Question

A projectile is fired at an angle of ${45}^o$ 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. ${45}^o$
• C. ${60}^o$
• D. ${\tan }^{-1}\frac{1}{2}$

Answer 1

The correct option is D ${\tan }^{-1}\frac{1}{2}$

Let $\alpha$ be the elevation angle
$\Rightarrow$ $\tan \alpha =\frac{H}{R/2}$
But when we take the ratio of maximum height with range of projectile we get,
$\frac{H}{R}=\frac{\tan \theta }{4}$
$\Rightarrow \tan \alpha =\frac{2\tan {45}^o}{4}$

$\Rightarrow \alpha ={\tan }^{-1}\left(\frac{1}{2}\right)$