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(D) The maximum possible acceleration of the person on the conveyor belt is given by the net force along the incline divided by mass:$a_{\max} = \frac

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$\sqrt{\frac{6 h}{g}}$

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(D) The maximum possible acceleration of the person on the conveyor belt is given by the net force along the incline divided by mass:$a_{\max} = \frac{\mu m g \cos 	heta - m g \sin 	heta}{m} = g(\mu \cos 	heta - \sin 	heta)$Given $	heta = 30^{\circ}$,$\mu = \frac{5}{3 \sqrt{3}}$,and $\sin 30^{\circ} = 0.5$,$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$:$a_{\max} = g \left( \frac{5}{3 \sqrt{3}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{2} ight) = g \left( \frac{5}{6} - \frac{3}{6} ight) = \frac{g}{3}$The length of the conveyor belt $L$ is related to height $h$ by $L = \frac{h}{\sin 30^{\circ}} = 2h$.The initial velocity $u$ of the person relative to the ground is the speed of the belt,$u = \sqrt{\frac{g h}{6}}$.Using the equation of motion $S = ut + \frac{1}{2} a t^2$:$2h = \left( \sqrt{\frac{g h}{6}} ight) t + \frac{1}{2} \left( \frac{g}{3} ight) t^2$$2h = \frac{t \sqrt{gh}}{\sqrt{6}} + \frac{gt^2}{6}$Multiplying by $6$:$12h = t \sqrt{6gh} + gt^2$$gt^2 + t \sqrt{6gh} - 12h = 0$Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$t = \frac{-\sqrt{6gh} + \sqrt{6gh - 4(g)(-12h)}}{2g} = \frac{-\sqrt{6gh} + \sqrt{6gh + 48gh}}{2g} = \frac{-\sqrt{6gh} + \sqrt{54gh}}{2g} = \frac{-\sqrt{6gh} + 3\sqrt{6gh}}{2g} = \frac{2\sqrt{6gh}}{2g} = \sqrt{\frac{6h}{g}}$

$A$ person climbs up a conveyor belt with a constant acceleration. The speed of the belt is $\sqrt{\frac{g h}{6}}$ and the coefficient of friction is $\frac{5}{3 \sqrt{3}}$. The time taken by the person to reach from $A$ to $B$ with the maximum possible acceleration is

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