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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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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)

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JEE Main Maths, Physics With 12 years exp.

t2<t1
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