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)

Options

A. `(4v^2)/(5g)`
B. `(4g)/(5v^2)`
C. `v^2/g`
D. `(4v^2)/(sqrt5g)`

Answer

A

Explanation

We know , R= 4H cot `theta rArr cot theta =1/2`
From triangle we can say that
sin `theta=2/sqrt15 , cos theta =1/sqrt5`
`therefore` Range of projectile `R=(2v^2 sin theta cos theta)/g`
`=(2v^2)/g xx 2/sqrt5 xx 1/sqrt5 =(4v^2)/(5g)`