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(C) Let $H$ be the maximum height and $R$ be the horizontal range of the projectile.From the geometry of the trajectory,the highest point is at a hori

Vedclass · Accepted Answer

$	an \phi = \frac{1}{2} 	an 	heta$

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(C) Let $H$ be the maximum height and $R$ be the horizontal range of the projectile.From the geometry of the trajectory,the highest point is at a horizontal distance of $R/2$ from the launch point.The elevation angle $\phi$ is defined as the angle whose tangent is the ratio of the maximum height to the horizontal distance of the highest point from the launch point.Therefore,$	an \phi = \frac{H}{R/2}$.Substituting the standard formulas $H = \frac{u^2 \sin^2 	heta}{2g}$ and $R = \frac{u^2 \sin 2	heta}{g}$,we get:$	an \phi = \frac{u^2 \sin^2 	heta / 2g}{(u^2 \sin 2	heta / g) / 2} = \frac{u^2 \sin^2 	heta / 2g}{u^2 \sin 2	heta / 2g} = \frac{\sin^2 	heta}{\sin 2	heta}$.Using the identity $\sin 2	heta = 2 \sin 	heta \cos 	heta$,we have:$	an \phi = \frac{\sin^2 	heta}{2 \sin 	heta \cos 	heta} = \frac{1}{2} \frac{\sin 	heta}{\cos 	heta} = \frac{1}{2} 	an 	heta$.

$A$ projectile is fired from level ground at an angle $\theta$ above the horizontal. The elevation angle $\phi$ of the highest point as seen from the launch point is related to $\theta$ by the relation

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