A projectile is fired at an angle of 45^circ | vedclass

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(C) Let $\phi$ be the elevation angle of the projectile at its highest point as seen from the point of projection $O$,and $	heta$ be the angle of pro

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $\phi$ be the elevation angle of the projectile at its highest point as seen from the point of projection $O$,and $	heta$ be the angle of projection with the horizontal.From the figure,$	an \phi = \frac{H}{R/2} = \frac{2H}{R}$.In projectile motion,the maximum height $H$ is given by $H = \frac{u^2 \sin^2 	heta}{2g}$ and the horizontal range $R$ is given by $R = \frac{u^2 \sin 2	heta}{g}$.Substituting these values into the expression for $	an \phi$:$	an \phi = \frac{2 \left( \frac{u^2 \sin^2 	heta}{2g} ight)}{\left( \frac{u^2 \sin 2	heta}{g} ight)} = \frac{\sin^2 	heta}{\sin 2	heta} = \frac{\sin^2 	heta}{2 \sin 	heta \cos 	heta} = \frac{1}{2} 	an 	heta$.Given $	heta = 45^\circ$,we have $	an \phi = \frac{1}{2} 	an 45^\circ = \frac{1}{2} 	imes 1 = \frac{1}{2}$.Therefore,$\phi = 	an^{-1}\left(\frac{1}{2}ight)$.

$A$ projectile is fired at an angle of $45^\circ$ with the horizontal. The elevation angle of the projectile at its highest point as seen from the point of projection is:

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