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When an object is shot from the bottom of a long smooth inclined plane kept at an angle $60^{\circ}$ with horizontal, it can travel a distance $x_{1}$ along the plane. But when the inclination is decreased to $30^{\circ}$ and the same object is shot with the same velocity, it can travel $x_{2}$ distance. Then $x_{1}$: $x_{2}$ will be:
Explanation
When object shot from inclined plane kept at $60^{\circ}$ and velocity say u, then distance travel $x_{1} = u^{2}/(2g \sin 60^{\circ})$ [add image] When object shot from inclined plane kept at $30^{\circ}$ and same velocity u, then distance travel $x_{2} = u^{2}/(2g \sin 30^{\circ})$ $\therefore x_{1}/x_{2} = (u^{2}/(2g \sin 60^{\circ}))/(u^{2}/(2g \sin 30^{\circ})) = \sin 30^{\circ}/\sin 60^{\circ} = (1 \times 2)/(2 \times \sqrt{3}) = 1:\sqrt{3}$ [add image]