Chapter 4.3, Problem 17PPSE

Solution

a.

To Find: The quadratic model for the data.

The quadratic model for the data is $y=-16x^2+33x+46$

Given:

A man throws a ball off the top of a building and records the height of the ball at different times, as shown in the table.

Time $\left(s\right)$	Height $\left(ft\right)$
$0$	$46$
$1$	$63$
$2$	$48$
$3$	$1$

Explanation:

Given the points are: $\left(\text{0,46}\right)\left(\text{1,63}\right)\left(\text{2,48}\right)$ and $\left(3,1\right)$ :

Find the quadratic function using the standard form of $y=a{x}^2+bx+c$ then substitute each point $\left(x,y\right)$ to the standard form:

Substitute $\left(\text{0,46}\right)$ :

  $\begin{array}{c}y=a{x}^2+bx+c\\ 46=a{\left(0\right)}^2+b\left(0\right)+c\\ 46=c\to \left(1\right)\end{array}$

Substitute $\left(1,63\right)$ :

  $\begin{array}{c}y=a{x}^2+bx+c\\ 63=a{(}^1+b(1)+c\\ 63=a+b+c\to \left(2\right)\end{array}$

Substitute $\left(2,48\right)$ :

  $\begin{array}{c}y=a{x}^2+bx+c\\ 48=a{\left(2\right)}^2+b\left(2\right)+c\\ 48=4a+2b+c\to \left(3\right)\end{array}$

Substitute $\left(3,1\right)$ :

  $\begin{array}{l}y=a{x}^2+bx+c\\ 1=a{\left(3\right)}^2+b\left(3\right)+c\\ 1=9a+3b+c\to \left(4\right)\end{array}$

Eliminate the equation between $\left(2\right)$ and $\left(3\right)$ :

  $\begin{array}{l}63=a+b+c\\ \underset{¯}{48=4a+2b+c}\\ 111=5a+3b+2c\to \left(5\right)\end{array}$

To get the value of $b$ simplify $0=2b$ :

  $\begin{array}{l}0=2b\\ 0=b\end{array}$

Multiply $\left(-1\right)$ to the equation $\left(4\right)$ :

  $\begin{array}{c}\left(-1\right)1=\left(9a+3b+c\right)-1\\ -1=-9a-3b-c\to \left(6\right)\end{array}$

Solve the equations $\left(5\right)$ and $\left(6\right)$ :

  $\begin{array}{l}111=5a+3b+2c\\ \underset{¯}{-1=-9a-3b-c}\\ 110=-4a+c\end{array}$

Substitute the value of $c=46$ in the above equation:

  $\begin{array}{c}110=-4a+c\\ 110=-4a+46\\ 64=-4a\\ -16=a\end{array}$

To find the value of $b$ , substitute $a=-16$ and $c=46$ into the equation $2$ :

  $\begin{array}{c}63=a+b+c\\ 63=-16+b+46\\ 63=30+b\\ 33=b\end{array}$

Now, substitute $a=-16,b=33$ and $c=46$ into the equation $y=a{x}^2+bx+c$ :

  $y=-16x^2+33x+46$

Here $x$ represents the seconds in time after release and $y$ represents the height.

b.

To Estimate: The height of the ball at $2.5s$ by using the model.

The estimated height of the ball at $2.5s$ by using the model is $28.5ft$ .

Given:

A man throws a ball off the top of a building and records the height of the ball at different times, as shown in the table.

Time $\left(s\right)$	Height $\left(ft\right)$
$0$	$46$
$1$	$63$
$2$	$48$
$3$	$1$

Explanation:

Estimate the height of the ball at $2.5s$ by using the model:

From part $\left(a\right)$ , the equation is: $y=-16x^2+33x+46$

Substitute $x=2.5$ to estimate the height of the ball at $2.5s$ :

  $\begin{array}{l}y=-16{\left(2.5\right)}^2+33\left(2.5\right)+46\\ y=-100+82.5+46\\ y=28.5\end{array}$

Thus, the estimated height of the ball at $2.5s$ by using the model is $28.5ft$ .

c.

To Determine: The maximum height of the ball.

The maximum height of the ball is $\text{63}\text{.02 }f\text{t}$ .

Given:

A man throws a ball off the top of a building and records the height of the ball at different times, as shown in the table.

Time $\left(s\right)$	Height $\left(ft\right)$
$0$	$46$
$1$	$63$
$2$	$48$
$3$	$1$

Explanation:

Find the maximum height of the ball,

From part $\left(a\right)$ the equation is $y=-16x^2+33x+46$

The maximum value of the function denoted by maximum height the ball.

For any quadratic function, the maximum value $y_0$ at the value of $x_0$ is determined by the formula $A\left(x_0,y_0\right)=\left(\frac{-b}{2a},\frac{4ac-b^2}{4a}\right)\to \left(1\right)$ .

Here $a,b,c$ represent the parameters of the quadratic equation.

From the equation $y=-16x^2+33x+46$ , the values are:

  $a=-16,b=33,c=46$

Substitute the above data into the formula $\left(1\right)$ :

  $\begin{array}{c}A\left(x_0,y_0\right)=\left(\frac{-33}{2\left(-16\right)},\frac{4\left(-16\right)\left(46\right)-{33}^2}{4\left(-16\right)}\right)\\ =\left(\frac{33}{32},\frac{4033}{64}\right)\\ =\left(1.03125,63.015625\right)\end{array}$

Thus, the maximum height of the ball is $\text{63}\text{.02 }f\text{t}$