A) \[\sqrt{\frac{2}{3}}\]
B) \[\frac{2}{\sqrt{3}}\]
C) \[\sqrt{\frac{3}{2}}\]
D) \[\frac{\sqrt{3}}{2}\]
Correct Answer: C
Solution :
Hmax \[=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] According to problem \[\,\,\,\frac{{{u}_{1}}^{2}{{\sin }^{2}}45{}^\circ }{2g}=\frac{{{u}_{2}}^{2}{{\sin }^{2}}60{}^\circ }{2g}\] \[\Rightarrow \,\,\frac{{{u}_{1}}^{2}}{{{u}_{2}}^{2}}=\frac{{{\sin }^{2}}60{}^\circ }{{{\sin }^{2}}45{}^\circ }\,\,\Rightarrow \,\,\frac{{{u}_{1}}}{{{u}_{2}}}=\frac{\sqrt{3}/2}{1/\sqrt{2}}=\sqrt{\frac{3}{2}.}\]
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