Question

If $R$ and $H$ represent horizontal range and maximum height of the projectile, then the angle of projection with the horizontal is

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

Answer 1

Answer: (C)

$tan^{-1}\left(\frac{4H}{R}\right)$

Answer 2

Maximum height, $H=\frac{u^{2}\,sin^{2}\,\theta}{2g}\,...\left(i\right)$ Horizontal range, $R=\frac{u^{2}\,sin^{2}\,2\theta }{g}$ $=\frac{2u^{2}\,sin\,cos\,\theta}{g}\,...\left(ii\right)$ Divide $\left(i\right)$ by $\left(ii\right)$, we get $\frac{H}{R}=\frac{tan\,\theta}{4}$ or $tan\,\theta =\frac{4H}{R}$ or $\theta=tan^{-1}\left(\frac{4H}{R}\right)$