Answer
$v(t)=-32t+16$
Work Step by Step
Define "up" as the positive direction of motion.
Acceleration is the derivative of velocity
$a(t)=v'(t)=-32\ \ \mathrm{f}\mathrm{t}/\mathrm{s}^{2}$
$v(t)=\displaystyle \int(-32)dt=-32t+C\ \ \mathrm{f}\mathrm{t}/\mathrm{s}$
Given that $ v(0)=+16\qquad$(upward)
$v(0)=0+C$
$16=C$
Thus,
$v(t)=-32t+16$
