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  4. An elevator car, whose floor to ceiling distance is equal to 2.7 m, starts ascending with constant acceleration of 1.2 ms 2 .2 sec after the start, a bolt begins fallings from the ceiling of the car. The free fall time of the bolt is

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Q. An elevator car, whose floor to ceiling distance is equal to $2.7 m$, starts ascending with constant acceleration of $1.2 ms ^{2} .2 sec$ after the start, a bolt begins fallings from the ceiling of the car. The free fall time of the bolt is

Motion in a Straight Line

Solution:

As $u =0$ and lift is moving upward with acceleration $a$, the time taken is given by:
$$ t =\sqrt{\frac{2 h }{( g + a )}}=\sqrt{\frac{2 \times 2.7}{(9.8+1.2)}}=\sqrt{\frac{5.4}{11}}=\sqrt{0.49}=0.7 sec $$