Answer
(a) $4060\,\mathrm{m}$
(b) $33.2\,\mathrm{s}$
(c) $862\,\mathrm{s}$
Work Step by Step
During the upward acceleration of the rocket,
$v_0=0$, $a=30.0\,\mathrm{m/s^2}$ and $\Delta x=1000\,\mathrm{m}$
The speed reached by the rocket when the engines stop is
\begin{align*}
v^2&=v_0^2+2a\Delta x\\
&=0+2(30.0)(1000)\\
&=60000\\
v&=\sqrt{60000}\\
v&=245\,\mathrm{m/s}
\end{align*}
(a) Once the engines stop, the rocket continues to ascend but with a deceleration of $a=9.8\,\mathrm{m/s^2}$. For this part of the asend, $v_0=245\,\mathrm{m/s}$ and $v=0$.
The vertical displacement during this phase is:
\begin{align*}
v^2&=v_0^2+2a\Delta x\\
0&=(245)^2+2(-9.8)\Delta x\\
\Delta x&={60000\over 19.6}\\
&= 3060\,\mathrm{m}
\end{align*}
Maximum altitude reached is the sum of the altitudes climbed with the engines on and with the engines off. I.e., $1000+3060=4060\,\mathrm{m}$
(b) To compute the time taken during the ascend with engines on we have:
$v_0=0$, $\Delta x=1000\,\mathrm{m}$ and $a=30.0\,\mathrm{m/s^2}$.
Using the kinematic equation:
\begin{align*}
\Delta x&=v_0t+{1\over 2} at^2\\
1000&=0+{1\over 2}(30.0)t_1^2\\
t_1^2&={1000\over 15.0}\\
t_1^2&=66.7\\
t_1&=\sqrt{66.7}\\
t_1&=8.2\,\mathrm{s}
\end{align*}
And for time taken to ascend while the engines are off:
$v_0=245\,\mathrm{m/s}$, $v=0$ and $a=-9.8\,\mathrm{m/s^2}$.
Then,
\begin{align*}
v&=v_0+at\\
0&=245+(-9.8)t_2\\
t_2&={245\over 9.8}\\
t_2&=25\,\mathrm{s}
\end{align*}
Total time for the rocket to reach its maximum height is $t_1+t_2=8.2+25=33.2\,\mathrm{s}$.
(c) During the initial free-fall,
$v_0=0$, $a=9.8\,\mathrm{m/s^2}$ and $t=0.500\,\mathrm{s}$
We first calculate the distance through which the rocket falls in this time:
\begin{align*}
\Delta x &=v_0t+{1\over 2}at^2\\
&=0+{1\over 2}(9.8)(0.500)^2\\
&=1.23\,\mathrm{m}
\end{align*}
The distance that the rocket falls after the parachute is deployed is then $\Delta x=4060-1.23=4059\,\mathrm{m}$.
Further the final speed during the initial free-fall will be the initial speed for the descend with the parachute and is:
\begin{align*}
v&=v_0+at\\
&=0+(9.8)(0.500)\\
&=4.9\,\mathrm{m/s}
\end{align*}
Hence for the final descend with the parachute we have,
$\Delta x=4059\,\mathrm{m}$, $v=4.9\,\mathrm{m/s}$ and $a=0$. The time taken for this is:
\begin{align*}
t&={\Delta x\over v}\\
&={4059\over 4.9}\\
&=828.3\,\mathrm{s}
\end{align*}
The time for the total trip is then $t=828.3+0.500+33.2=862\,\mathrm{s}$.
