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(A) Given that the range $R$ is twice the maximum height $H$,so $R = 2H$.We know the formulas for range $R = \frac{v^2 \sin(2	heta)}{g} = \frac{2v^2

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$\frac{4v^2}{5g}$

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(A) Given that the range $R$ is twice the maximum height $H$,so $R = 2H$.We know the formulas for range $R = \frac{v^2 \sin(2	heta)}{g} = \frac{2v^2 \sin	heta \cos	heta}{g}$ and maximum height $H = \frac{v^2 \sin^2	heta}{2g}$.Substituting these into the given condition $R = 2H$:$\frac{2v^2 \sin	heta \cos	heta}{g} = 2 \left( \frac{v^2 \sin^2	heta}{2g} ight)$$\frac{2v^2 \sin	heta \cos	heta}{g} = \frac{v^2 \sin^2	heta}{g}$$2 \cos	heta = \sin	heta$$	an	heta = 2$.From the right-angled triangle with opposite side $2$ and adjacent side $1$,the hypotenuse is $\sqrt{2^2 + 1^2} = \sqrt{5}$.Thus,$\sin	heta = \frac{2}{\sqrt{5}}$ and $\cos	heta = \frac{1}{\sqrt{5}}$.Now,substitute these into the range formula:$R = \frac{2v^2 \sin	heta \cos	heta}{g} = \frac{2v^2}{g} \left( \frac{2}{\sqrt{5}} ight) \left( \frac{1}{\sqrt{5}} ight) = \frac{4v^2}{5g}$.

$A$ particle is projected with a velocity $v$ such that its range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is (where $g$ is acceleration due to gravity)

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