1. The engines of a rocket launched vertically upward from the Earth's surface, providing the rocket with an acceleration of $30 \mathrm{~m} / \mathrm{c}^{2}$, suddenly stopped working 20 seconds after launch. To what maximum height will the rocket rise? Can this rocket pose a danger to an object located at an altitude of 20 km? Air resistance is not taken into account, and the acceleration due to gravity is considered to be $10 \mathrm{~m} / \mathrm{c}^{2}$.
Answer: a) 24 km; b) yes. Solution. Let $a=30 \mathrm{m} / \mathrm{s}^{2}, \tau=20$ s. During the first part of the motion, when the engines were working, the speed and the height gained are respectively: $V=a t, y=\frac{a t^{2}}{2}$. Therefore, at the moment the engines stop: $V_{0}=a \tau, y_{0}=\frac{a \tau^{2}}{2}$ - this will be the "zero" moment for the second part.

During the second part: $V=V_{0}-g t=a \tau-g t, y=y_{0}+V_{0} t-\frac{g t^{2}}{2}=\frac{a \tau^{2}}{2}+a \tau t-\frac{g t^{2}}{2}$. At the maximum height, $V=a \tau-g t=0$, so the time when the rocket will be at the maximum height is: $t=\frac{a \tau}{g}$. Therefore, the maximum height the rocket will reach is: $y_{\max }=\frac{a \tau^{2}}{2}+a \tau \frac{a \tau}{g}-\frac{g}{2} \frac{a^{2} \tau^{2}}{g^{2}}=\frac{a \tau^{2}(g+a)}{2 g}$. Substituting the numbers gives $y_{\max }=24$ km.

Grading criteria: 20 points: correct solution and correct answers to both questions; 15 points: correct solution and correct answer to the first question, incorrect or omitted answer to the second question; 10 points: correct equations and logic, but incorrect answer due to arithmetic error; 5 points: correct equations and correct answer for the case where the acceleration on the first stage is reduced by $g$.