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}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{6}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{0.7}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1 s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\sqrt{0.7}$ 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

Suggest Corrections

Similar questions

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

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

Q. An elevator car whose floor to ceiling distance is equal to 2 m starts ascending with constant acceleration 2

m / s^{2}

5 s e c

after the start, a bolt begins falling from ceiling of car. Time after which bolt hits floor of elevator starting from fall.

(g = 10 m / s^{2})

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.

Q. An elevator car whose floor to ceiling distance is equal to 2 m starts ascending with constant acceleration 2

m / s^{2}

5 s e c

after the start, a bolt begins falling from ceiling of car. Time after which bolt hits floor of elevator starting from fall.

(g = 10 m / s^{2})

Join BYJU'S Learning Program

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 √0.7 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 $\sqrt{0.7}$ 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