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A projectile is fired at an angle of $45^\circ$ with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection is
A
$\tan^{-1}\left(\frac{\sqrt{3}}{2}\right)$
B
$45^\circ$
C
$60^\circ$
D
$\tan^{-1}\frac{1}{2}$
Detailed Solution
At the highest point the projectile is at height H and at horizontal distance $\frac{R}{2}$ from the point of projection.
$H = \frac{u^2\sin^2\theta}{2g}$ and $R = \frac{2u^2\sin\theta\cos\theta}{g}$
Elevation angle $\phi$: $\tan\phi = \frac{H}{R/2} = \frac{u^2\sin^2\theta/2g}{u^2\sin\theta\cos\theta/g}$
$\tan\phi = \frac{\tan\theta}{2}$
With $\theta = 45^\circ$: $\tan\phi = \frac{\tan45^\circ}{2} = \frac{1}{2}$
$\phi = \tan^{-1}\frac{1}{2}$
$H = \frac{u^2\sin^2\theta}{2g}$ and $R = \frac{2u^2\sin\theta\cos\theta}{g}$
Elevation angle $\phi$: $\tan\phi = \frac{H}{R/2} = \frac{u^2\sin^2\theta/2g}{u^2\sin\theta\cos\theta/g}$
$\tan\phi = \frac{\tan\theta}{2}$
With $\theta = 45^\circ$: $\tan\phi = \frac{\tan45^\circ}{2} = \frac{1}{2}$
$\phi = \tan^{-1}\frac{1}{2}$