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A lift whose floor to ceiling distance is 2.7 m,starts from rest and ascend with a constant acceleration of 1.2 m s^-2.two seconds later,a particle drops from the ceiling of the lift.caalculate the time taken by the partcle to hit the floor of the lift.
1.5 sec
The time taken will be approx. 1 second.
1.5
let us solve the question from the frame of reference of the lift.
the net downward acceleration of the particle realtive to the lift is g + 1.2 m/s2.
then we can write 2.7 = (g + 1.2)t2/2.
we take g = 9.8 m/s2. this gives t = 0.7s.
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