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. `(4v^(3))/(5g^(2))`
D. `(4v)/(5g^(2))(4v)/(5g^(2))`

Answer

A

Explanation

`R=2H rArr tan theta=(4)/(2)=2`
`R=(2V^(3))/(g) sin theta cos theta`
`R=(2v^(2))/(g)Xx(2)/sqrt(5)xx(1)/sqrt(5)=(4v^(2))/(5g)`