First slide

Projection Under uniform Acceleration

Question

The time taken by a particle to slide down a smooth inclined plane is double the time it would take in falling down through a height equal to the vertical height of the plane. The inclination of the plane with horizontal is:

Moderate

A

30°
B

45°
C

60°
D

90°

Solution

$time taken along the inclined planet = \sqrt{\frac{2 L}{gsinθ}} from the same height if it falls t' = \sqrt{\frac{2 h}{g}} given t = 2 t' but h = Lsinθ \sqrt{\frac{2 L}{gsinθ}} = 2 \sqrt{\frac{2 h}{g}} \frac{2 L}{sinθg} = 4 \frac{2 Lsinθ}{g} \sin^{2} θ = \frac{1}{4}; θ = 30^{0}$

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