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

Answer 1

Answer: (D)

${\tan }^{-1}\frac{1}{2}$


Let $\alpha$ be the elevation angle
$\Rightarrow \tan \alpha =\frac{H}{R/2}.......\text{(i)}$
By using the relation, $4H=R\tan \theta$
$\frac{H}{R}=\frac{\tan \theta }{4}.........\text{(ii)}$
Using $\text{(i)}$ and $\text{(ii)}$, we get
$\tan \alpha =\frac{\tan \theta }{2}$
$\Rightarrow \tan \alpha =\frac{\tan {45}^{\circ }}{2}$
$\alpha ={\tan }^{-1}\left(\frac{1}{2}\right)$