Physics - NLM Question with Solution | TestHub

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