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)`)

Options

A. `(5)/(3)s`
B. `(3)/(10)s`
C. `(10)/(3)s`
D. `(3)/(5)s`

Answer

C

Explanation

Here, `v=20ms^(-1), a=2ms^(-2), g=10ms^(-2)`
The coin will fall back into the person's hand after t s. `therefore t=(2v)/(a+g)=(2xx20ms^(-1))/((2+10)ms^(-2))=(40)/(12)s=(10)/(3)s`