In the frame of elevator,
$a=$ acceleration of the particle with respect to the elevator
$\begin{aligned} \therefore \quad m \sin 30(g+2) & =m a \\ a & =(g+2) \sin 30^{\circ} \\ & =(10+2) \cdot \frac{1}{2} \\ & =6 \mathrm{~m} / \mathrm{s}^2\end{aligned}$
$\therefore$ The distance travelled by the particle from the top to the bottom,
$\begin{aligned} d & =\frac{4}{\cos 30^{\circ}}=\frac{4}{\sqrt{3} / 2}=\frac{8 \sqrt{3}}{3} \mathrm{~m} \\ \text { Using } s & =u t+\frac{1}{2} a t^2,\end{aligned}$
$\begin{aligned} & & \frac{8 \sqrt{3}}{3} & =0 \times t+\frac{1}{2} 6 t^2 \\ \Rightarrow & & t^2 & =\frac{16 \sqrt{3}}{3 \times 6} \\ \therefore & & t & =\frac{4}{3} \sqrt{\frac{\sqrt{3}}{2}} \mathrm{~s}\end{aligned}$