question_answer

A particle is projected with a velocity \[\nu \] 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) \[\frac{4{{v}^{2}}}{5\,g}\]                  

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 \[\sin \,\theta \,\,=\,\,\frac{2}{\sqrt{5}},\,\,\cos \,\,\theta =\frac{1}{\sqrt{5}}\] \[\therefore \,\,Range\,\,of\,\,projectile\,\,R=\frac{2{{\nu }^{2}}\sin \theta \,\cos \theta }{g}\] \[=\,\,\frac{2{{\nu }^{2}}}{g}\times \frac{2}{\sqrt{5}}\times \frac{1}{\sqrt{5}}=\frac{4{{\nu }^{2}}}{5g}\]