Speed Time And Distance

To determine the ratio of the distances traveled by an object on an inclined plane at two different angles, we must understand how the inclination angle affects the range along the plane.

The range of a projectile on an inclined plane is affected by the inclination of the plane and the initial velocity of projection. The range on an inclined plane is given by:

R = \frac{v^2 \sin(2\theta)}{g \cos\alpha}

where:

• v is the initial velocity of projection.
• \theta is the angle of projection with respect to the inclined plane.
• \alpha is the angle of the inclined plane with the horizontal.
• g is the acceleration due to gravity.

In this question, the incline angles are and . We need to find the ratio of distances traveled, and , when the same object is projected with the same initial velocity on planes inclined at these angles.

For \alpha = 60^\circ :

{x_1} = \frac{u^2 \sin(2\theta)}{g \cos{60^\circ}}

For \alpha = 30^\circ :

{x_2} = \frac{u^2 \sin(2\theta)}{g \cos{30^\circ}}

Now, divide the two equations to get the ratio:

\frac{x_1}{x_2} = \frac{\cos{30^\circ}}{\cos{60^\circ}} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}

Thus, the ratio is 1:\sqrt{3} .

Therefore, the correct answer is: