Small Salah

The height of the ball above a point on the ground $x \text{ m}$ from the corner is given by $f(x;a) = -ax^2 + \tan(\theta) x$ where $a > 0$ is some parameter and $\theta$ is the angle of elevation of the pitch.

Along the line segment from the corner to the penalty spot, we find that it is $\sqrt{11^2 + 34^2} = \sqrt{1277} \text{ m}$ to the penalty spot and [by similar triangles] $\frac{14}{34}\sqrt{1277} = \frac{7}{17}\sqrt{1277} \text{ m}$ to the edge of the penalty area. Therefore, in order to make it over Kompany and to Salah, we must have $f\left(\sqrt{1277};a\right) \leq 1.75, \qquad f\left(\frac{7}{17}\sqrt{1277};a\right) \geq 1.93$

In order to make the following steps easier, we will uniformly scale everything by a factor of $\frac{17}{\sqrt{1277}}$ . This changes $a>0$ to some other $A > 0$ , but it doesn't change $\tan\theta\text{.}$ The result is $f\left(17;A\right) \leq \frac{17 \cdot 1.75}{\sqrt{1277}}, \qquad f\left(7;A\right) \geq \frac{17\cdot 1.93}{\sqrt{1277}}$ and by writing these out: $-17^2A + 17\tan\theta \leq \frac{17 \cdot 1.75}{\sqrt{1277}} \implies -17A + \tan\theta \leq \frac{1.75}{\sqrt{1277}} \\ -49A + 7\tan\theta \geq \frac{17\cdot 1.93}{\sqrt{1277}}$ Multiplying the first inequality by $-49$ and the second by $17$ gives $833A + -49\tan\theta \geq \frac{-49 \cdot 1.75}{\sqrt{1277}} \\ -833A + 119\tan\theta \geq \frac{289\cdot 1.93}{\sqrt{1277}}$ Adding them: $70\tan\theta \geq \frac{289\cdot 1.93 - 49 \cdot 1.75}{\sqrt{1277}}$ and then solving for $\theta$ : $\theta \geq \tan^{-1}\left(\frac{289\cdot 1.93 - 49 \cdot 1.75}{70\sqrt{1277}}\right) \approx 10.69^\circ$

Rounded to the nearest degree, this gives an answer of $\boxed{11}$ .