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My first point is that you consider the question again because the acceleration of the lift is more than gravitational acceleration and moreover the unit is incorrect. If it is really having acceleration greater than g then the coin will hit the upper part of the lift.
a(rel)=accln. of lift - accln. of coin
a(rel)=11-10=1 m/s(2)
now distance of coin from upper roof= 8m
using s=ut+(1/2)at(square)
so u=0 as both are in the same velocity initially.
8=(1/2)*1*t(square)
t(square)=16
t=4.(ans)
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