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a coin is released inside a lift at a height of 2m from the floor of the lift. The height of the ligt is 10m. The lift is moving with an acceleration 11m/s. Downwards. The time after which the coin will strike with the lift is.

Ankit Kumar
18 Points
7 years ago

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)

Chaitanya
20 Points
3 years ago
The above ans is correct except for the fact that using the formula of relative motion x`(a with respect to b )= 8-10=-2 X=x`+0+1/2*t^2 0=-2 +1/2 t^2t^2=4T=4 sec