A) \[45{}^\circ \]
B) \[60{}^\circ \]
C) \[90{}^\circ \]
D) \[30{}^\circ \]
Correct Answer: D
Solution :
Let u be initial velocity of projection at angle 9 with the horizontal. Then, horizontal range, \[R=\frac{{{u}^{2}}\sin 2\theta }{g}\] and maximum height, \[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] Given, \[R=4\sqrt{3}H\] \[\therefore \] \[\frac{{{u}^{2}}\sin 2\theta }{g}=4\sqrt{3}.\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] \[\therefore \] \[2\sin \theta \cos \theta =2\sqrt{3}{{\sin }^{2}}\theta \] Or \[\frac{\cos \theta }{\sin \theta }=\sqrt{3}\] Or \[\cot \theta =\sqrt{3}=\cot {{30}^{o}}\]You need to login to perform this action.
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