Answer
a) The solution set for the system of equations involving $a,{{v}_{0}},{{s}_{0}}$ is $\left( -32,56,0 \right)$.
b) The height of the ball is $0\text{ feet}$ after $3.5\text{ seconds}$.
c) The maximum height is $49\text{ feet}$ and we attain it after $1.75\text{ seconds}$.
Work Step by Step
(a)
Consider the provided graph .
The point $\left( 1,40 \right)$ indicates the height $\left( s \right)$ of the ball is $40\text{ feet}$ after $1\text{ second}$.
Similarly, $\left( 2,48 \right)$ indicates the height $\left( s \right)$ of the ball is $48\text{ feet}$ after $2\text{ seconds}$ and $\left( 3,24 \right)$ indicates the height $\left( s \right)$ of the ball is $40\text{ feet}$ after $3\text{ seconds}$.
Now, substitute these values in the provided position function,
$s\left( t \right)=\frac{1}{2}a{{t}^{2}}+{{v}_{0}}t+{{s}_{0}}$
For the point $\left( 1,40 \right)$ ,
$\begin{align}
& 40=\frac{1}{2}a\times {{(1)}^{2}}+{{v}_{0}}\times 1+{{s}_{0}} \\
& =\frac{1}{2}a+{{v}_{0}}+{{s}_{0}}
\end{align}$
For the point $\left( 2,48 \right)$ ,
$\begin{align}
& 48=\frac{1}{2}a\times {{(2)}^{2}}+{{v}_{0}}\times 2+{{s}_{0}} \\
& =\frac{4}{2}a+2{{v}_{0}}+{{s}_{0}} \\
& =2a+2{{v}_{0}}+{{s}_{0}}
\end{align}$
For the point $\left( 3,24 \right)$ ,
$\begin{align}
& 24=\frac{1}{2}a\times {{(3)}^{2}}+{{v}_{0}}\times 3+{{s}_{0}} \\
& =\frac{9}{2}a+3{{v}_{0}}+{{s}_{0}} \\
& =\frac{9}{2}a+3{{v}_{0}}+{{s}_{0}}
\end{align}$
Thus, the system of equations is:
$\begin{align}
& \frac{1}{2}a+{{v}_{0}}+{{s}_{0}}=40 \\
& 2a+2{{v}_{0}}+{{s}_{0}}=48 \\
& \frac{9}{2}a+3{{v}_{0}}+{{s}_{0}}=24 \\
\end{align}$
Now write the augmented matrix,
$\left[ \begin{matrix}
\frac{1}{2} & 1 & 1 & 40 \\
2 & 2 & 1 & 48 \\
\frac{9}{2} & 3 & 1 & 24 \\
\end{matrix} \right]$
Now, reduce the matrix to row echelon form by using row operation
${{R}_{1}}\to 2{{R}_{1}}$ ,
$\left[ \begin{matrix}
1 & 2 & 2 & 80 \\
2 & 2 & 1 & 48 \\
\frac{9}{2} & 3 & 1 & 24 \\
\end{matrix} \right]$
${{R}_{2}}\to {{R}_{2}}-2{{R}_{1}}$ ,
$\left[ \begin{matrix}
1 & 2 & 2 & 80 \\
0 & -2 & -3 & -112 \\
\frac{9}{2} & 3 & 1 & 24 \\
\end{matrix} \right]$
${{R}_{3}}\to {{R}_{3}}-\frac{9}{2}{{R}_{1}}$ ,
$\left[ \begin{matrix}
1 & 2 & 2 & 80 \\
0 & -2 & -3 & -112 \\
0 & -6 & -8 & -336 \\
\end{matrix} \right]$
${{R}_{2}}\to \frac{-1}{2}{{R}_{2}}$ ,
$\left[ \begin{matrix}
1 & 2 & 2 & 80 \\
0 & 1 & \frac{3}{2} & 56 \\
