frac{1}{2} s$

$\frac{1}{\sqrt{2}} s$

1 s

answer is B.

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Detailed Solution

$d_{rel} = \frac{1}{2} a_{rel} t^{2} \Rightarrow 1.5 = \frac{1}{2} (10 + 2) t^{2}$

$1.5 = \frac{1}{2} \times 12 \times t^{2} \Rightarrow t = \frac{1}{2} s$

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