A) \[\frac{4}{3}\]
B) \[2\sqrt{3}\]
C) \[4\sqrt{3}\]
D) \[\frac{3}{4}\]
Correct Answer: C
Solution :
Given, \[u\cos \theta =\frac{\sqrt{3}u}{2}\] \[\Rightarrow \,\,\cos \theta =\frac{\sqrt{3}}{2}\Rightarrow \theta =30{}^\circ \] Range (R) \[\frac{{{u}^{2}}\sin 2\theta }{g}=\frac{{{u}^{2}}\sin 60{}^\circ }{g}=\frac{\sqrt{3}{{u}^{2}}}{2g}\] Maximum height \[=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}=\frac{{{u}^{2}}{{\sin }^{2}}30{}^\circ }{2g}=\frac{{{u}^{2}}}{8g}\] Now, Range \[=P\times H\] \[\Rightarrow \,\,\frac{\sqrt{3}{{u}^{2}}}{2g}=P\times \frac{{{u}^{2}}}{8g}\Rightarrow P=4\sqrt{3}\]
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