The problem requires the stone to continuously move farther away from the thrower, who is at the origin. Let the stone's position vector be $\vec{r}(t)$ and its velocity vector be $\vec{v}(t)$. The distance from the origin is $D(t) = |\vec{r}(t)|$. The condition that the distance is always increasing or constant is equivalent to stating that the radial component of the velocity is always non-negative. This can be expressed using the dot product:

$$\vec{r}(t) \cdot \vec{v}(t) \ge 0$$

for the entire duration of the flight, $t > 0$.

Let the initial velocity be $v_0$ and the launch angle be $\theta$. The position and velocity vectors of the projectile are:

$$\vec{r}(t) = (v_0 t \cos\theta)\hat{i} + \left(v_0 t \sin\theta - \frac{1}{2}gt^2\right)\hat{j}$$ $$\vec{v}(t) = (v_0 \cos\theta)\hat{i} + (v_0 \sin\theta - gt)\hat{j}$$

Now, we compute the dot product:

$$\vec{r} \cdot \vec{v} = (v_0 t \cos\theta)(v_0 \cos\theta) + \left(v_0 t \sin\theta - \frac{1}{2}gt^2\right)(v_0 \sin\theta - gt)$$ $$\vec{r} \cdot \vec{v} = v_0^2 t \cos^2\theta + v_0^2 t \sin^2\theta - v_0 gt^2 \sin\theta - \frac{1}{2}v_0 gt^2 \sin\theta + \frac{1}{2}g^2t^3$$

Combining terms and using $\cos^2\theta + \sin^2\theta = 1$:

$$\vec{r} \cdot \vec{v} = v_0^2 t - \frac{3}{2}v_0 gt^2 \sin\theta + \frac{1}{2}g^2t^3$$

Applying the condition $\vec{r} \cdot \vec{v} \ge 0$ for $t > 0$, we can divide the expression by $\frac{t}{2}$ (which is positive):

$$2v_0^2 - 3v_0 gt \sin\theta + g^2t^2 \ge 0$$

Rearranging this into a standard quadratic form in the variable $t$:

$$g^2t^2 - (3gv_0\sin\theta)t + 2v_0^2 \ge 0$$

This is a quadratic inequality for $t$. The expression on the left is a parabola opening upwards. For this inequality to hold for all $t > 0$, the parabola must have either no real roots or a single repeated real root. This means its discriminant, $\Delta$, must be less than or equal to zero.

$$\Delta = b^2 - 4ac \le 0$$

Here, $a = g^2$, $b = -3gv_0\sin\theta$, and $c = 2v_0^2$.

$$\Delta = (-3gv_0\sin\theta)^2 - 4(g^2)(2v_0^2) \le 0$$ $$9g^2v_0^2\sin^2\theta - 8g^2v_0^2 \le 0$$

Dividing by the positive term $g^2v_0^2$:

$$9\sin^2\theta - 8 \le 0$$ $$9\sin^2\theta \le 8$$ $$\sin^2\theta \le \frac{8}{9}$$

Since the launch angle $\theta$ is in the range $(0, \pi/2]$, $\sin\theta$ is positive. Taking the square root:

$$\sin\theta \le \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$$

The maximum angle $\theta_{max}$ corresponds to the equality:

$$\sin\theta_{max} = \frac{2\sqrt{2}}{3}$$ $$\theta_{max} = \arcsin\left(\frac{2\sqrt{2}}{3}\right)$$