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Asked by Prabhash Last Modified
Answer 
Achal Bhavsar
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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