A) \[{{\tan }^{-1}}\left( \sqrt{(3)} \right)\]
B) \[{{\tan }^{-1}}\left( 4 \right)\]
C) \[{{\tan }^{-1}}\left( \sqrt{2} \right)\]
D) \[{{\tan }^{-1}}\left( \frac{1}{\sqrt{3}} \right)\]
Correct Answer: B
Solution :
The maximum height of projectile is \[h=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] ?(i) Horizontal range of the projectile \[R=\frac{{{u}^{2}}\sin 2\theta }{g}=\frac{2{{u}^{2}}\sin \theta \cos \theta }{g}\] ?(iii) Horizontal range of projectile = Horizontal range \[\frac{2{{u}^{2}}\sin \theta \cos \theta }{g}=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] or \[\tan \theta =4\] Hence, \[\theta ={{\tan }^{-1}}4\]
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