If the horizontal range of projectile be (a) and the maximum height attained by it is (b) then prove that the velocity of projection is
` [ 2 g (b+ a^2 /(16 b)) ] ^(1//2)`.

Explanation

Let (u) be the velocity of projection of a projectile and ` theta` be its angle of projection with the horozontal direction from ground .
Maximum height ` =b = ( u^2 sin^2 theta)/(2 g) `
or ` sin^2 theta = (2 bg)/u^2` ….(i)`
Horizontal range
` =a= ( u^2 sin 2 theta )/g = ( 2 u^2 sin theta cos thet )/g`
or ` 2 sin theta cos theta = a g//u^2`
or `4 isn^2 cos^2 theta = a^2 g^2 //u^4`
or ` 4 sin ^2 theta (1- sin^2 theta) = a^2 g^2 //u^4`
or ` ( 2 bg)/u^2 ) [ 1 - (2 bg)/u^2 ] = (a^2 g^2)/u^4 ` [ From (i) ]`
or (8 bg) /u^2 - (16 b^2 g^2)/u^4) = (A^2 g^2) /u^41 `
or ` a^2 g^2 + 16 b^2 g^2 = u^2 8 bg`
or ` u^2 = ( a^2 g^2 + 16 b^2 g^2 )/(8 bg) = (a^2 g+ 16b^2 g)/(8b)`
` = 2 g [ a^2 /( 16 b0 + d] `
or ` u= [ 2 g (a + a^2 /( 16b)) ]^(1//2)` . Proved.