A person in an elevator accelerating upwards with an acceleration of `2 m/s^(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

Acceleration of elevator is =2 `m//s^(2)` in upward direction
Resultant acceleration g = g + `a_(p)`
`=10 + 2`
`=12 m//s^(2)`
Initial velocity of coin is 20 `m//s^(2)`
`t = (20)/(12) = (5)/(3)`
Coin will take equal time to come back total time takes coin to fall back into
Hand `=(5 )/( 3) + (5 )/(3 ) = (10)/(3 ) = 3.33 sec `