5. The star "tau" in the constellation Cetus has a planetary system. On the third planet from the star, there is a very unusual gravity: the acceleration due to gravity is $g=10 \mathrm{~m} / \mathrm{c}^{2}$ up to a height of $h=10$ m from the surface, but above that, it is half as much. An astronaut throws a body straight up from the surface of the planet with a speed such that on Earth it would rise to a height of $H=20$ m. To what height above the surface of this planet will the body rise?
5. Since the height of the body's rise on Earth is $H$, the initial velocity of the thrown body is

$$
v_{0}=\sqrt{2 g H}
$$

On the planet from the "tau Ceti" system, the body will move to a height $h$ with acceleration $g$, and then with acceleration $g / 2$. Therefore, above this boundary, the body will rise to a height

$$
h_{0}=\frac{v_{1}^{2}}{2(g / 2)}=\frac{v_{1}^{2}}{g}
$$

where $v_{1}$ is the speed the body had at height $h$ from the surface. Let's find this speed. While the body was moving with acceleration $g$, the laws of its motion had the form

$$
\begin{aligned}
& x(t)=v_{0} t-\frac{g t^{2}}{2} \\
& v_{x}(t)=v_{0}-g t
\end{aligned}
$$

The body will rise to height $h$ in such a time $t_{1}$, which is determined from the equation

$$
\frac{g t_{1}^{2}}{2}-v_{0} t_{1}+h=0
$$

Solving this quadratic equation, we get

$$
\left(t_{1}\right)_{1,2}=\frac{v_{0} \pm \sqrt{v_{0}^{2}-2 g h}}{g}
$$

(we need the smaller root - with the minus sign). Substituting this time into the second equation (*), we find the speed of the body at the moment of acceleration change

$$
v_{1}=v_{x}\left(t_{1}\right)=\sqrt{v_{0}^{2}-2 g h}
$$

Substituting this speed into expression (*), we find the height of the body's rise above the level where its acceleration changes

$$
h_{0}=\frac{v_{0}^{2}-2 g h}{g}=\frac{v_{0}^{2}}{g}-2 h
$$

Therefore, above the planet's surface, the body will rise to a height

$$
H_{1}=h+h_{0}=\frac{v_{0}^{2}}{g}-h
$$

Substituting now this value of the initial velocity into this formula, we get finally

$$
H_{1}=2 H-h=30 \mathrm{~m}
$$

Criteria for evaluating the problem (maximum score for the problem - 2 points)