question_answer

A coin is released inside a lift at a height of 2 m from the floor of the lift. The height of the lift is 10 m. The lift is moving with an acceleration of \[9\,m/{{s}^{2}}\] downwards. The time after which the coin will strike with the lift is:\[(g=10m/{{s}^{2}})\]

A) 4 s

B) 2 s

C) \[\frac{4}{\sqrt{21}}s\]

D) \[\frac{2}{\sqrt{11}}s\]

Correct Answer: B

Solution :

Relative to lift initial velocity and acceleration of coin are 0 m/s and \[1\,\,m/{{s}^{2}}\]downward

\[\therefore \]      \[2=\frac{1}{2}(1)\,{{t}^{2}}\] or         \[t=2\] second