A elevator car whose floor to distance is 2.7 m starts ascending with a constant acceleration of 1.2 m / s 2, 2 s after a bolt is begin to fall from the ceiling of the car. The free fall time of the bolt is (g=9.8 m / s e)

Question

A elevator car whose floor to distance is 2.7 m starts ascending with a constant acceleration of 1.2 m / s 2, 2 s after a bolt is begin to fall from the ceiling of the car. The free fall time of the bolt is (g=9.8 m / s e) (A)  √(2.7/9.8)s  (B)  √(5.4/9.8)s  (C)  √(5.4/8.6)s  (D)  √(5.4/11)s

Tardigrade Admin · Accepted Answer

Net acceleration on the elevator car is more than acceleration due to gravity.
Since elevator car is ascending upwards, from Newtons second
law, the net force is acting upwards hence resultant acceleration is

a_{r} = (g + a)

= (9.8 + 1.2) = 11 m / s^{2}

Relative velocity of observer to elevator is

u_{r} = 0

from the equation

s = u_{r} t + \frac{1}{2} a, t^{2}

2.7 = 0 + \frac{1}{2} \times 11 \times t^{2}

t^{2} = \frac{5.4}{11}

t = \frac{5.4}{11} s

Q. A elevator car whose floor to distance is $2.7 m$ starts ascending with a constant acceleration of $1.2 m / s^{2}, 2$ s after a bolt is begin to fall from the ceiling of the car. The free fall time of the bolt is $(g = 9.8 m / s^{e})$

$\frac{2.7}{9.8} s$

$\frac{5.4}{9.8} s$

$\frac{5.4}{8.6} s$

$\frac{5.4}{11} s$

Q. A elevator car whose floor to distance is 2.7m starts ascending with a constant acceleration of 1.2m/s2,2 s after a bolt is begin to fall from the ceiling of the car. The free fall time of the bolt is (g=9.8m/se)

9.82.7​​s

9.85.4​​s

8.65.4​​s

115.4​​s

Q. A elevator car whose floor to distance is $2.7 m$ starts ascending with a constant acceleration of $1.2 m / s^{2}, 2$ s after a bolt is begin to fall from the ceiling of the car. The free fall time of the bolt is $(g = 9.8 m / s^{e})$

$\frac{2.7}{9.8} s$

$\frac{5.4}{9.8} s$

$\frac{5.4}{8.6} s$

$\frac{5.4}{11} s$