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

Explanation

Greatest height attained
` h = (u^(2) sin ^(2) theta )/(2g) `
Horizontal range
` R = (u^(2) sin 2 theta )/ (g) = (2u^(2) sin theta cos theta )/(g)`
Given that R `=2h `
`implies (2u ^(2) sintheta cos theta )/ (g)= (2u^(2) sin ^(2) theta )/ (2g)`
` tan theta =2`
Hence `sin theta = 2 //sqrt5,cos theta = 1//sqrt5,`
`:.` From `(B)` R ` = 4u^(2) //5g`