Question

the height, h (in feet) of a model rocket launched from the roof of a building at t seconds is given by h = s(t)= - 16(t + 2)(t - 6)

f. what does the value of c in the formula s(t)=at²+bt + c tell us about the object?
a. c is the maximum height
b. c is the initial height of the object
c. c is the velocity when the height is maximum
d. c is the initial velocity and the slope of the tangent line to the function at t = 0
g. when does it hit the ground? the object hits the ground at t = sec

Explanation:

Step1: Recall the position - velocity - acceleration relationship

The general position function is $s(t)=at^{2}+bt + c$. When $t = 0$, $s(0)=a\times0^{2}+b\times0 + c=c$. So $c$ represents the initial position (height in this case) of the object.

Step2: Expand the given function

We start with $h=s(t)=-16(t + 2)(t - 6)=-16(t^{2}-6t + 2t-12)=-16(t^{2}-4t - 12)=-16t^{2}+64t + 192$.

Step3: Find when the object hits the ground

The object hits the ground when $s(t)=0$. So we set $-16(t + 2)(t - 6)=0$. Using the zero - product property, if $AB = 0$, then either $A = 0$ or $B = 0$. Here, $t+2 = 0$ gives $t=-2$ and $t - 6=0$ gives $t = 6$. Since time $t\geq0$ in the context of the problem (we are talking about time after launch), we discard $t=-2$.

Answer:

f. B. c is the initial height of the object
g. 6