A girl is standing in an elevator that is moving upward at a velocity of $5\mathrm{m}/\mathrm{s}$ and acceleration $2\mathrm{m}/{\mathrm{s}}^2$, when she drops her handbag. If she was originally holding the bag at a height of $1.5\mathrm{m}$ above the elevator floor, how long will it take the bag to hit the floor.

Solution In such problems, generally called elevator problems we can use the concept of relative acceleration. Here we solve the problem with respect to elevator or we assume that we are observing from inside of elevator. The concept of relative motion is discussed in detail in section $1.11$ of this chapter. That much of detail is not required here. Imagine the situation, if you are standing in an elevator accelerating up with an acceleration $a$. If you are holding a box in your hand, it is also accelerating up with the same acceleration. If you release it free, it falls with acceleration $g$ towards the elevator floor, which is coming up with acceleration $a$. Here we can say that the approach acceleration of box towards the elevator floor is $(g+a)$, and we assume that elevator floor is at rest and box is going down with respect to floor with this acceleration called relative acceleration. Similarly if elevator is going down, we take relative acceleration $(g-a)$. In this problem, elevator is going up with a velocity $5\mathrm{m}/\mathrm{s}$, and an acceleration $2\mathrm{m}/{\mathrm{s}}^2$, when the bag is dropped. Bag's relative velocity with respect to elevator floor is zero as both at that instant have the same velocity but relative acceleration of the bag is taken as $10+2=12\mathrm{m}/{\mathrm{s}}^2$. The distance fallen by the bag is $1.5\mathrm{m}$, as we are assuming elevator is at rest. Thus time required is $=\sqrt{\frac{2H}{a}}=\sqrt{\frac{2\times 1.5}{12}}=0.5\mathrm{s}$