A) \[\frac{4}{3}\]
B) \[2\sqrt{3}\]
C) \[4\sqrt{3}\]
D) \[\frac{3}{4}\]
Correct Answer: C
Solution :
[c] : Let u be the initial speed and \[\theta \]is the angle of the projection. As per question, Speed at the maximum height \[{{v}_{H}}=u\cos \theta =\frac{\sqrt{3}}{2}u\] \[\therefore \]\[\cos \theta =\frac{\sqrt{3}}{2}\] \[\theta ={{\cos }^{-1}}\left( \frac{\sqrt{3}}{2} \right)={{30}^{o}}\] Range, \[R=\frac{{{u}^{2}}\sin 2\theta }{g}\] Maximum height, \[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] As R = PH (Given) \[\therefore \]\[\frac{{{u}^{2}}\sin 2\theta }{g}=P\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] \[\Rightarrow \]\[2\sin \theta \cos \theta =\frac{P}{2}{{\sin }^{2}}\theta \Rightarrow \tan \theta =\frac{4}{P}\] Or \[P=\frac{4}{\tan {{30}^{o}}}=4\sqrt{3}\]
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