Q33 of 42 Page 34

A person in an elevator accelerating upwards with an acceleration of 2 m s-2, tosses a coin vertically upwards with a speed of 20 m s1. After how much time will the coin fall back into his hand? (g = 10 m s-2)

Here, the initial velocity of the coin is, u = 20 m s1

And the acceleration of the elevator is, a = 2 m s-2


Here, let’s think this problem from the elevator reference frame. For the person inside the elevator going up with acceleration 2 m s-2 will experience a net acceleration of (g+a) which is 12 m s-2.


And as the coin return back to its original position (because we are in the elevator reference frame), the net displacement (s) is zero.


So we can use the laws of motion to solve the problem.



Here, a’=a+g = 12 m s-2




So the time of flight of the coin is 3.33 seconds.


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