A very broad elevator is going up vertically with a constant acceleration of `2m//s^(2)`. At the instant when its velocity is `4m//s` a ball is projected form the floor of the lift with a speed of `4m//s` relative to the floor at an elevation of `30^(@)`. Time taken by the ball to return the floor is `(g=10ms^(2))`

Options

A. `(1)/(2)s`
B. `(1)/(3)s`
C. `(1)/(4)s`
D. 1s

Answer

B

Explanation

Ley us see the motion relative to elevator,
`a_(r) = a_(b) = (-10) -(+2) =- 12 ms^(-2)`
Now, `T = (2u_(y))/(a_(r)) = (2 xx u sin theta)/(a_(r)) = (2 xx 4 xx sin 30^(@))/(12) = (1)/(3)s`