Question

In case of a projectile if the maximum vertical height H is equal to horizontal range R then angle of projection $\theta$ is

• A. ${\tan }^{-1}4$
• B. ${\tan }^{-1}\left(\frac{1}{4}\right)$
• C. ${\tan }^{-1}2$
• D. ${\tan }^{-1}\left(\frac{1}{2}\right)$

Answer 1

The correct option is A ${\tan }^{-1}4$

Lets consider, $u=$ initial velocity

$\theta =$ angle of projection

In projectile motion,

Maximum vertical height, $H=\frac{u^{2}si{n}^2\theta }{2g}$

Horizontal range, $R=\frac{2u^{2}sin\theta cos\theta }{g}$

According to question,

$R=H$

$\frac{2u^{2}sin\theta cos\theta }{g}=\frac{u^{2}si{n}^2\theta }{2g}$

$\frac{sin\theta }{cos\theta }=4$

$tan\theta =4$

$\theta =ta{n}^{-1}\left(4\right)$

The correct option is A.