A) \[1:1\]
B) \[2:1\]
C) \[4:1\]
D) \[3:2\]
E) \[4:3\]
Correct Answer: C
Solution :
\[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\]or\[\frac{{{H}_{1}}}{{{H}_{2}}}=\frac{{{u}^{2}}{{\sin }^{2}}{{\theta }_{1}}}{{{u}^{2}}{{\sin }^{2}}{{\theta }_{2}}}\] Or \[\frac{3}{1}=\frac{{{\sin }^{2}}{{\theta }_{1}}}{{{\sin }^{2}}{{\theta }_{2}}}\]or\[\frac{\sin {{\theta }_{1}}}{\sin {{\theta }_{2}}}=\frac{\sqrt{3}}{1}\] Logically, we can conclude that \[{{\theta }_{1}}={{60}^{o}},{{\theta }_{2}}={{30}^{o}}\] Again \[R=\frac{{{u}^{2}}\sin 2\theta }{g}\] Or \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{4{{u}^{2}}\sin 2{{\theta }_{1}}}{{{u}^{2}}\sin 2{{\theta }_{2}}}\] Or \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{4\sin 2({{60}^{o}})}{\sin 2(30{}^\circ )}=\frac{4\sin 120{}^\circ }{\sin 60{}^\circ }\] Or \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{4\times \frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}}=4\]
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