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 = 6 sec
h. the velocity upon impact is - 128 ft per sec
i. find the time t when the velocity of the rocket is 0 ft per sec at t = sec

Explanation:

Step1: Expand the height - function

First, expand $s(t)=-16(t + 2)(t - 6)$. Using the FOIL method, $(t + 2)(t - 6)=t^{2}-6t+2t - 12=t^{2}-4t - 12$. Then $s(t)=-16(t^{2}-4t - 12)=-16t^{2}+64t + 192$. In the general form $s(t)=at^{2}+bt + c$, here $a=-16$, $b = 64$, and $c = 192$.

Step2: Recall the meaning of $c$

When $t = 0$, $s(0)=a\times0^{2}+b\times0 + c=c$. So $c$ represents the initial height of the object.

Step3: Find the time when velocity is 0

The velocity function $v(t)$ is the derivative of the height - function $s(t)$. Since $s(t)=-16t^{2}+64t + 192$, using the power rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we have $v(t)=s^\prime(t)=-32t+64$.
Set $v(t)=0$, then $-32t + 64 = 0$.
Solve for $t$:
\[

$$\begin{align*} -32t&=-64\\ t& = 2 \end{align*}$$

\]

Answer:

f. B. c is the initial height of the object.
i. 2