3. The sledge run consists of a straight slope $AB$ and a horizontal section $BC$. Point $A$ is 5 m away from the nearest point $H$ on the horizontal ground surface. The distance $HC$ is 3 m, and point $B$ lies on the segment $HC$. Find the distance from point $H$ to point $B$ so that the time of the sledge's motion from rest along the broken line $ABC$ is minimized. Assume a uniform gravitational field, neglect friction, air resistance, and any change in the magnitude of the sledge's velocity vector at the junction point $B$. The acceleration due to gravity is considered to be $10 \mathrm{m} / \mathrm{c}^{2}$.
Answer: $\frac{5 \sqrt{3}}{3}$

m. Solution. Let $S$ be the distance $HC$, $H$ be the distance $AH$, and $x$ be the required distance. Then the time of fall $t_{\text {AH }}=\sqrt{\frac{2 H}{g}}$, the time of descent from point $A$ to point $B$ is $t_{A B}=\frac{t_{A H}}{\sin \angle A B H}$, where $\sin \angle A B H=\frac{H}{\sqrt{H^{2}+x^{2}}}$, and the speed at point $B$ is $V=\sqrt{2 g H}$. Therefore, the time of movement of the sleds is: $t_{A B C}=t_{A B}+t_{B C}=\frac{2 \sqrt{H^{2}+x^{2}}+S-x}{\sqrt{2 g H}}$.

We need to find the minimum of the function $f(x)=2 \sqrt{H^{2}+x^{2}}-x$. The derivative of this function $f^{\prime}(x)=\frac{2 x}{\sqrt{H^{2}+x^{2}}}-1$ equals 0, so $2 x=\sqrt{H^{2}+x^{2}} \Rightarrow 4 x^{2}=H^{2}+x^{2} \Rightarrow x=\frac{H}{\sqrt{3}}$. It is not difficult to show that this will be a minimum point. For $H=5$ we get: $x=\frac{5}{\sqrt{3}}$.

Answer to variant 162: $\sqrt{3}$ m.

Answer to variant 163: $\frac{4 \sqrt{3}}{3} \mathrm{m}$.

Answer to variant 164: $\frac{2 \sqrt{3}}{3}$ m.