5. A 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 $v_{0}=10 \mathrm{M} / \mathrm{c}$. The normal acceleration due to gravity $g=10 \mathrm{~m} / \mathrm{c}^{2}$. 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 $S=3 \sqrt{3}$ m from the Stalker. Determine the acceleration due to gravity inside the anomaly. (15 points)
Answer: $250 \mu / c^{2}$

Solution. Equations of motion before the anomaly: $x=v_{0} \cos \alpha \cdot t=5 \sqrt{3} \cdot t$ (2 points), $y=v_{0} \sin \alpha \cdot t-\frac{g t^{2}}{2}=5 \cdot t-5 \cdot t^{2}$ (2 points), $v_{x}=v_{0} \cos \alpha=5 \sqrt{3}$ (2 points), $v_{y}=v_{0} \sin \alpha-g t=5-10 t$ (2 points). From this, we obtain that for the highest point of the flight: $t_{6}=0.5 s, x_{b}=2.5 \sqrt{3} m$, $y_{b}=1.25$ m (1 point). Equations of motion inside the anomaly: $S=x_{b}+v_{x} t=2.5 \sqrt{3}+5 \sqrt{3} t$ (2 points), $y=0=y_{b}-\frac{g_{a} t^{2}}{2}=1.25-\frac{g_{a} t^{2}}{2}$ (2 points).

Solving this system, we get: $g_{a}=250 m / s^{2}$ (2 points).