Answer
a) $\rm 54.8\; km$
b) $227.6\;\rm s$
c) See the figure below.
Work Step by Step
The rocket undergoes three stages of motion. The first stage starts from rest when it speeds up while moving up at a constant acceleration of $30\;\rm m/s^2$ for 30 seconds. The second stage starts when its fuel runs out and then it moves up at the free-fall acceleration $-9.8\;\rm m/s^2$ which means it is still moving up but slowing down until it stops. And the third stage starts when it starts to fall from rest toward the ground under the free-fall acceleration as well.
a)
To find the rocket's maximum altitude, we need to find the vertical distance traveled on the first and second stages.
$$ y_{max}=\Delta y_1+\Delta y_2\tag 1$$
In the first stage, the rocket accelerates from rest
$$\Delta y_1=\overbrace{ y_i}^{0} +\overbrace{ v_{iy}t}^{0} +\frac{1}{2}a_yt^2=\frac{1}{2}\cdot 30\cdot 30^2=\bf 13500\;\rm m$$
$$\Delta y_1=\color{blue}{\bf 13500}\;\rm m \tag 2$$
Now we need to find the vertical distance traveled in the second stage.
$$v_{fy,2}^2=v_{iy,2}^2+2a_{y2}\Delta y_2$$
Thus,
$$\Delta y_2=\dfrac{v_{fy,2}^2-v_{iy,2}^2}{2a_{y2}}$$
Noting that:
- $v_{fy,2}=0\;\rm m/s$ since it stops before it falls back again.
- $v_{iy,2}$ is the final velocity of the first stage
- $a_{y2}=-g=-9.8\;\rm m/s^2$
$$\Delta y_2=\dfrac{ -v_{iy,2}^2}{-2\cdot 9.8}$$
$$\Delta y_2=\dfrac{v_{iy,2}^2}{19.6}\tag 3$$
Now we need to find the final velocity of the first stage which is the initial velocity of the second stage.
$$v_{fy,1}=v_{iy,1}+a_yt_1=0+30\cdot 30=\bf 900\;\rm m/s$$
Plugging into (3);
$$\Delta y_2=\dfrac{900^2}{19.6} =\color{blue}{\bf 41326.5}\;\rm m\tag 4$$
Plugging (2) and (4) into (1);
$$ y_{max}= 13500+41326.5=54826\;\rm m\approx \color{red}{\bf54.8}\;\rm km $$
b)
The time it takes the rocket to hit the ground is the sum of the three stages' time intervals.
The first stage takes 30 s and hence we need to find the time interval of the second and third stages.
$$t_1=\bf 30\;\rm s\tag A$$
$$v_{fy,2}=v_{iy,2}-gt_2$$
$$t_2=\dfrac{v_{fy,2}-v_{iy,2}}{-g}=\dfrac{0-900}{-9.8}=\bf 91.8 \;\rm s\tag B$$
$$\overbrace{ y_{f3}}^{0}=\overbrace{ y_{i3}}^{y_{max}} +\overbrace{ v_{iy,3}t_3}^{0} -\frac{1}{2}gt_3^2$$
The initial velocity of the third stage is zero since it falls back to the ground from rest. And the initial height is the maximum height while the final height of this stage is the ground at which $y=0$.
$$0=y_{max}-4.9t_3^2$$
$$t_3=\sqrt{\dfrac{-y_{max}}{-4.9}}=\sqrt{\dfrac{54826}{4.9}}=\bf 105.8\;\rm s\tag C$$
From A, B and C, we can see that the time interval of the whole rocket's trip in the air is given by
$$t_{tot}=t_1+t_2+t_3=30+91.8 +105.8=\color{red}{\bf 227.6}\;\rm s$$
c)
See the velocity versus time graph of the rocket for the whole trip.
