Always moving away!

The square of distance $X = x^2$ from the point of projection as a function of time is given as :

$X = u^2 \cos^2 \theta t^2 + (u \sin \theta t - \frac{1}{2} g t^2)^2$

Now, since $x$ always increases, hence $X$ always increases. Hence, $\frac{dX}{dt} \geq 0$

Hence, $2 u^2 \cos^2 \theta t + 2(u \sin \theta t - \frac{1}{2} gt^2)(u \sin \theta - gt) \geq 0$

Rearrange to get :

$g^2t^2 - 3gu \sin \theta t + 2u^2 \geq 0$

Now, the value of $t$ at which above function attains minimum is $\frac{3 u \sin \theta }{2g}$ which is in the interval of time of flight.

Hence, discriminant = $9 g^2 u^2 \sin^2 \theta - 8u^2 g^2 \leq 0 \Rightarrow \sin \theta \leq \frac{2 \sqrt{2}}{3}$ , corresponding to $70.5^{\circ}$

Another solution can be, which does not use calculus is: The distance will stop increasing when the angle between the position vector of projectile at any point during flight with its velocity vector is 90 or less, as long as this angle is greater than 90 it will always move away from the point of projection so all we need to do is to make dot product of those two vectors and the condition at which angle is 90 degree gives us the required angle, Position vector = (u(\cos \theta) t)x + (u(\sin \theta) t-0.5 g t^2)y And velocity vector is = (u\cos \theta)x + (u(\sin \theta-g*t)y When you take dot product of these two vectors you gete same condition as you would have got by using calculus which is-

for angle to be 90 degree 2 u^2-3 (u(\sin \theta) g t+g^2*t^2 is greater than or equal to zero you get same answer sorry for my presentation above, but the method is finally what matters

if distance is always increasing then angle between velocity vector and position vector is acute