Answer
Maximum angle of projection can be $70.52^0$.
Work Step by Step
Let point of projection is $P (0, 0)$ and projectile is projected with speed $u$ and an angle $\theta$ up the horizontal.
Then at any time t, its x-coordinate is given as, $x=ucos\theta t$ and y-coordinate is given as, $y=usin\theta t-\frac{1}{2}gt^2$.
So, at time t, its distance from P is $r=\sqrt{x^2+y^2}=\sqrt{(ucos\theta t)^2+(usin\theta t-\frac{1}{2}gt^2)^2}$
$\Rightarrow r = \sqrt{u^2t^2+\frac{1}{4}g^2t^4-gusin\theta t^3)^2}$
Since distance of the particle from point P is always increasing, therefore $\frac{dr}{dt}>0$
So, $\frac{dr}{dt}=\frac{1}{2}\frac{2u^2t+g^2t^3-3gusin\theta t^2}{\sqrt{u^2t^2+\frac{1}{4}g^2t^4-gusin\theta t^3)^2}}>0\Rightarrow t(g^2t^2-3gusin\theta t+2u^2)>0$
Since t is always positive, therefore $g^2t^2-3gusin\theta t+2u^2>0$
This is a quadratic equation in time t (upward parabola) and has no real root, so discriminant $D<0$.
So, $D=b^2-4ac=9g^2u^2sin^2\theta-4\times g^2\times 2u^2<0\Rightarrow sin\theta
