let u be the velocity of the projection of the body,
then at the point of projection the total mechanical energy of the particle is 1/2 mu^2 + mgh
let v be the velocity at lowest point then 1/2 mv^2 = 1/2 mu^2 + mgh
v^2 = u^2 +2gh
also we know that horizontal velocity remains constant , so ucos30 = vcos45,
sqrt ( u^2 +2gh) / sqrt 2 = u*sqrt(3) /2
3u^2 / 2 = u^2 +2gh ,
u =2 * sqrt(gh)
let $ be the angle it makes with horizontal
again the total velocity remains the same , and the horizontal velocity remains the same
u cos60 = v cos $
u / sqrt ( u^2 + 2gh ) * 1/2 = cos $
2*sqrt(gh) / sqrt ( 6gh) * 1/2 = cos $
cos $ = 1/ sqrt (6) ,
so initially i got wrong answer but now this one's correct :)