A coin is dropped in a lift. It takes time t1 to reach the floor when lift is stationary. It takes time t2 when lift is moving up with constant acceleration. Then relate t1 and t2.

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Om Prakash Mishra

JEE Main Maths, Physics With 12 years exp.

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If the height of the coin from which it is dropped is h then in stationery state of the lift we have h =(1/2)g t1^2 Or. t1=(2h/g)^(0.5)…………. (1) In the second case if the acceleration of the lift is a then acceleration of the coin for the observer in the inertial frame (effective...
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If the height of the coin from which it is dropped is h then in stationery state of the lift we have h =(1/2)g t1^2 Or. t1=(2h/g)^(0.5)…………. (1) In the second case if the acceleration of the lift is a then acceleration of the coin for the observer in the inertial frame (effective acceleration ) is =acceleration due to gravity (g) - acceleration of the lift wrt the inertial frame( - a) =(g+a) Here the directions of g and a are opposite to each other and they have directions downward and upward respectively. In this case the height h would be covered with acceleration (g+a).Thus h=(1/2)(g+a)( t2)^2 Or t2=(2h/(g+a))^(0.5)……………. (2) From relations (1) and (2) we have t1/t2=[(g+a)/g]^(0.5) read less
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