The motion of the water is described by projectile motion equations. With the origin at the nozzle, the horizontal ($x$) and vertical ($y$) positions at time $t$ are:

$$x(t) = (v_0 \cos\alpha) t$$ $$y(t) = (v_0 \sin\alpha) t - \frac{1}{2}gt^2$$

First, we derive the trajectory equation by eliminating time $t$. From the $x$-equation, $t = x/(v_0 \cos\alpha)$. Substituting this into the $y$-equation gives:

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

Using the trigonometric identity $1/\cos^2\alpha = 1 + \tan^2\alpha$, the trajectory equation becomes:

$$y = x \tan\alpha - \frac{gx^2}{2v_0^2}(1 + \tan^2\alpha)$$

For the water to just clear the wall, the trajectory must pass through the point $(x, y) = (d, h)$.

$$h = d \tan\alpha - \frac{gd^2}{2v_0^2}(1 + \tan^2\alpha)$$

This equation relates the initial speed $v_0$ and the launch angle $\alpha$. To find the minimum speed, we can treat this as a quadratic equation for $\tan\alpha$. Let $T = \tan\alpha$:

$$\left(\frac{gd^2}{2v_0^2}\right)T^2 - d T + \left(h + \frac{gd^2}{2v_0^2}\right) = 0$$

For a given $v_0$, a real solution for $T$ (and thus a real launch angle $\alpha$) exists only if the discriminant of this quadratic equation is non-negative ($\Delta \ge 0$). The minimum initial speed $v_{0,min}$ corresponds to the case where there is exactly one solution for $\tan\alpha$, which occurs when the discriminant is zero ($\Delta = 0$).

The discriminant is $\Delta = B^2 - 4AC$:

$$\Delta = (-d)^2 - 4\left(\frac{gd^2}{2v_0^2}\right)\left(h + \frac{gd^2}{2v_0^2}\right) = 0$$ $$d^2 = \frac{2gd^2}{v_0^2}\left(h + \frac{gd^2}{2v_0^2}\right)$$

Dividing by $d^2$ (since $d eq 0$):

$$1 = \frac{2gh}{v_0^2} + \frac{g^2d^2}{v_0^4}$$

Multiplying by $v_0^4$ gives a quadratic equation for $v_0^2$:

$$v_0^4 - (2gh)v_0^2 - g^2d^2 = 0$$

Solving for $v_0^2$ using the quadratic formula:

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

Since $v_0^2$ must be positive and $\sqrt{h^2+d^2} > h$, we must take the positive sign. This gives the square of the minimum speed:

$$v_{0,min}^2 = g(h + \sqrt{h^2+d^2})$$

[Q1] Minimum initial speed $v_0$ Substituting the given values $d=3$ m, $h=4.0$ m, and $g=10$ m/s²:

$$\sqrt{h^2+d^2} = \sqrt{(4.0)^2 + (3)^2} = \sqrt{16+9} = \sqrt{25} = 5 \text{ m}$$ $$v_{0,min}^2 = (10 \text{ m/s²})(4.0 \text{ m} + 5 \text{ m}) = 90 \text{ (m/s)²}$$ $$v_{0,min} = \sqrt{90} \text{ m/s} = 3\sqrt{10} \text{ m/s}$$

[Q2] Corresponding launch angle $\alpha$ When the discriminant is zero, the unique solution for $T = \tan\alpha$ is given by $T = -B/(2A)$:

$$\tan\alpha = \frac{-(-d)}{2\left(\frac{gd^2}{2v_{0,min}^2}\right)} = \frac{d}{\frac{gd^2}{v_{0,min}^2}} = \frac{v_{0,min}^2}{gd}$$

Substitute the expression for $v_{0,min}^2$:

$$\tan\alpha = \frac{g(h+\sqrt{h^2+d^2})}{gd} = \frac{h+\sqrt{h^2+d^2}}{d}$$

Substituting the numerical values:

$$\tan\alpha = \frac{4.0 \text{ m} + 5 \text{ m}}{3 \text{ m}} = \frac{9}{3} = 3$$

The corresponding angle is $\alpha = \arctan(3)$.