If the range of a projectile is twice its max | vedclass

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Question

(A) The formula for the range $R$ is $R = \frac{u^2 \sin(2	heta)}{g} = \frac{2u^2 \sin 	heta \cos 	heta}{g}$.The formula for maximum height $H$ is

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

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(A) The formula for the range $R$ is $R = \frac{u^2 \sin(2	heta)}{g} = \frac{2u^2 \sin 	heta \cos 	heta}{g}$.The formula for maximum height $H$ is $H = \frac{u^2 \sin^2 	heta}{2g}$.Given that the range is twice the maximum height,$R = 2H$.Substituting the formulas: $\frac{2u^2 \sin 	heta \cos 	heta}{g} = 2 \left( \frac{u^2 \sin^2 	heta}{2g} ight)$.Simplifying,we get $2 \sin 	heta \cos 	heta = \sin^2 	heta$,which implies $	an 	heta = 2$.If $	an 	heta = 2$,then $\sin 	heta = \frac{2}{\sqrt{5}}$ and $\cos 	heta = \frac{1}{\sqrt{5}}$.Substituting these into the range formula: $R = \frac{2u^2}{g} \left( \frac{2}{\sqrt{5}} ight) \left( \frac{1}{\sqrt{5}} ight) = \frac{4u^2}{5g}$.

If the range of a projectile is twice its maximum height,what is the range?

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