A) \[{{30}^{o}}\]
B) \[{{45}^{o}}\]
C) \[{{90}^{o}}\]
D) \[{{40}^{o}}\]
Correct Answer: A
Solution :
When body is projected with an initial velocity u, at an angle 6 with the horizontal, then \[R=\frac{{{u}^{2}}\sin 2\theta }{g}\] \[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2\,g}\] Given, \[R=4\,\sqrt{3}\,H\] \[\therefore \] \[\frac{{{u}^{2}}(2\sin \theta \cos \theta )}{g}=4\sqrt{3}\,\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2\,g}\] \[\Rightarrow \] \[\frac{\sin \theta }{\cos \theta }=\frac{1}{\sqrt{3}}\] \[\Rightarrow \] \[\tan \theta =\frac{1}{\sqrt{3}}\] \[\Rightarrow \] \[\theta ={{30}^{o}}\]. Note: Horizontal range is same whether the body is projected at \[{{30}^{o}}\] or \[({{90}^{o}}-{{30}^{o}}={{60}^{o}})\].You need to login to perform this action.
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