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
B
500 m
C
750 m
D
1000 m

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=0So, $$0-u^2=2(-gsin30^{\circ})s$$$$s=\dfrac{u^2}{g}$$$$s=500m$$Physics

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