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

Question

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 (A) √0.54 s (B) √6 s (C) 0.7 s (D) 1 s

Tardigrade Admin · Accepted Answer

As u =0 and lift is moving upward with acceleration a, the time taken is given by: t =√(2 h /( g + a ))=√(2 × 2.7/(9.8+1.2))=√(5.4/11)=√0.49=0.7 sec

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

$0.54 s$

$6 s$

$0.7 s$

$1 s$

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

Q. An elevator car, whose floor to ceiling distance is equal to 2.7m, starts ascending with constant acceleration of 1.2ms2.2sec after the start, a bolt begins fallings from the ceiling of the car. The free fall time of the bolt is

0.54​s

6​s

0.7s

1s