Question

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. $45^{\circ }$
• B. $60^{\circ }$
• C. ${\tan }^{-1}\frac{1}{2}$
• D. ${\tan }^{-1}\left(\frac{\sqrt{3}}{2}\right)$

Answer 1

Answer: (C)

Let $\varphi$ be elevation angle of the projectile at its highest point as seen from the point of projection $O$ and $\theta$ be angle of projection with the horizontal.
From figure, $\tan \varphi =\frac{H}{R/2}$...(i)
In case of projectile motion
Maximum height, $H=\frac{u^2{\sin }^2\theta }{2g}$
Horizontal range, $R=\frac{u^2\sin 2\theta }{g}$
Substituting these values of $H$ and $R$ in (i), we get
$\tan \varphi =\frac{\frac{u^2{\sin }^2\theta }{2g}}{\frac{u^2\sin 2\theta }{2g}}$
$\tan \varphi =\frac{{\sin }^2\theta }{\sin 2\theta }=\frac{{\sin }^2\theta }{2\sin \theta \cos \theta }=\frac{1}{2}\tan \theta$
$\tan \varphi =\frac{1}{2}\tan 45^{\circ }=\frac{1}{2}$
Here, $\theta =45^{\circ }$
$\therefore \tan \varphi =\frac{1}{2}\tan 45^{\circ }=\frac{1}{2}\left(\because \tan 45^{\circ }=1\right)$
$\varphi ={\tan }^{-1}\left(\frac{1}{2}\right)$