The motion of the water stream is described by the equations of projectile motion:

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

Eliminating time $t$ yields the trajectory equation. 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 identity $\sec^2\alpha = 1/\cos^2\alpha = 1 + \tan^2\alpha$, we get:

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

The stream must just clear the wall, meaning its trajectory passes through the point $(x, y) = (d, h)$:

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

Let $T = \tan\alpha$. We can rearrange this into a quadratic equation for $T$:

$$(gd^2)T^2 - (2dv_0^2)T + (2hv_0^2 + gd^2) = 0$$

For a given $v_0$, this equation yields the required angle(s) $\alpha$. For a real solution for $\alpha$ to exist, the quadratic equation for $T$ must have real roots. This requires the discriminant $(\Delta)$ to be non-negative:

$$\Delta = (-2dv_0^2)^2 - 4(gd^2)(2hv_0^2 + gd^2) \ge 0$$ $$4d^2v_0^4 - 8ghd^2v_0^2 - 4g^2d^4 \ge 0$$

Dividing by $4d^2$ (since $d eq 0$):

$$v_0^4 - 2ghv_0^2 - g^2d^2 \ge 0$$

The minimum initial velocity, $v_{0,min}$, is the smallest $v_0$ that satisfies this condition. This occurs when the expression is equal to zero, which corresponds to the discriminant being zero (a single real solution for $\tan\alpha$).

$$v_0^4 - 2ghv_0^2 - g^2d^2 = 0$$

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

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

Since $v_0^2$ must be positive, we take the positive root:

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

[Q1] Minimum initial velocity $v_0$

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

Substituting the given values $d=3.0$ m, $h=4.0$ m, and $g=10$ m/s$^2$:

$$\sqrt{h^2+d^2} = \sqrt{(4.0)^2 + (3.0)^2} = \sqrt{16+9} = \sqrt{25} = 5.0 \text{ m}$$ $$v_{0,min} = \sqrt{10(4.0 + 5.0)} = \sqrt{90} = 3\sqrt{10} \text{ m/s}$$

[Q2] Corresponding elevation angle $\alpha$ When the discriminant is zero, the quadratic equation for $T = \tan\alpha$ has a single root, given by $T = -b/(2a)$:

$$\tan\alpha = \frac{-(-2dv_0^2)}{2(gd^2)} = \frac{v_0^2}{gd}$$

Substituting 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:

$$\tan\alpha = \frac{4.0 + 5.0}{3.0} = \frac{9.0}{3.0} = 3.0$$ $$\alpha = \arctan(3.0)$$