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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.

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.

Grade:12

4 Answers

mohit yadav
54 Points
7 years ago

1.5 sec

Madhukar Thalore
58 Points
7 years ago

The time taken will be approx. 1 second.

FITJEE
43 Points
7 years ago

1.5

SOURAV MISHRA
37 Points
7 years ago

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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