An elevator car whose floor to ceiling distance is equal to $2.7\mathrm{m}$ starts ascending with constant acceleration $1.2\mathrm{m}/{\mathrm{s}}^2$. Two seconds after the start a bolt begins falling from the ceiling of the car. Find :
(a) the time after which bolt hits the floor of the elevator.
(b) the net displacement and distance traveled by the bolt, with respect to earth. (Take $g=9.8\mathrm{m}/{\mathrm{s}}^2$ )

(a) If we consider elevator at rest, then relative acceleration of the bolt is
$a_r=9.8+1.2=11\mathrm{m}/{\mathrm{s}}^2$
Initial velocity of the bolt is $2.4\mathrm{m}/\mathrm{s}$ and it is getting retarded with $11\mathrm{m}/{\mathrm{s}}^2$. With respect to elevator initial velocity of bolt is 0 and it has to travel $2.7\mathrm{m}$ with $11\mathrm{m}/{\mathrm{s}}^2$. Thus time taken can be directly given as
$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2\times 2.7}{11}}=0.7\text{ seconds }$
(b) Velocity of the elevator after $2\sec$ is $v=at=1.2\times 2=2.4\mathrm{m}/\mathrm{s}$ upwards.
In $0.7$ seconds distance traveled by the elevator is
$\begin{array}{rl}l& =2.4\times 0.7+0.5\times 1.2\times 0.49\\ & =1.98\mathrm{m}\end{array}$
The final displacement of the bolt is $2.7-1.98=0.72\mathrm{m}$.
We can observe the actual motion of the bolt, it is shown in figure-1.8. It first goes up and then gets back down and hit the floor of the elevator at a distance $0.72\mathrm{m}$ below the starting point. To find the distance we find the distance it travels before coming to rest at the topmost point of its trajectory. At the time of detachment of the bolt, its velocity is $2.4\mathrm{m}/\mathrm{s}$.
So its distance up to the top $=\frac{u^2}{2g}=\frac{(2.4{)}^2}{19.6}=0.293\mathrm{m}$
Total distance traveled by the bolt is $=0.293\times 2+0.72=1.31\mathrm{m}$