Kinemax it

The horizontal and vertical coordinates of the projectile are $x=v_0 t \cos \theta$ and $y=v_0 t \sin \theta -\frac{1}{2} gt^2$ . The distance to the thrower is

$d^2=x^2+y^2=v_0 t^2 \left[1+\frac{1}{4}\left(\frac{gt}{v_0}\right)^2-\frac{gt}{v_0}\sin \theta\right]$

Assume we wanted to find the maximum distance, and try to determine what is the time when a maximum is reached. We make the derivative with respect to time zero:

$0= \left(\frac{gt}{v_0}\right)^2-3\frac{gt}{v_0} \sin\theta +2$ , or

$0= x^2-3x \sin\theta +2$

with $x=\frac{gt}{v_0}$ . In fact, we are looking for the possibility that this equation has no real solution at all. This quadratic equation has no real solution only if

$(3 \sin\theta)^2 - 4*2 <0$

We solve this for the critical angle

$\sin^2\theta =8/9$

Note that we saved some calculations by maximizing $d^2$ , instead of $d$