A) \[\frac{4{{v}^{2}}}{5\,g}\]
B) \[\frac{4g}{5{{v}^{2}}}\]
C) \[\frac{{{v}^{2}}}{g}\]
D) \[\frac{4{{v}^{2}}}{\sqrt{5}\,g}\]
Correct Answer: A
Solution :
\[R=2H\,\,given\] We know \[\,R = 4\,H\,\,\cot \,\theta \,\,\,\Rightarrow \, cot\,\theta = \frac{1}{2}\] From triangle we can say that \[\sin \,\theta \,\,=\,\,\frac{2}{\sqrt{5}},\,\,\cos \,\,\theta =\frac{1}{\sqrt{5}}\] \[\therefore \,\,Range\,\,of\,\,projectile\,\,R=\frac{2{{\nu }^{2}}\sin \theta \,\cos \theta }{g}\] \[=\,\,\frac{2{{\nu }^{2}}}{g}\times \frac{2}{\sqrt{5}}\times \frac{1}{\sqrt{5}}=\frac{4{{\nu }^{2}}}{5g}\]You need to login to perform this action.
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