Answer
(a) $6.12 \mathrm{s}$
(b) $183.67 \mathrm{m}$
(c) $6.12 s$
(d) $-59.95 m/ s$
Work Step by Step
Using Formula (15) and Formula (16) from the book:
\[
\begin{array}{l}
s_{0}+v_{0} t-\frac{g}{2} t^{2} =s(t) \\
v_{0}-g t= v(t)\\
g=9.8 \mathrm{m} / \mathrm{s}^{2}
\end{array}
\]
The maximum height will be reached at a critical point from the position function and therefore at the zero of the velocity function.
(a) $v(t)=60-9.8 t=0$
\[
\Rightarrow t=\frac{-60}{-9.8}=6.12 \mathrm{s}
\]
$60=v_{0}$
(b) $s(6.12)= 6.12^{2} \cdot (-4.9)+6.12 \cdot 60=183.67 \mathrm{m}$
Determine the position at $t=6.12 \mathrm{s}$
\[
s_{0}=0
\]
So it takes the rocket $12.24-6.12=6.12$ s to drop back to the ground from its highest point.
The projectile hits the ground at:
(c) $s(t)=60 t-4.9 t^{2}=0$
$\Rightarrow t=\frac{-60}{-4.9}=12.24 \mathrm{s}$
Velocity at $t=12.24 \mathrm{s}$:
(d) $v(12.24)=60-9.8 \cdot 12.24=-59.95 \mathrm{m} / \mathrm{s}$
