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Topic: Uniformly accelerated motion relations· Kinematics

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

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 :

1. \(1:\sqrt{2}\)

2. \(\sqrt{2}:1\)

3. \(1:\sqrt{3}\)

4. \(1:2\sqrt{3}\)

1 answer

Best answer

Answer: 3. \(1:\sqrt{3}\)

On a smooth incline the retardation is \(g\sin\theta\). For the same launch speed \(v\), the distance up the plane is \(x=\dfrac{v^2}{2g\sin\theta}\), so \(x\propto\dfrac1{\sin\theta}\).

\(\dfrac{x_1}{x_2}=\dfrac{\sin30^\circ}{\sin60^\circ}=\dfrac{1/2}{\sqrt3/2}=\dfrac1{\sqrt3}\).

\(x_1:x_2=1:\sqrt3\).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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