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Question

The maximum range of projectile is $$500\ m$$. If the particle is thrown up a plane, which is inclined at an angle of $$30^{o}$$ with the same speed, the distance covered by it along the inclined plane will be:


A
250 m
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B
500 m
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C
750 m
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D
1000 m
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Solution

The correct option is B $$500\ m$$
Range = $$\dfrac{u^2 \sin 2 \theta}{g}$$ ..(1)
and 
$$R_{max} = \dfrac{4^2}{g}$$ ... (ii)
From (1) , max rang at $$\theta = 45^{\circ}$$
$$500 = \dfrac{u^2 \times  \sin 90}{g}$$
$$\dfrac{u^2}{g} = 500$$
Now, 
The distance covered along inclined plane is 
$$v^2-u^2=2as$$
Here $$a=-gsin30^{\circ}$$ and v=0
So, 
$$0-u^2=2(-gsin30^{\circ})s$$
$$s=\dfrac{u^2}{g}$$
$$s=500m$$


Physics

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