### Rocket

The rocket travels a distance: $x_{1}=\frac{1}{2}at_{1}^{2}=\frac{1}{2}(20\frac{m}{s^{2}})(60s)^{2}=36,000m$ during the time when it accelerates.
The rocket velocity at the end of this time is $v=at=(20\frac{m}{s^{2}})(60s)=1200\frac{m}{s}.$
The second half of its path can be found by writing the kinematic equations you have learned, and plugging in $y_{0}=36,000m$ and $v_{0y}=1200\frac{m}{s.}$
(a) The highest point of the path is when $v_{y}=0,$ so $v_{y} = v_{0y}-gt$ => $0 = (1200\frac{m}{s})-(9.8\frac{m}{s^{2}})t$ => $122s = t.$
We then use this time to help find how high the rocket goes up. $y = y_{0}+v_{0y}t-\frac{1}{2}gt^{2} = (36000m)+(1200\frac{m}{s})(122s)-\frac{1}{2}(9.8\frac{m}{s^{2}})(122s)^{2}=110000m.$
(b) The easiest way to find the total time of flight is to solve the equation for $y(t)$ at the time when the rocket has returned to the ground:
$y = y_{0}+v_{0y}t-\frac{1}{2}gt^{2}$ $0 = (36000m)+(1200\frac{m}{s})t-\frac{1}{2}(9.8\frac{m}{s^{2}})t^{2}.$
This is a quadratic equation and has TWO solutions for t. But one is negative and so physically impossible.
The other is $t=270s.$ This is the free fall part of the problem. To find the total time we need to add to this also the 60 seconds when there was acceleration.
The total time is then 330 seconds.