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 \[9\text{ }m/{{s}^{2}}\] down wards. The time after which the coin will strike with the lift is: \[(g=10\text{ }m/{{s}^{2}}):\]

A) 4 s

B) 2 s

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

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

Correct Answer: B

Solution :

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

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