Answer
a. $-32t$ ft/s, $32t$ ft/s, and $-32ft/s^2$.
b. $3.3$ sec.
c. $-107$ ft/s downward.
Work Step by Step
a. Given $s(t)=179-16t^2$, we can find the velocity as $v(t)=s'(t)=-32t$ ft/s, speed $|v(t)|=32t$ ft/s, and acceleration $a(t)=v'(t)=-32ft/s^2$.
b. When the ball hits the ground, $s(t)=0$, we have $179-16t^2=0$, which gives $t=\sqrt {179/16}\approx3.3$ sec.
c. At $t=\sqrt {179/16}$, the velocity $v(t)=-32(\sqrt {179/16})\approx-107$ ft/s (negative means downward).
