Answer
See explanation.
Work Step by Step
Applying Formula (15) and Formula (16) from the book:
\[
\begin{array}{l}
v_{0} t+s_{0} -\frac{g}{2} t^{2}=s(t)\\
v_{0}-g t=v(t) \\
g=9.8 \mathrm{m} / \mathrm{s}^{2}
\end{array}
\]
(a) The final and initial speed seem to be identical
(b) $0=s(t)=v_{0} t-4.9 t^{2}$
$\Rightarrow t=\frac{v_{0}}{4.9} \mathrm{s}$ is the moment that the rockets return to ground level.
\[
\begin{array}{l}
\qquad s_{0}=0 \text { and } s(t)=0 \\
v_{0} -9.8 t=v(t)\\
Velocity =\frac{v_{0}}{4.9}:\\
v\left(\frac{v_{0}}{4.9}\right)=-9.8 \cdot \frac{v_{0}}{4.9}+v_{0}=-2 v_{0}+v_{0}=-v_{0}
\end{array}
\]
So the speed is $\left|v\left(\frac{v_{0}}{4.9}\right)\right|=v_{0}$
