average rate of change suppose a ball thrown in the air has a height h(t) = -16t² + 40t + 6 feet after t…

Explanation:

Step1: Recall average - rate - of - change formula

The average rate of change of a function $y = h(t)$ from $t=a$ to $t = b$ is $\frac{h(b)-h(a)}{b - a}$. Here, $a = 0$, $b = 2$, and $h(t)=-16t^{2}+40t + 6$.

Step2: Calculate $h(0)$

Substitute $t = 0$ into $h(t)$: $h(0)=-16(0)^{2}+40(0)+6=6$.

Step3: Calculate $h(2)$

Substitute $t = 2$ into $h(t)$: $h(2)=-16(2)^{2}+40(2)+6=-16\times4 + 80+6=-64 + 80+6=22$.

Step4: Calculate the average rate of change

$\frac{h(2)-h(0)}{2 - 0}=\frac{22 - 6}{2}=\frac{16}{2}=8$.

Answer:

B. 8 feet/second