Question

# 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 $$30^o$$ and the same object is shot with the same velocity, it can travel $$x_2$$ distance. Then $$x_1 : x_2$$ will be :

A
1:2
B
2:1
C
1:3
D
1:23

Solution

## The correct option is A $$1 : \sqrt{3}$$(Stopping distance) $$x_1 = \dfrac{u^2}{2g sin 60^o}$$(Stopping distance) $$x_2 = \dfrac{u^2}{2g sin 30^o}$$$$\Rightarrow \dfrac{x_1}{x_2} = \dfrac{sin 30^o}{sin 60^o} = \dfrac{1 \times 2}{2 \times \sqrt{3}} = 1 : \sqrt{3}$$Physics

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