A particle is projected with a velocity v so that its horizontal range twice the greatest height attained. The horizontal range is

Options

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

Answer

D

Explanation

`R= 2H rArr (v^(2)sin 2 theta_(0))/(g)= (2v^(2) sin^(2) theta_(0))/(2g) rArr cos theta_(0)= (1)/(2)`
`rArr cos theta_(0)= (1)/(sqrt5)`
`sin theta_(0)= (2)/(sqrt5) therefore R= (2v^(2) sin theta_(0)cos theta_(0))/(g)= (2v^(2) xx (1)/(sqrt5) (2)/(sqrt5))/(g)= (4v^(2))/(5g)`