7. (15 points) A stone with a mass of $m=8002$ was thrown at an angle $\alpha$ to the horizon 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 $k=0.2(H \cdot s) / M)$, the maximum height reached by the stone is $h=10 \mathrm{m}$. The work done by the resistance force over this time interval is $A=40$ Joules. Determine the acceleration of the stone at the highest point of the trajectory. The acceleration due to gravity $g=10 \mathrm{M} / \mathrm{s}^{2}$.
Answer: $\approx 10.3 \frac{M}{c^{2}}$

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:

$$
v_{\kappa}=\sqrt{v_{0}^{2}-2 g h-\frac{2 A}{m}}=\sqrt{20^{2}-2 \cdot 10 \cdot 10-\frac{2 \cdot 40}{0.8}}=10 \frac{M}{c}
$$

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: $a_{x}=\frac{F_{\text {resist }}}{m}=\frac{k v_{\kappa}}{m}=\frac{0.2 \cdot 10}{0.8}=2.5 \frac{\mathrm{M}}{c^{2}}$.

(3 points)

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

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

The total acceleration: $a=\sqrt{a_{x}^{2}+a_{y}^{2}}=\sqrt{2.5^{2}+10^{2}} \approx 10.3 \frac{M}{c^{2}}$.