A) \[{{45}^{o}}\]
B) \[{{60}^{o}}\]
C) \[{{90}^{o}}\]
D) \[{{30}^{o}}\]
Correct Answer: D
Solution :
Let u be initial velocity of projection at angle \[\theta \]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 \[\cot \theta =\sqrt{3}=\cot \,{{30}^{o}}.\] \[\Rightarrow \] \[\theta ={{30}^{o}}\]You need to login to perform this action.
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