Answer
a) $27\;\rm m/s^2$
b) $392.8\;\rm m/s$
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 m/s$^2$ for 16 seconds. The second stage starts when its fuel runs out and then it moves up at the free-fall acceleration −9.8m/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 acceleration of the rocket, we need to find the final height at the moment the motor stops.
We know that the rocket moves under a constant upward acceleration for 16 s and then it moves at the free-fall acceleration.
After 20 s from launch, it reaches 5100 m.
The velocity of the rocket after the motor stops is given by
$$v_{y1}=v_{iy,1}+a_{y1}t_1=0+16a_{y1}$$
$$v_{y1}= 16a_{y1}\tag 1$$
Hence, the vertical distance traveled in this stage is given by
$$y_1=\overbrace{ y_{i1}}^{0} +\overbrace{ v_{iy}t_1}^{0} +\frac{1}{2}a_{y1}t_1^2 =\frac{1}{2}a_{y1}\cdot 16^2$$
$$y_1=128a_{y1}\tag 2$$
Now we know that the total vertical distance is given by
$$y_2=y_1+v_{i2}t_2+\frac{1}{2}a_{y2}t_2^2$$
In the second stage, its initial velocity is the final velocity of the first stage. Also note that $a_{y2}=-g$.
$$y_2=y_1+v_{y1}t_2+\frac{1}{2}a_{y2}t_2^2$$
Plugging from (1) and (2);
$$y_2= 128a_{y1}+ 16a_{y1}t_2 -\frac{1}{2}gt_2^2$$
Plugging the known and solving for $a_{y1}$;
$$5100= 128a_{y1}+ 16a_{y1}\cdot 4-\frac{9.8}{2} \cdot 4^2$$
$$5100= 192a_{y1}-78.4$$
Therefore,
$$a_{y1}=\color{red}{\bf 27}\;\rm m/s^2$$
b)
We need to find the speed of the rocket at 5100 m above the ground.
$$v_{y2}=v_{y1}-gt_2$$
Plugging from (1) and plugging the known;
$$v_{y2}=16a_{y1}-g t_2=(16\cdot 27) -(9.8\cdot 4)$$
$$v_{y2}=\color{red}{\bf 392.8}\;\rm m/s$$
