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

Answer

D

Explanation

`R = 2H_("max")`
`(v^(2) sin 2theta)/g = 2 xx (v^(2) sin^(2) theta)/(2g)`
On solving
`theta = tan^(-1)(2)` or `theta = 63.43^(@)`
Now, Range, `R = (v^(2) sin [2 xx 63.43])/g = 4/5 v^(2)/g`