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} \bigg( \frac{\sqrt 3}{2}\bigg)$

Answer 1

Answer: (C)

Let $\phi$ 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, $\tan \phi=\frac{H}{R / 2}\,\,\,\,$...(i)
In case of projectile motion
Maximum height, $H=\frac{u^{2} \sin ^{2} \theta}{2 g}$
Horizontal range, $R=\frac{u^{2} \sin 2 \theta}{g}$
Substituting these values of $H$ and $R$ in (i), we get
$\tan \phi= \frac{\frac{u^{2} \sin ^{2} \theta}{2 g}}{\frac{u^{2} \sin 2 \theta}{2 g}} $
$\tan \phi= \frac{\sin ^{2} \theta}{\sin 2 \theta}=\frac{\sin ^{2} \theta}{2 \sin \theta \cos \theta}=\frac{1}{2} \tan \theta $
$\tan \phi= \frac{1}{2} \tan 45^{\circ}=\frac{1}{2}$
Here, $\theta=45^{\circ}$
$\therefore \tan \phi =\frac{1}{2} \tan 45^{\circ}=\frac{1}{2} \left(\because \tan 45^{\circ}=1\right)$
$\phi =\tan ^{-1}\left(\frac{1}{2}\right)$