question_answer

A particle is projected with a velocity v such that its range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is (where g is acceleration due to gravity)

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 \[R=\frac{2{{v}^{2}}\sin \theta \cos \theta }{g}\]\[=\frac{2{{v}^{2}}}{g}\times \frac{2}{\sqrt{5}}\times \frac{1}{\sqrt{5}}=\frac{4{{v}^{2}}}{5g}\]