5. Gavriila was traveling in Africa. On a sunny and windy day, at noon, when the rays from the Sun fell vertically, the boy threw a ball from behind his head at a speed of 5 m/s against the wind at an angle to the horizon. After 1 second, the ball hit him in the stomach 1 m below the point of release. Determine the greatest distance the shadow of the ball moved away from Gavriila's feet. The force acting on the ball from the air is directed horizontally and does not depend on the position and speed. The acceleration due to gravity \( g \) is \( 10 \, \text{m} / \text{s}^2 \).
Answer: 75 cm.

Solution. In addition to the force of gravity, a constant horizontal force $F=m \cdot a$ acts on the body, directed opposite. In a coordinate system with the origin at the point of throw, the horizontal axis $x$ and the vertical axis $y$, the law of motion has the form:

$$
\begin{aligned}
& x(t)=V \cdot \cos \alpha \cdot t-\frac{a t^{2}}{2} \\
& y(t)=V \cdot \sin \alpha \cdot t-\frac{g t^{2}}{2}
\end{aligned}
$$

From the condition, the coordinates of the point at time $\tau$ are known:

$$
\begin{gathered}
0=V \cdot \cos \alpha \cdot \tau-\frac{a \tau^{2}}{2} \\
-H=V \cdot \sin \alpha \cdot \tau-\frac{g \tau^{2}}{2}
\end{gathered}
$$

From this system of two equations with two unknowns, we find the acceleration created by the force acting from the air and the angle at which the throw was made. From the second equation: $\sin \alpha=\frac{1}{V}\left(\frac{g \tau}{2}-\frac{H}{\tau}\right)=\frac{1}{5}\left(\frac{10 \cdot 1}{2}-\frac{1}{1}\right)=\frac{4}{5}$. Therefore, $\cos \alpha=\frac{3}{5}$, and from the first equation: $a=\frac{2 V \cdot \cos \alpha}{\tau}=\frac{2 \cdot 5 \cdot 3}{5 \cdot 1}=6 \mathrm{~m} / \mathrm{c}^{2}$.

The maximum distance of the shadow corresponds to the maximum value of $x$, which is achieved at $t=\frac{V \cos \alpha}{a}$ (the vertex of the parabola). This value is $L=\frac{V^{2} \cos ^{2} \alpha}{2 a}$ $=\frac{5^{2} \cdot 3^{2}}{5^{2} \cdot 2 \cdot 6}=\frac{3}{4} \mathrm{m}$.

For reference: the answer in general form $L=\frac{\tau}{4} \sqrt{V^{2}-\left(\frac{g \tau}{2}-\frac{H}{\tau}\right)^{2}}$ (in variants 171 and 173), $L=\frac{\tau}{4} \sqrt{V^{2}-\left(\frac{g \tau}{2}+\frac{H}{\tau}\right)^{2}}$ (in variants 172 and 173).

Answer to variant 172: 1.6 meters.

Answer to variant 173: 60 cm.

Answer to variant 174: 2 meters.