Chapter 7.4, Problem 94E

a.

To determine

To Find: the equation of parabola that passes through the points.

Answer to Problem 94E

The equation of parabola is

  $y=\frac{-1}{250}{x}^2+\frac{11}{30}x+5$

Explanation of Solution

Given information:

A video of the path of ball thrown by a baseball player was analyzed with a grid covering the TV screen. The video was paused three times, and the position of the ball was measured each timethe coordinates obtained are shown in the table (x and y are measured in feet)

  $\begin{array}{cc}\text{Horizontal distance,}x& \text{Height},y\\ 0& 5.0\\ 15& 9.6\\ 30& 12.4\end{array}$

Calculation:

Given the equation of parabola: $y=a{x}^2+bx+c$

Putting $x=0$ and $y=5.0$ in $y=a{x}^2+bx+c$

  $c=5$

Putting $x=15$ and $y=9.6$ in $y=a{x}^2+bx+c$

  $225a+15b+c=9.6$

Putting $x=30$ and $y=12.4$ in $y=a{x}^2+bx+c$

  $900a+30b+c=12.4$

Equation 1: $c=5$

Equation 2: $225a+15b+c=9.6$

Equation 3: $900a+30b+c=12.4$

Now solving this system of equation

Augmented matrix: $\left(\begin{array}{ccccc}0& 0& 1& ⋮& 5\\ 225& 15& 1& ⋮& 9.6\\ 900& 30& 1& ⋮& 12.4\end{array}\right)$

Reducing this augmented matrix in echelon form

  $\begin{array}{l}\left(\begin{array}{ccccc}0& 0& 1& ⋮& 5\\ 225& 15& 1& ⋮& 9.6\\ 900& 30& 1& ⋮& 12.4\end{array}\right)\\ \frac{1}{900}{R}_3\leftrightarrow R_1\\ \left(\begin{array}{ccccc}1& \frac{1}{30}& \frac{1}{900}& ⋮& \frac{12.4}{900}\\ 225& 15& 1& ⋮& 9.6\\ 0& 0& 1& ⋮& 5\end{array}\right)\\ R_2\to R_2-225R_1\\ \left(\begin{array}{ccccc}1& \frac{1}{30}& \frac{1}{900}& ⋮& \frac{12.4}{900}\\ 0& \frac{15}{2}& \frac{3}{4}& ⋮& \frac{13}{2}\\ 0& 0& 1& ⋮& 5\end{array}\right)\end{array}$

Now system of linear equation according to echelon matrix

  $\begin{array}{l}a+\frac{b}{30}+\frac{c}{900}=\frac{12.4}{900}\\ \frac{15b}{2}+\frac{3c}{4}=\frac{13}{2}\\ c=5\\ b=\frac{11}{30}\\ a=-\frac{1}{250}\end{array}$

Hence the equation of parabola is

  $y=\frac{-1}{250}{x}^2+\frac{11}{30}x+5$

b.

To determine

To Graph: use a graphing utility to graph the parabola

Explanation of Solution

Given information:

  $y=\frac{-1}{250}{x}^2+\frac{11}{30}x+5$

Graph:

  

c.

To determine

To Find: maximum height of the ball and the point at which the ball strikes the ground.

Answer to Problem 94E

Maximum height of the ball is: $13.40ft$

The point at which the ball strikes the ground: $103.71ft$

Explanation of Solution

Graph:

  

Interpretation:

From graph it is clear that

Maximum height of the ball is: $13.40ft$

The point at which the ball strikes the ground: $103.71ft$

d.

To determine

To Find: Algebraically, the maximum height of the ball and the point at which the ball strikes the ground

Answer to Problem 94E

The maximum height of the ball is $y=\frac{965}{72}=13.40ft$

The point at which the ball strikes the ground is $x=103.72ft$

Explanation of Solution

Given:

  $y=\frac{-1}{250}{x}^2+\frac{11}{30}x+5$

Calculation:

The equation of parabola is $y=\frac{-1}{250}{x}^2+\frac{11}{30}x+5$

Maximum height of parabola occurs at its vertex so converting the equation into standard form that is $y=a{\left(x-h\right)}^2+k$

Where $\left(h,k\right)$ is the vertex of parabola

  $\begin{array}{l}y=\frac{-1}{250}{x}^2+\frac{11}{30}x+5\\ y=\frac{-1}{250}\left(x^2-\frac{275}{3}x-1250\right)\\ y=\frac{-1}{250}\left(x^2-\frac{275}{3}x+{\left(\frac{275}{6}\right)}^2-{\left(\frac{275}{6}\right)}^2-1250\right)\\ y=\frac{-1}{250}\left({\left(x-\frac{275}{6}\right)}^2-\frac{120,625}{36}\right)\\ y=\frac{-1}{250}{\left(x-\frac{275}{6}\right)}^2+\frac{965}{72}\end{array}$

Therefore, Maximum height of parabola is $y=\frac{965}{72}=13.40ft$

  $y-\text{coordinate}$ of parabola is $0$ at the point at which the ball strikes the ground

Therefore putting $y=0$ in $y=\frac{-1}{250}{x}^2+\frac{11}{30}x+5$

  $\begin{array}{l}\frac{-1}{250}{x}^2+\frac{11}{30}x+5=0\\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x=\frac{-\left(\frac{11}{30}\right)\pm \sqrt{{\left(\frac{11}{30}\right)}^2-4\times \frac{-1}{250}\times 5}}{2\left(\frac{-1}{250}\right)}\\ x=103.72ft\end{array}$

Reject $-12.06$

Therefore, the ball strikes the ground at $x=103.72ft$