By the Work-Energy Theorem, the work done by the player on the ball (initially at rest) is equal to the initial kinetic energy imparted to it:

$$W = K_0 = \frac{1}{2}mv_0^2$$

To find the minimum work $W_{min}$, we must find the minimum initial speed $v_0$ required for the ball to reach the target point $(x,y) = (s,h)$.

The trajectory of a projectile is given by:

$$y = x \tan\theta - \frac{gx^2}{2v_0^2 \cos^2\theta}$$

Using the identity $1/\cos^2\theta = 1 + \tan^2\theta$, we can rewrite this as a quadratic equation for $u = \tan\theta$:

$$\left(\frac{gx^2}{2v_0^2}\right)u^2 - (x)u + \left(y + \frac{gx^2}{2v_0^2}\right) = 0$$

For a point $(x,y)$ to be reachable, there must be a real solution for $\tan\theta$. This requires the discriminant of the quadratic equation ($\Delta = B^2 - 4AC$) to be non-negative:

$$(-x)^2 - 4\left(\frac{gx^2}{2v_0^2}\right)\left(y + \frac{gx^2}{2v_0^2}\right) \ge 0$$

Dividing by $x^2$ (for $x eq 0$) and rearranging gives the condition for reachability:

$$v_0^4 - 2gyv_0^2 - g^2x^2 \ge 0$$

The minimum speed required to reach a point occurs when that point lies on the boundary of the reachable region (the "parabola of safety"), where the equality holds. Substituting $(x,y) = (s,h)$:

$$(v_0^2)^2 - (2gh)v_0^2 - (g^2s^2) = 0$$

This is a quadratic equation for $v_0^2$. Solving for $v_0^2$ yields:

$$v_0^2 = \frac{2gh \pm \sqrt{(2gh)^2 - 4(1)(-g^2s^2)}}{2} = gh \pm g\sqrt{h^2+s^2}$$

Since $v_0^2$ must be positive, and $\sqrt{h^2+s^2} > h$, we must take the positive root:

$$(v_0^2)_{min} = g(h + \sqrt{h^2+s^2})$$

The minimum work is then found by substituting this minimum initial speed squared into the work-energy equation:

$$W_{min} = \frac{1}{2}m(v_0^2)_{min} = \frac{1}{2}mg(h + \sqrt{h^2+s^2})$$

Now, we substitute the given values: $m = 0.5$ kg, $s = 11$ m, $h = 2.5$ m, and $g \approx 9.8$ m/s$^2$.

$$\sqrt{h^2+s^2} = \sqrt{(2.5)^2 + (11)^2} = \sqrt{6.25 + 121} = \sqrt{127.25} \approx 11.28 \text{ m}$$ $$W_{min} = \frac{1}{2}(0.5 \text{ kg})(9.8 \text{ m/s}^2)(2.5 \text{ m} + 11.28 \text{ m})$$ $$W_{min} = (2.45 \text{ N})(13.78 \text{ m}) \approx 33.76 \text{ J}$$