Answer
(b) 0.5 s
(a) 1 s
Work Step by Step
\[
\begin{array}{l}
s(t)=-\frac{g}{2} t^{2}+v_{0} t+s_{0} \\
v(t)=-g t+v_{0}
\end{array}
\]
$g=32 \mathrm{ft} / \mathrm{s}^{2}$
Using Formula (15) and Formula (16) from the book:
(a) $0=s(t)=-16 t^{2}+16 t$
$\Rightarrow t=\frac{-16}{-16}=1 \mathrm{s}$
$v_{0}=16$ and $s_{0}=0$ and $s(t)=0$
(b) $v(t)=-32 t+16=0$
$t=\frac{-16}{-32}=0.5 \mathrm{s}$
The maximal height will be reached at a critical point of the position function and therefore at the zero of the velocity function. After reaching the maximum height, the projectile stops moving up.
