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

A person in an elevator accelerating upwards with an acceleration of 2

1.	A person in an elevator accelerating upwards with an acceleration of 2 ms^(2) tosses a coin vertically upwards with a speed of 20ms^(-2) After how much time will the coin fall back into his hand ? (g = 10 ms^(-2)).
Answer» Solution :Here, `U=20 ms^(-1) , v = 0` Effective acceleration , ` a = - (g + a) = - (10 + 2) = - 12 m//s^(2)` If t is time of eacent of the coin, then from `upsilon = u + "at"` `0 = 20 - 12 t, t = (20)/(12) = (5)/(3)s` Time of descent = time of ASCENT ` = (5)/(3)s ` `:.`Total time after which the coin will fall black into HAND ` = (5)/(3) + (5)/(3) = (10)/(3)s = 3.33s` .