Question

An elevator whose floor-to-ceiling distance is $2.50m$, start ascending with a constant acceleration of 1.25 $m{s}^{-2}$. One second after the start, a bolt begins falling from the elevator.
Calculate:
a. Free fall time of the bolt
b. The displacement and distance covered by the bolt during the free fall in the reference frame of ground

Answer 1

a. $t=\sqrt{\frac{2h}{g+a}=}\sqrt{\frac{2\times 2.5}{10+1.25}=}\frac{2}{3}s$
b. Velocity of lift after $1s$
$v=0+1.25\times 1=1.25{ms}^{-1}=5/4{ms}^{-1}$
This will be the initial velocity of bolt
Distance moved up by lift in $2/3s$
$x=\frac{5}{4}\times \frac{2}{3}+\frac{1}{2}1.25{\left(\frac{2}{3}\right)}^2=\frac{10}{9}m$
Displacement of bolt $=2.5-x=2.5-\frac{10}{9}=\frac{25}{18}m$
Maximum height attained by bolt above the point of dropping
$\frac{v^2}{2g}=\frac{{\left(5/4\right)}^2}{2\times 10}=\frac{5}{64}m$
So distance travelled $=2\times \frac{5}{64}+\frac{25}{18}=\frac{445}{288}m$