A gody is projected from the ground at an angle of $\tan ^{-1}\left(\frac{8}{7}\right)$ with the horizontal.…

A gody 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
  1. $8: 7$
  2. $4: 7$
  3. $2: 7$
  4. $1: 7$

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, $\theta=\tan ^{-1} \frac{8}{7} \Rightarrow \tan \theta=\frac{8}{7}=\frac{\frac{8}{7}}{4}=\frac{2}{7}$

Asked in: AP EAMCET 2018 (22 Apr Shift 1)

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