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question mark

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.

Keerthana Mohan , 12 Years ago
Grade 12
anser 4 Answers
mohit yadav

1.5 sec

Last Activity: 12 Years ago
Madhukar Thalore

The time taken will be approx. 1 second.

Last Activity: 12 Years ago
FITJEE

1.5

Last Activity: 12 Years ago
SOURAV MISHRA

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.

Last Activity: 12 Years ago
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