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(C) Using the third equation of motion,$v^{2} = u^{2} - 2as$. At the highest point,the final velocity $v = 0$.Therefore,$0 = u^{2} - 2as$,which gives

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$1: \sqrt{3}$

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(C) Using the third equation of motion,$v^{2} = u^{2} - 2as$. At the highest point,the final velocity $v = 0$.Therefore,$0 = u^{2} - 2as$,which gives $s = \frac{u^{2}}{2a}$.For an object on a smooth inclined plane,the acceleration $a = g \sin 	heta$.Thus,the distance traveled is $s = \frac{u^{2}}{2g \sin 	heta}$.Since $u$ and $g$ are constant,$s \propto \frac{1}{\sin 	heta}$.Therefore,$\frac{x_{1}}{x_{2}} = \frac{\sin 	heta_{2}}{\sin 	heta_{1}} = \frac{\sin 30^{\circ}}{\sin 60^{\circ}}$.Substituting the values,$\frac{x_{1}}{x_{2}} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}}$.Hence,the ratio $x_{1}: x_{2}$ is $1: \sqrt{3}$.

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