Question

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

• A. $\sqrt{0.54}s$
• B. $\sqrt{6}s$
• C. $0.7s$
• D. $1s$

Answer 1

The correct option is C

$0.7s$

Step 1. Given data:

The distance between the elevator floor and ceiling, $S=2.7m$

Acceleration of the elevator, $a=1.2m/s^2$

Acceleration of the bolt relative to the elevator $(a\text{'})=(g+a)$

$=(9.8+1.2)$

$=11m/s^2$

Relative to the elevator the initial velocity of the bolt $\left(u\right)=0$

Step 2. Finding free fall of the bolt

From the equation of motion,

$S=ut+\left(\frac{1}{2}\right)a\text{'}{t}^2$

Putting all values in the given equation,

$\begin{array}{rcl}2.7& =& 0+\left(\frac{1}{2}\right)x11x{t}^2\\ & \Rightarrow & t^2=\frac{2.7\times 2}{11}\\ & \Rightarrow & t^2=\frac{5.4}{11}\\ & \Rightarrow & t=\sqrt{\frac{5.4}{11}}\\ \therefore t& =& 0.7sec\end{array}$

The free fall time of the bolt, $\begin{array}{rcl}t& =& 0.7sec\end{array}$

Hence, correct option is (C).