Question

a ball is straight upward. a position function ( s(t)=\frac{1}{2} g t^{2}+v_{0} t+s_{0} ) can be used to describe the balls height, ( s(t) ), in feet, after ( t ) seconds. a graph of the balls height is shown to the right. complete parts a through c below. a. use the points labeled in the graph to find the values of ( g, v_{0} ), and ( s_{0} ). solve the system of linear equations involving ( g, v_{0} ), and ( s_{0} ) using matrices. ( g= ) ( v_{0}= ) ( s_{0}= )

Explanation:

Step1: Substitute the points into the position function

Given \(s(t)=\frac{1}{2}gt^{2}+v_{0}t + s_{0}\).
When \(t = 1\), \(s(1)=50\), so \(\frac{1}{2}g(1)^{2}+v_{0}(1)+s_{0}=50\), which simplifies to \(\frac{1}{2}g + v_{0}+s_{0}=50\).
When \(t = 2\), \(s(2)=96\), so \(\frac{1}{2}g(2)^{2}+v_{0}(2)+s_{0}=96\), which simplifies to \(2g+2v_{0}+s_{0}=96\).
When \(t = 3\), \(s(3)=72\), so \(\frac{1}{2}g(3)^{2}+v_{0}(3)+s_{0}=72\), which simplifies to \(\frac{9}{2}g+3v_{0}+s_{0}=72\).

Step2: Write the system of equations in matrix form

The system of equations is:
\(

$$\begin{cases}\frac{1}{2}g + v_{0}+s_{0}=50\\2g+2v_{0}+s_{0}=96\\\frac{9}{2}g+3v_{0}+s_{0}=72\end{cases}$$

\)
In matrix form \(AX = B\), where \(A=

$$\begin{bmatrix}\frac{1}{2}&1&1\\2&2&1\\\frac{9}{2}&3&1\end{bmatrix}$$

\), \(X=

$$\begin{bmatrix}g\\v_{0}\\s_{0}\end{bmatrix}$$

\), \(B=

$$\begin{bmatrix}50\\96\\72\end{bmatrix}$$

\)

Step3: Find the inverse of matrix \(A\)

First, find the determinant of \(A\):
\(\text{det}(A)=\frac{1}{2}(2\times1 - 3\times1)-1(2\times1-\frac{9}{2}\times1)+1(2\times3 - 2\times\frac{9}{2})\)
\(=\frac{1}{2}(2 - 3)-1(2-\frac{9}{2})+1(6 - 9)\)
\(=-\frac{1}{2}+\frac{5}{2}-3=-1\)

The cofactor matrix of \(A\) is:
\(C=

$$\begin{bmatrix}-1&\frac{5}{2}&-3\\-1&-4&3\\0&-\frac{1}{2}&-1\end{bmatrix}$$

\)
The adjoint of \(A\) is \(C^{T}=

$$\begin{bmatrix}-1&-1&0\\\frac{5}{2}&-4&-\frac{1}{2}\\-3&3&-1\end{bmatrix}$$

\)
The inverse of \(A\) is \(A^{-1}=\frac{1}{\text{det}(A)}C^{T}=

$$\begin{bmatrix}1&1&0\\-\frac{5}{2}&4&\frac{1}{2}\\3&-3&1\end{bmatrix}$$

\)

Step4: Solve for \(X\)

\(X = A^{-1}B\)
\(

$$\begin{bmatrix}g\\v_{0}\\s_{0}\end{bmatrix}$$

=

$$\begin{bmatrix}1&1&0\\-\frac{5}{2}&4&\frac{1}{2}\\3&-3&1\end{bmatrix}$$

$$\begin{bmatrix}50\\96\\72\end{bmatrix}$$

\)
\(g=1\times50+1\times96+0\times72 = - 32\)
\(v_{0}=-\frac{5}{2}\times50 + 4\times96+\frac{1}{2}\times72=96\)
\(s_{0}=3\times50-3\times96 + 1\times72 = 0\)

Answer:

\(g=-32\), \(v_{0}=96\), \(s_{0}=0\)