A particle with a velcoity (u) so that its horizontal ange is twice the greatest height attained. Find the horizontal range of it.

Options

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

Answer

A

Explanation

Given, a particle having horizontal range is twice the greatest height attained by it, i.e.

`R = 2H`
`(v^(2)sin theta)/(g) = (2v^(2)sin^(2)theta)/(2g)`
`rArr 2 sin theta cos theta = sin^(2) theta rArr tan theta = 2`
`rArr sin theta = (2)/(sqrt(5)) rArr cos theta = (1)/(sqrt(5))`
`:. R = (v^(2)sin 2 theta)/(g) = (v^(2) xx 2 sin theta cos theta)/(g)`
`rArr R = v^(2) xx (2)/(g) xx (2)/(sqrt(5)) xx (1)/(sqrt(5))`
i.e., `R = (4v^(2))/(5g)`