A ball is projected with a velocity 'v' so that its range on a horizontal plane is twice the greatest height attained by the projectile. Then, its range is

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Solution

The correct option is C

R = 2H

tan $θ$ = $\frac{4}{2}$ = 2 , Now R = $\frac{u^{2} 2 s i n θ c o s θ}{g}$

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