When an object is shot from the bottom of a long, smooth inclined plane kept at an angle of 60$^{\circ}$ with

Question

When an object is shot from the bottom of a long, smooth inclined plane kept at an angle of 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:

Accepted Answer

According to the principle of conservation of mechanical energy, the total energy of a body moving under an external conservative force (like gravity) remains constant . When an object is shot up a smooth incline, its initial kinetic energy at the bottom is entirely converted into gravitational potential energy ($V = mgh$) at the highest point reached .

1.  **Energy Relation**: For a mass $m$ with projection velocity $v$, the maximum height $h$ reached is given by $\frac{1}{2}mv^2 = mgh$, which simplifies to $h = \frac{v^2}{2g}$.
2.  **Incline Trigonometry**: On an inclined plane, the vertical height $h$ is related to the distance $x$ travelled along the plane by $h = x \sin	heta$.
3.  **Equating Both Cases**: Since the object is shot with the same velocity $v$ in both cases, the vertical height $h$ reached remains the same irrespective of the angle .
    $$x_1 \sin 60^{\circ} = x_2 \sin 30^{\circ}$$
    $$x_1 \left(\frac{\sqrt{3}}{2}ight) = x_2 \left(\frac{1}{2}ight)$$
    $$\frac{x_1}{x_2} = \frac{1}{\sqrt{3}}$$

Thus, the ratio $x_1:x_2$ is **1:$\sqrt{3}$**.

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