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Q. A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is

AP EAMCETAP EAMCET 2018

Solution:

For a projectile projected at angle $\theta$;
Maximum height, $H_{\max }=\frac{u^{2} \sin ^{2} \theta}{2 g}$
and range, $R=\frac{u^{2} \sin 2 \theta}{g}$
$\therefore $ Ratio $=\frac{H_{\max }}{R}=\frac{\left(\frac{u^{2} \sin ^{2} \theta}{2 g}\right)}{\left(\frac{u^{2} \sin 2 \theta}{g}\right)}=\frac{\tan \theta}{4}$
Here, $\theta=\tan ^{-1} \frac{8}{7}$
$\Rightarrow \tan \theta=\frac{8}{7}=\frac{\frac{8}{7}}{4}=\frac{2}{7}$