A person in an elevator accelerating upwards with an acceleration of ` 2ms^(-2)` , tosses a coin vertically upwards with a speed of `20 ms^(-1)` . After how much time will the coin fall back into his hand ? (g = 10 `ms^(-2)`)

Explanation

Here , initial speed of the coin (u) = 20 m/s
Acceleration of the elevator (a) = 2 m/ `s^(2) " " ` (upwards)
Acceleration due to gravity (g) = 10 `m//s^(2)`
`therefore` Effective acceleration a' = g + a = 10 + 2 = 12 `m//s^(2)` (here , acceleration is w.r.t. the lift)
If the time of ascent of the coins is t , then
`v = u + at `
`0 = 20 + (-12) xxt `
or `" " t = (20)/(12) = (5)/(3) s `
Time of ascent = Time of desent
`therefore` Total time after which the coin fall back into hand = `((5)/(3) + (5)/(3))s = (10)/(3) s = 3.33s `