A very broad elevator is going up vertically with a constant acceleration of $2 m/s^2$. At the instant when its velocity is $4m/s$ a ball is projeted from the floor of the lift with a speed of $4 m/s$ relative to the floor at an elevation of $30 ^{\circ}$. The time taken by the ball to return to the floor is $(g=10 m/s^2)$ - Clay6.com, a Free resource for your JEE, AIPMT and Board Exam preparation

A very broad elevator is going up vertically with a constant acceleration of $2 m/s^2$. At the instant when its velocity is $4m/s$ a ball is projeted from the floor of the lift with a speed of $4 m/s$ relative to the floor at an elevation of $30 ^{\circ}$. The time taken by the ball to return to the floor is $(g=10 m/s^2)$

\[(a)\;\frac{1}{2} s \quad (b)\;\frac{1}{3} s \quad (c)\;\frac{1}{4}s \quad (d)\;1 s\]