A shell fired from the base of a mountain jus | vedclass

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(C) Let the maximum height of the projectile be $H$ and the horizontal range be $R$. The shell clears the summit,which is at the peak of the trajector

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$	an^{-1}(\frac{1}{2} 	an \alpha)$

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(C) Let the maximum height of the projectile be $H$ and the horizontal range be $R$. The shell clears the summit,which is at the peak of the trajectory.The angular elevation $\beta$ of the summit from the base is given by $	an \beta = \frac{H}{R/2} = \frac{2H}{R}$.Substituting the standard formulas for maximum height $H = \frac{u^2 \sin^2 \alpha}{2g}$ and horizontal range $R = \frac{u^2 \sin(2\alpha)}{g} = \frac{2u^2 \sin \alpha \cos \alpha}{g}$:$	an \beta = \frac{2 \left( \frac{u^2 \sin^2 \alpha}{2g} ight)}{\left( \frac{2u^2 \sin \alpha \cos \alpha}{g} ight)}$$	an \beta = \frac{u^2 \sin^2 \alpha / g}{2u^2 \sin \alpha \cos \alpha / g} = \frac{\sin^2 \alpha}{2 \sin \alpha \cos \alpha} = \frac{	an \alpha}{2}$.Therefore,$\beta = 	an^{-1}\left(\frac{	an \alpha}{2}ight)$.

$A$ shell fired from the base of a mountain just clears it. If $\alpha$ is the angle of projection,then the angular elevation of the summit $\beta$ is

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