question_answer

What is the angle of projection for a projectile motion whose range R is n times the maximum height H?

Answer:

                Here    \[R=nH\]                 or            \[\frac{{{u}^{2}}\sin 2\theta }{g}=n.\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] or            \[\frac{{{\upsilon }^{2}}\times 2\sin \theta \cos \theta }{g}=n.\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\]\[or\tan \theta =\frac{4}{n}\] \[\Rightarrow \]               \[\theta ={{\tan }^{-1}}\left( \frac{4}{n} \right)\]