Answer
(a) $v=297\,\mathrm{m/s}$
(b) $a=3.66\,\mathrm{m/s^2}$
(c) $t=108\,\mathrm{s}$
Work Step by Step
(a) During the free-fall,
$\Delta x=8000\,\mathrm{m}$, $v_0=200\,\mathrm{m/s}$ and $a=3.00\,\mathrm{m/s^2}$.
To find the speed of the probe at the end of the free-fall, we use the kinematic equation:
\begin{align*}
v^2&=v_0^2+2a\Delta x\\
&=(200)^2+2(3.00)(8000)\\
&=40000+48000\\
&=88000\\
v&=\sqrt{88000}\\
&=297\,\mathrm{m/s}
\end{align*}
(b) To determine the minimum constant deceleration during the remainder $12000 \mathrm{m}$ fall, we use
$v=20.0\,\mathrm{m/s}$, $v_0=297\,\mathrm{m/s}$ and $\Delta x=12000\,\mathrm{m}$.
Using
\begin{align*}
v^2=v_0^2+2a\Delta x\\
(20.0)^2&=(297)^2+2a(12000)\\
400&=88200+a(24000)\\
a&=-{87800\over 24000}\\
a&=-3.66\,\mathrm{m/s^2}
\end{align*}
(c) Time taken for the free-fall depends on:
$v=297\,\mathrm{m/s}$, $v_0=200\,\mathrm{m/s}$ and $a=3.00\,\mathrm{m/s^2}$
Using the kinematic equation:
\begin{align*}
v&=v_0+at\\
297&=200+(3.00)t_1\\
t_1&={97\over 3.00}\\
t_1&=32\,\mathrm{s}
\end{align*}
Time-taken for the descend with the parachute depends on:
$v_0=297\,\mathrm{m/s}$, $a=-3.66\,\mathrm{m/s^2}$ and $\Delta x=12000\,\mathrm{m}$.
Hence,
\begin{align*}
\Delta x&=v_0t+{1\over 2} at^2\\
12000&=(297)t_2+{1\over 2}(-3.66)t_2^2\\
1.833t_2^2-297t_2+12000&=0\\
\mathrm{On \,\,solving},\\
t_2&=76\,\mathrm{s}
\end{align*}
Total time taken to land $=t_1+t_2=32+76=108\,\mathrm{s}$.
