Answer
a. $a=-32\ ft/s^2, v_0=128\ ft/sec, s_0=6\ ft$
b. $118\ ft$, see graph
c. $4\ sec$, $262\ ft$
Work Step by Step
a. Step 1. Rewrite the given equation
$s(t)=s_0+v_0t+\frac{1}{2}at^2$
Let
$x=s_0, y=v_0, z=a$
We have
$s(t)=x+yt+\frac{1}{2}zt^2$.
Step 2. Using the three points on the curve, we can set up a system of equations as
$\begin{cases} x+2y+2z=198 \\ x+5y+\frac{25}{2}z=246 \\ x+8y+32z=6 \end{cases}$ or $\begin{cases} x+2y+2z=198 \\ 2x+10y+25z=492 \\ x+8y+32z=6 \end{cases}$
Step 3. Write the system in matrix form; we have:
$\begin{bmatrix} 1 & 2 & 2 & | & 198 \\ 2 & 10 & 25 & | & 492 \\ 1 & 8 & 32 & | & 6 \end{bmatrix} \begin{array} ..\\-2R1+R2\to R2\\ -R1+R3\to R3 \end{array}$
Step 4. Perform row operations given to the right of the matrix:
$\begin{bmatrix} 1 & 2 & 2 & | & 198 \\ 0 & 6 & 21 & | & 96 \\ 0 & 6 & 30 & | & -192 \end{bmatrix} \begin{array} ..\\..\\ -R2+R3\to R3 \end{array}$
Step 5. Perform row operations given to the right of the matrix:
$\begin{bmatrix} 1 & 2 & 2 & | & 198 \\ 0 & 6 & 21 & | & 96 \\ 0 & 0 & 9 & | & -288 \end{bmatrix} \begin{array} ..\\..\\ .. \end{array}$
Step 6. The last equation gives $9z=-288$; thus $z=-32$. Using back-substitution, we can get $y=128, x=6$. Thus, the solutions are $a=-32\ ft/s^2, v_0=128\ ft/sec, s_0=6\ ft$
b. Find and interpret $s(7)$. Identify your solution as a point on the graph shown.
With the known equation $s(t)=6+128t-16t^2$, we have $s(7)=6+128(7)-16(7)^2=118\ ft$ which means that at $t=7 sec$, the height of the football is $118\ ft$; see graph for the point.
c. We can find the time for the maximum height as $t=-\frac{128}{2(-16)}=4\ sec$ and the maximum height as $s(4)=6+128(4)-16(4)^2=262\ ft$
