Answer
(a) The maximum height that the stage-two rocket reaches above the launch pad is $3090m$.
(b) It takes about $38.6s$ to fall back to the launch pad after the end of the stage-two firing.
(c) The speed of the rocket just as it reaches the launch pad is $246\frac{m}{s}$.
Work Step by Step
Let the origin of our system be at the launch pad with the (+)ve y-direction upwards.
We can consider the motion of the rocket to be of 3 stages, each stage with a constant acceleration. We can apply the kinematics equation in the 3 stages seperately.
(a)Let us calculate the distance travelled by the rocket in the 1st stage of its motion. This stage last for $t=25.0s$
Given: $a_{y}=3.50\frac{m}{s^2}$ and $v_{i}=0$
Now, $d_{1}=y-y_{0}=v_{i}t+\frac{1}{2}a_{y}t^2$ , where $d_{1}=$distance travelled by the rocket in the 1st stage of its motion
$y=\frac{1}{2}\times3.50\times25.0^2=1094m$
The rocket travels a distance of $1094m$ in the 1st stage of the motion.
The velocity of the rocket at the end of 1st stage is
$v_{f}=v_{i}+a_{y}t=3.50\times25.0=87.5\frac{m}{s}$
This velocity is the initial velocity of the rocket for the 2nd stage of its motion
The 2nd stage last for t=10.0s.
Let us calculate the distance travelled by the rocket in the 2nd stage of motion.
$d_{2}=y-y_{0}=\frac{v_{f}+v_{i}}{2}t=\frac{132.5+87.5}{2}\times10.0=1100m$, where $d_{2}=$distance travelled by the rocket in the 2nd stage of its motion
$y=1100+1094=2194m$
So, at the end of the 2nd stage the rocket travelled a distance of $2194m$ from the launch pad.
For the 3rd stage of the motion, the velocity at the end of the 2nd stage becomes the initial velocity of the 3rd stage and $a_{y}=-9.80\frac{m}{s^2}$
Now, let us calculate the distance travelled by the rocket till it reaches peak height in the 3rd stage of its motion.
$d_{3}=y-y_{0}=\frac{v_{f}^2-v_{i}^2}{2a_{y}}=\frac{0^2-132.5^2}{-2\times9.80}=895.73m$ ,where $d_{3}=$distance travlled by the rocket in the 3rd stage of its motion ,and $v_{f}=0$ at peak height
$y=2194+895.73=3090m$
The maximum height that the stage-two rocket reaches above the launch pad is $3090m$.
(b) Let us calculate the duration of the 3rd stage of the rocket's motion.
$y-y_{0}=v_{i}t+\frac{1}{2}a_{y}t^2$
$0-2194=132.5t-\frac{9.80}{2}t^2$
$\frac{9.80}{2}t^2-132.5t-2194=0$
Using quadratic formula, we get
$t=\frac{132.5+-\sqrt ((-132.5)^2+4\times\frac{9.80}{2}\times2194)}{2\times\frac{9.80}{2}}=38.63s=38.6s$
Thus, it takes about $38.6s$ to fall back to the launch pad after the end of the stage-two firing.
(c) Let us now calculate the velocity of the rocket at the end of the 3rd stage of its motion ,i.e, velocity when the rocket falls back to the launch pad.
$v_{f}=v_{i}+a_{y}t=132.5-9.80\times38.63=-246\frac{m}{s}$
The speed of the rocket just as it reaches the launch pad is $246\frac{m}{s}$.
