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

\sqrt{0.54} s

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\sqrt{6} s

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0.7 s

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1 s

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Solution

The correct option is B $0.7 s$
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

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An elevator car whose floor-to-ceiling distance is equal to 2.7 m starts ascending with constant acceleration 1.2 $m / s^{2}$ . 2s after the start, a bolt begins falling from the ceiling of the car. Find:
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An elevator car, whose floor to ceiling distance is equal to 2.7m, starts ascending with constant acceleration of 1.2ms2 . 2 sec after the start, a bolt begins fallings from the ceiling of the car. The free fall time of the bolt is

The correct option is B 0.7sAs u=0 and lift is moving upward with acceleration a ,the time taken is given by : t=√2h(g+a)=√2×2.7(9.8+1.2)=√5.411=√0.49=0.7 sec

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

The correct option is B $0.7 s$
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