6. The game of hockey is popular all over the world. Much in the game depends on the goalkeeper. To hone the skills of goalkeepers and ensure a training process that does not depend on other players, a puck launcher was created. The machine can be set to eject pucks at a specified frequency, speed, and angle. The goal line is 25 m from the central point $O$ of the hockey rink. Let the machine be installed at a distance $d=16 \text{ m}$ from point $O$ in the direction of the goal, and the ejection speed of the puck is $V_{0}=20 \text{ m} / \text{s}$. The shots are taken in a plane perpendicular to the ice surface and the goal line. In this case, for safety, the trajectory of the flying pucks should, on the one hand, not be higher than a straight line connecting the center of the ice rink $O$ with a point in the plane of the puck's flight, in the plane of the goal, and at a distance of one meter from the ice surface, and on the other hand - should intersect the plane of the goal on the descending branch of the trajectory.

Determine the maximum possible value of the tangent of the angle at which the pucks can be launched from the puck launcher, if the trajectory of the puck, considered as a material point, in the plane of its flight in a coordinate system with the center at $O$ and the x-axis directed along the ice surface, is described by the equations: $\left\{\begin{array}{l}x=d+V_{0} t \cos \alpha, \\ y=V_{0} t \sin \alpha-\frac{g t^{2}}{2} .\end{array}\right.$

For simplification of calculations, assume that the acceleration due to gravity $g=10 \text{ m} / \text{s}^{2}$.
Solution. Introduce a coordinate system with the center at point O. The x-axis is directed towards the goal line.

Express time from the first equation of the system and substitute into the second:
$y(x)=\frac{V_{0}(x-d)}{V_{0} \cos \alpha} \sin \alpha-\frac{g\left(\frac{x-d}{V_{0} \cos \alpha}\right)^{2}}{2}=(x-d) \operatorname{tg} \alpha-\frac{g}{2}\left(\frac{x-d}{V_{0} \cos \alpha}\right)^{2} \Rightarrow$ $y(x)=(x-d) \operatorname{tg} \alpha-\frac{g}{2} \frac{(x-d)^{2}}{V_{0}^{2}}\left(1+\operatorname{tg}^{2} \alpha\right)$.

To ensure the puck is below the conditional line for any value of $x$, the condition $y_{b}(x)=x \frac{1}{25} \geq y(x)$ or the inequality $y_{b}(x)-y(x) \geq 0$ must be satisfied for any $x \in[16 ; 25]$. Since the trajectory of the outgoing pucks must intersect the plane of the goal along the descending branch of the trajectory, the inequality $x \frac{1}{25}-(x-d) \operatorname{tg} \alpha+\frac{g}{2} \frac{(x-d)^{2}}{V_{0}^{2}}\left(1+\operatorname{tg}^{2} \alpha\right) \geq 0$ must be satisfied for all $x$.

Rewrite the inequality in a more convenient form and note that this inequality can only be satisfied with a non-positive discriminant:

$$
\frac{g}{2} \frac{(x-d)^{2}}{V_{0}^{2}}\left(1+\operatorname{tg}^{2} \alpha\right)-(x-d)\left(\operatorname{tg} \alpha-\frac{1}{25}\right)+\frac{d}{25} \geq 0 \Leftrightarrow D=\left(\operatorname{tg} \alpha-\frac{1}{25}\right)^{2}-4 \frac{d}{25} \frac{g}{2} \frac{1}{V_{0}^{2}}\left(1+\operatorname{tg}^{2} \alpha\right) \leq 0
$$

$\left(\operatorname{tg} \alpha-\frac{1}{25}\right)^{2}-\frac{4 d}{5 V_{0}^{2}}\left(1+\operatorname{tg}^{2} \alpha\right) \leq 0$

$\left(1-\frac{4 d}{5 V_{0}^{2}}\right) \operatorname{tg}^{2} \alpha-\frac{2}{25} \operatorname{tg} \alpha+\left(\frac{1}{25}\right)^{2}-\frac{4 d}{5 V_{0}^{2}} \leq 0$

$\frac{D}{4}=\left(\frac{1}{25}\right)^{2}-\left(1-\frac{4 d}{5 V_{0}^{2}}\right)\left(\frac{1}{25^{2}}-\frac{4 d}{5 V_{0}^{2}}\right)=\frac{4 d}{5 V_{0}^{2}} \frac{1}{25^{2}}+\frac{4 d}{5 V_{0}^{2}}\left(1-\frac{4 d}{5 V_{0}^{2}}\right)=\frac{4 d}{5 V_{0}^{2}} \frac{626}{625}-\left(\frac{4 d}{5 V_{0}^{2}}\right)^{2}$

For $\mathrm{d}=16 \mathrm{~m}, V_{0}=20 \mathrm{M} / \mathrm{c}$

$\frac{4 \cdot 16}{5 \cdot 400} \cdot \frac{626}{625}-\left(\frac{4 \cdot 16}{5 \cdot 400}\right)^{2}=\frac{4}{5 \cdot 25} \cdot \frac{626}{25^{2}}-\left(\frac{4}{5 \cdot 25}\right)^{2}=\frac{4}{25^{2} \cdot 25}\left(\frac{626}{5}-4\right)=\frac{4 \cdot 606 \cdot 2}{25^{2} \cdot 25 \cdot 5 \cdot 2}$

$\operatorname{tg} \alpha=\left(\frac{1}{25} \pm \frac{2}{25 \cdot 5} \sqrt{\frac{1212}{10}}\right) /\left(\frac{121}{125}\right)=\left(\frac{1}{25} \pm \frac{2}{25 \cdot 5} \sqrt{121.2}\right) \cdot \frac{125}{121}=\frac{5 \pm 2 \sqrt{121.2}}{121} \approx \frac{5 \pm 22}{121}=\left[\begin{array}{l}27 / 121 \\ -17 / 121\end{array}\right.$

$\operatorname{tg} \alpha \in\left[0 ; \frac{5+2 \sqrt{121.2}}{121}\right) \approx[0 ; 0.223)$

Answer: $\frac{5+2 \sqrt{121.2}}{121}$

## Solution for Variant #3 (Mathematics - 11th Grade)