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

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Solution

The correct option is C s
$t = \sqrt{\frac{2 h}{(g + a)}} = \sqrt{\frac{2 \times 2.7}{(9.8 + 1.2)}} = \sqrt{\frac{5.4}{11}} = \sqrt{0.49} = 0.7 s e c$
As $u = 0$ and lift is moving upward with acceleration

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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:
(a) the bolt's free fall time;
(b) the displacement and the distance covered by the bolt during the free fall in the reference frame fixed to the ground.

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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

The correct option is C st=√2h(g+a)=√2×2.7(9.8+1.2)=√5.411=√0.49=0.7sec As u=0 and lift is moving upward with acceleration

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 C s
$t = \sqrt{\frac{2 h}{(g + a)}} = \sqrt{\frac{2 \times 2.7}{(9.8 + 1.2)}} = \sqrt{\frac{5.4}{11}} = \sqrt{0.49} = 0.7 s e c$
As $u = 0$ and lift is moving upward with acceleration