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

Explanation

Max. height , `H= (u^2 sin^2 theta)/ (2g)`
Horizontal range,
` R= ( u^2 sin 2 theta ) /g = ( 2 u^2 sin theta cos theta)/( 2g)`
As per question, ` R= 2 H`
` ( 2u^2 sin theta cos theta)/g = ( 2 u^2 sin ^2 theta)/(2g)` ltbRgt Solving we get, ` tan theta =2 , therfore
` sin theta= ( 2/ sqrt2)` ltbRgt and ` cos theta = (1 /sqrt 4)` ltbRgt :. Horizontal range,
` R= (2 u^2)/g xx 2/(sqrt5) xx 1/(sqrt 5) = (4u^@)/(5g)`.