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