A) \[\frac{4{{v}^{2}}}{5g}\]
B) \[\frac{4g}{5{{v}^{2}}}\]
C) \[\frac{{{v}^{2}}}{g}\]
D) \[\frac{4{{v}^{2}}}{\sqrt{5}g}\]
Correct Answer: A
Solution :
[a] We know, |
\[R=4H\cot \theta \Rightarrow \cot \theta =\frac{1}{2}\] |
\[\text{From triangle we can say that }\]\[\sin \theta =\frac{2}{\sqrt{5}},\text{ cos}\theta \text{=}\frac{1}{\sqrt{5}}\] |
\[\therefore \] Range of projectile \[R=\frac{2{{v}^{2}}\sin \theta \cos \theta }{g}\]\[=\frac{2{{v}^{2}}}{g}\times \frac{2}{\sqrt{5}}\times \frac{1}{\sqrt{5}}=\frac{4{{v}^{2}}}{5g}\] |
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