A particle is released from a height of S above the surface of the earth. At a certain height, its kinetic energy is three times its potential energy. The distance from the earth's surface and the speed of the particle at that instant are respectively: 

1. $\frac{\mathrm{S}}{2}, \frac{\sqrt{3\mathrm{gS}}}{2}$

2. $\frac{\mathrm{S}}{4}, \sqrt{\frac{3\mathrm{gS}}{2}}$

3. $\frac{\mathrm{S}}{4}, \frac{3\mathrm{gS}}{2}$

4. $\frac{\mathrm{S}}{4}, \frac{\sqrt{3\mathrm{gS}}}{2}$

When an object is shot from the bottom of a long, smooth inclined plane kept at an angle of 60$°$ with horizontal, it can travel a distance $x_1$ along the plane. But when the inclination is decreased to 30$°$ and the same object is shot with the same velocity, it can travel $x_2$ distance. Then $x_1$:$x_2$ will be:

1. $1:2\sqrt{3}$

2. $1:\sqrt{2}$

3. $\sqrt{2}:1$

4. $1:\sqrt{3}$