To analyze the height of an object thrown upward, modeled by the polynomial function $h(t) = -16t^2 + 64t + 80$, we can find its zeros, the time it reaches a specific height, and its height at a specific time. First, to find the zeros which indicate when the object hits the ground, set $h(t) = 0$. Factoring the equation $-16t^2 + 64t + 80 = 0$ gives $-16(t^2 - 4t - 5) = 0$, which factors further to $-16(t - 5)(t + 1) = 0$. This yields $t = 5$ and $t = -1$. Since time cannot be negative, we discard $t = -1$, concluding the object hits the ground at 5 seconds. Second, to find when the object is at a specific height of 80 feet, set $h(t) = 80$. Subtracting 80 from both sides gives $-16t^2 + 64t = 0$. Factoring out the greatest common factor gives $-16t(t - 4) = 0$, yielding $t = 0$ and $t = 4$. Thus, the object is at 80 feet at 0 and 4 seconds. Third, to find the height at a specific time, such as $t = 2$ seconds, substitute 2 into the function. The result is $h(2) = -16(2)^2 + 64(2) + 80 = 144$. The object is at 144 feet after 2 seconds.