Question

A lift is coming down With a downward acceleration 2m/s^2 .A boy standing in the lift throws a stone vertically upwards with speed 5m / s with respect to himself. The time after which he will catch the stone

Answer 1

Answer:

Sl = Vl t + 1/2 a t^2           distance lift travels in time t

Ss = (Vl - Vs) t + 1/2 g t^2      distance stone travels in time t    (down is +)

Sl = Ss

Vs t = 1/2 (g - a)t^2

t = 2 Vs / (g - a) = 2 * 5 / 7.8 = 1.28 sec

Note: From the viewpoint of the lift the boy experiences  a net

acceleration of g - a.

tu = 5 / 7.8 = .64 sec      time stone travels upwards

and the total time is 2 * .64 = 1.28 sec

One has to be careful in using Newton's Laws in a non-inertial frame!

In this case the boy on the lift can be considered to be in an inertial frame that experiences a gravitational acceleration of (g - a)