A) \[\frac{4{{\nu }^{2}}}{5\,g}\]
B) \[\frac{4\,g}{5\,{{\nu }^{2}}}\]
C) \[\frac{{{\nu }^{2}}}{g}\]
D) \[\frac{4{{\nu }^{2}}}{\sqrt{5}g}\]
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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