Answer: a) 75 km; b) yes. Solution. Let $a=20$ m/s$^2$, $\tau=50$ 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 $y_{\max }=75$ 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 $g$ is subtracted from the acceleration during the first stage.