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)

$\tan \theta=\frac{ H }{\frac{ R }{2}}=\frac{\frac{ u ^{2} \sin ^{2} 45^{\circ}}{2\, g }}{\frac{2 u ^{2} \sin 45^{\circ} \cos 45^{\circ}}{2\, g }}$
$\frac{\tan 45^{\circ}}{2}=\frac{1}{2} $
$\Rightarrow \theta=\tan ^{-1}\left(\frac{1}{2}\right)$