A person in an elevator accelerating upwards with an acceleration of
2 ms-2, tosses a coin vertically upwards with a speed of 20 m/s. After
how much time will the coin fall back into his hand? (g = 10 ms-2)


Hint: Apply the concept of relative motion.
Step 1: Find geff.

Here, acceleration of the elevator (a) = 2 m/s2                      (upwards)
Acceleration due to gravity (g) = 10 m/s2
Effective acceleration a' = g + a = 10+ 2 = 12 m/s2     (here, acceleration is w.r.t. the lift)

Step 2: Find the time after which the coin falls back into the hand.

Initial speed of the coin (u) = 20 m/s

     v=u+ator  0=20+(12)×t       t=2012=53s

Time of ascent = Tirne of desent

 Total time after which the coin falls back into the hand =(53+53)s=103s=3.33s