what is the angle at which the maximum height reached by the particle and the horizontal range are equal?

we can derive the angle by simple calculations.

                  let the angle of projection be 'A' and the initial velocity be 'u'

then, horizontal range=(u²sin2A)/g

and the maximum height =(u²sin²A)/2g

on equating both the relation and solving, we get A=tan^-14, which is the angle of projection at which both the range and the maximum height are equal.