0 & -6 & -8 & -336 \\
\end{matrix} \right]$
${{R}_{3}}\to {{R}_{3}}+6{{R}_{2}}$ ,
$\left[ \begin{matrix}
1 & 2 & 2 & 80 \\
0 & 1 & \frac{3}{2} & 56 \\
0 & 0 & 1 & 0 \\
\end{matrix} \right]$
${{R}_{2}}\to {{R}_{2}}-\frac{3}{2}{{R}_{3}}$ ,
$\left[ \begin{matrix}
1 & 2 & 2 & 80 \\
0 & 1 & 0 & 56 \\
0 & 0 & 1 & 0 \\
\end{matrix} \right]$
${{R}_{1}}\to {{R}_{1}}-2{{R}_{3}}$ ,
$\left[ \begin{matrix}
1 & 2 & 0 & 80 \\
0 & 1 & 0 & 56 \\
0 & 0 & 1 & 0 \\
\end{matrix} \right]$
${{R}_{1}}\to {{R}_{1}}-2{{R}_{2}}$ ,
$\left[ \begin{matrix}
1 & 0 & 0 & -32 \\
0 & 1 & 0 & 56 \\
0 & 0 & 1 & 0 \\
\end{matrix} \right]$
Thus, $a=-32,{{v}_{0}}=56,{{s}_{0}}=0$
Hence, the solution set for the system of equations involving $a,{{v}_{0}},{{s}_{0}}$ is $\left( -32,56,0 \right)$.
(b)
Consider the provided function, $s\left( t \right)=\frac{1}{2}a{{t}^{2}}+{{v}_{0}}t+{{s}_{0}}$
Substitute $t=3.5$ in above equation,
That is,
$\begin{align}
& s\left( 3.5 \right)=\frac{1}{2}a{{(3.5)}^{2}}+{{v}_{0}}(3.5)+{{s}_{0}} \\
& =\frac{1}{2}\times 12.25\times a+3.5{{v}_{0}}+{{s}_{0}} \\
& =6.125a+3.5{{v}_{0}}+{{s}_{0}}
\end{align}$
Now, substitute the values $a=-32,{{v}_{0}}=56,{{s}_{0}}=0$ obtained in part (a),
$\begin{align}
& s\left( 3.5 \right)=6.125a+3.5{{v}_{0}}+{{s}_{0}} \\
& =6.125\times -32+3.5\times 56+0 \\
& =-196+196 \\
& =0
\end{align}$
Therefore, $s\left( 3.5 \right)=0$
Hence, the height of the ball is $0\text{ feet}$ after $3.5\text{ seconds}$.
(c)
The provided graph represents a parabola opening downwards.
Therefore, the maximum height will be the vertex of parabola.
The vertex of parabola is given by,
$\left( -\frac{{{v}_{0}}}{2\left( \frac{1}{2}a \right)},f\left( -\frac{{{v}_{0}}}{2\left( \frac{1}{2}a \right)} \right) \right)$
Here, $-\frac{-{{v}_{0}}}{a}$ represents the time and $f\left( -\frac{{{v}_{0}}}{a} \right)$ represents the maximum height of the ball.
Therefore,
$\begin{align}
& -\frac{{{v}_{0}}}{a}=-\frac{56}{-32} \\
& =1.75
\end{align}$
Substitute $t=1.75$ in $s\left( t \right)=\frac{1}{2}a{{t}^{2}}+{{v}_{0}}t+{{s}_{0}}$ Thus,
$\begin{align}
& s(1.75)=\frac{1}{2}(-32){{(1.75)}^{2}}+56(1.75)+0 \\
& =-49+98 \\
& =49
\end{align}$
Therefore, the ball attains the max height after $1.75\text{ seconds}$ and the maximum height is $49\text{ feet}$.
Hence, the maximum height is $49\text{ feet}$ and we attain it after $1.75\text{ seconds}$
