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

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(B) Let the angle of projection be $	heta = 45^{\circ}$.At the highest point $P$, the height is $H = \frac{u^2 \sin^2 	heta}{2g}$ and the horizontal

Vedclass · Accepted Answer

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

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(B) Let the angle of projection be $	heta = 45^{\circ}$.At the highest point $P$, the height is $H = \frac{u^2 \sin^2 	heta}{2g}$ and the horizontal distance from the point of projection $O$ is $R/2 = \frac{u^2 \sin 	heta \cos 	heta}{g}$.The elevation angle $\alpha$ is the angle subtended by the highest point $P$ at the point of projection $O$ with the horizontal.In the right-angled triangle $	riangle POM$, we have:$	an \alpha = \frac{H}{R/2} = \frac{\frac{u^2 \sin^2 	heta}{2g}}{\frac{u^2 \sin 	heta \cos 	heta}{g}} = \frac{\sin 	heta}{2 \cos 	heta} = \frac{1}{2} 	an 	heta$.Substituting $	heta = 45^{\circ}$:$	an \alpha = \frac{1}{2} 	an 45^{\circ} = \frac{1}{2} (1) = \frac{1}{2}$.Therefore, $\alpha = 	an^{-1}\left(\frac{1}{2}ight)$.

$A$ 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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