7. (15 points) A stone with a mass of $m=6002$ was thrown at an angle $\alpha$ to the horizontal with an initial velocity of $v_{0}=20 \mathrm{M} / \mathrm{s}$. In the presence of air resistance proportional to the stone's velocity (proportionality coefficient $k=0.1(\mathrm{H} \cdot \mathrm{s}) / \mathrm{m})$, the maximum height reached by the stone is $h=10$ m. The work done by the resistance force over this time interval is $A=30$ Joules. Determine the acceleration of the stone at the highest point of the trajectory. The acceleration due to gravity is $g=10 \mathrm{M} / \mathrm{s}^{2}$.
Answer: $\approx 10.1 \frac{M}{c^{2}}$

Solution. The law of conservation of energy in this situation is as follows: $\frac{m v_{0}^{2}}{2}=\frac{m v^{2}}{2}+m g h+A$.

As a result, we find the velocity at the highest point:

$$
v_{\kappa}=\sqrt{v_{0}^{2}-2 g h-\frac{2 A}{m}}=\sqrt{20^{2}-2 \cdot 10 \cdot 10-\frac{2 \cdot 30}{0.6}}=10 \frac{M}{c}
$$

The acceleration associated with the resistance force is directed against the velocity, i.e., parallel to the $O X$ axis, and it is equal to: $a_{x}=\frac{F_{\text {resist }}}{m}=\frac{k v_{\kappa}}{m}=\frac{0.1 \cdot 10}{0.6}=\frac{5}{3} \frac{M}{c^{2}}$.

The acceleration associated with the gravitational force is directed vertically downward:

$a_{y}=g=10 \mathrm{M} / c^{2}$.

The total acceleration: $a=\sqrt{a_{x}^{2}+a_{y}^{2}}=\sqrt{\left(\frac{5}{3}\right)^{2}+10^{2}} \approx 10.1 \frac{\mathcal{M}}{c^{2}}$.