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)  \[\frac{4{{v}^{2}}}{\sqrt{{}}5g}\]

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