Distance on Inclined Plane – Rankers Physics

Uncategorized: Practice Problem & Solution

When an object is shot from the bottom of a long smooth inclined plane kept at an angle \(60°text{ with horizontal, it can travel a distance }x_1text{ along the plane. But when the inclination is decreased to }30°text{ and the same object is shot with the same velocity, it can travel }x_2text{ distance. Then }x_1text{ : }x_2text{ will be: (2019)}\
\(1:sqrt{2}\)
\(sqrt{2}:1\)
\(1:sqrt{3}\)
\(1:2sqrt{3}\)

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Uncategorized. Understanding the underlying formula is key to arriving at the correct answer below:

Concept: Motion on an inclined plane with gravity.
Formula: Distance traveled up the incline \(x = u^2 / (2g sin alpha)\).
For \(alpha_1 = 60°\), \(x_1 = u^2 / (2g sin 60°) = u^2 / (gsqrt{3})\).
For \(alpha_2 = 30°\), \(x_2 = u^2 / (2g sin 30°) = u^2 / g\).
Ratio \(x_1 : x_2 = (u^2 / (gsqrt{3})) : (u^2 / g) = 1 : sqrt{3}\).

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