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) `= 2 m//s^(2)`
Acceleration of the elevator (a) `= 2 m//s^(2)`
Acceleration due to gravity (g) `= 10 m//s^(2)`
`:.` 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 coin is t, then v = u + at 0 = 20 + - 12 `xx` t
or `t = (20)/(12) = (5)/(3) s`
Time of ascent = Time of descent
`:.` Total time after which the coin fall back into hand `= ((5)/(3) + (5)/(3)) s = (10)/(3) s = 3.33 s`
Note: While calculating net acceleration we should be aware that if lift is going upward net acceleration is (g + a) and for downward net acceleration is (g -a).