Question

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 1

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` .