5. The Stalker, to detect a gravitational anomaly (an area where the acceleration due to gravity changes sharply in magnitude), throws a small nut from the surface of the Earth at an angle $\alpha=30^{\circ}$ to the horizontal with a speed of $v_{0}=20 \mathrm{M} / \mathrm{c}$. The normal acceleration due to gravity is $g=10 \mathrm{~m} / \mathrm{c}^{2}$. At the highest point of its trajectory, the nut enters the anomaly zone and continues to move within it. As a result, the nut falls to the ground at a distance of $S=15 \sqrt{3}$ m from the Stalker. Determine the acceleration due to gravity inside the anomaly. (15 points)
Answer: $40 \mathrm{M} / \mathrm{c}^{2}$

Solution. Equations of motion before the anomaly: $x=v_{0} \cos \alpha \cdot t=10 \sqrt{3} \cdot t$ (2 points), $y=v_{0} \sin \alpha \cdot t-\frac{g t^{2}}{2}=10 \cdot t-5 \cdot t^{2} \quad\left(2\right.$ points), $v_{x}=v_{0} \cos \alpha=10 \sqrt{3}$ (2 points), $v_{y}=v_{0} \sin \alpha-g t=10-10 t$ (2 points). From this, we obtain that for the highest point of the flight: $t_{\varepsilon}=1 s, x_{\varepsilon}=10 \sqrt{3} \mathrm{M}$, $y_{\varepsilon}=5$ m (1 point). Equations of motion inside the anomaly: $S=x_{\varepsilon}+v_{x} t=10 \sqrt{3}+10 \sqrt{3} t$ (2 points), $y=0=y_{s}-\frac{g_{a} t^{2}}{2}=5-\frac{g_{a} t^{2}}{2}$ (2 points). Solving this system, we get: $g_{a}=40 \mathrm{M} / \mathrm{c}^{2}$ (2 points).