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) \[\frac{4{{\nu }^{2}}}{5\,g}\]              

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