#### Answer

a) $v(t)=-32t(m/s)$, $a(t)=-32(m/s^2)$ and the ball's speed is $32t(m/s)$.
b) It takes the ball about $3.345$ seconds to hit the ground.
c) The ball's velocity at the moment of impact is $-107.04$ feet/second.

#### Work Step by Step

$$s(t)=179-16t^2$$
a) The ball's velocity at time $t$: $$v(t)=\frac{ds}{dt}=-32t(m/s)$$
The ball's speed at time $t$: $$|v(t)|=|-32t|=32t(m/s)$$ (because $t$ is the time so $t\ge0$)
The ball's acceleration at time $t$: $$a(t)=\frac{dv}{dt}=-32(m/s^2)$$
b) The ball would hit the ground when $$s(t)=0$$ $$179-16t^2=0$$ $$t^2=\frac{179}{16}$$ $$t=\frac{\sqrt{179}}{4}\approx3.345(sec)$$ (because $t\ge0$, we do not consider the negative values of $t$ here)
So it takes the ball about $3.345$ seconds to hit the ground.
c) At the moment of impact, which is $t=3.345$, the ball's velocity is $$-32\times3.345=-107.04(ft/sec)$$