Answer
(a) $-48 \mathrm{ft} / \mathrm{s},16 \mathrm{ft} / \mathrm{s}$
(b) $196 \mathrm{ft}$
(c) $112 \mathrm{ft} / \mathrm{s}$
Work Step by Step
Using Formula (15) and Formula (16) from the book:
$s(t)=s_{0}+v_{0} t-\frac{g}{2} t^{2}$
$v(t)=v_{0}-g t$
$g=32 \mathrm{ft} / \mathrm{s}^{2}$
The maximum height will be reached at a critical point from the position function, and therefore at zero for the velocity function.
(a) $v(t)=112-32 t$
$v(3)=3 \cdot -32+112=16 \mathrm{ft} / \mathrm{s}$
$v(5)=5 \cdot -32+112=-48 \mathrm{ft} / \mathrm{s}$
$v_{0}=112$ , $s_{0}=0$
(b) $v(t)=112-32 t=0$
$t=\frac{-112}{-32}=3.5$
$s(3.5)=-16 \cdot 3.5^{2}+3.5 \cdot 113=196 \mathrm{ft}$
Determine when the projectile hits the ground.
$(\mathrm{c}) s(t)=112 t-16 t^{2}=0$
\[
\Rightarrow t=\frac{-112}{-16}=7
\]
Speed at $t=7$:
$|v(7)|=|112-32 \cdot 7|=|-112|=112 \mathrm{ft} / \mathrm{s}$
