A particle is projected with a velocity `v` so that its range on a horizontal plane is twice the greatest height attained. If `g` is acceleration due to gravity, then its range is

Options

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

Answer

a

Explanation

`R=2H` given
We know `R=2H cot theta`
`implies cot theta=1/2`
From triangle we can say that
`sin theta=2/(sqrt(5)), cos theta=1/(sqrt(5))`
`:.` Range of projectile `R=(2v^(2)sin theta cos theta)/(g)`
`=(2v^(2))/gxx2/(sqrt(5))xx1/(sqrt(5))=(4v^(2))/(5g)`