Chapter 2, Problem 16CR

To determine

To find: The function that models the given data and to find how long is the object in the air.

Answer to Problem 16CR

The function that models the given data is $y=-16x^2+150$ and the object is $3.06$ seconds in the air.

Explanation of Solution

Given information: A table of set of data is given.

$\text{Time(seconds)},x$	$0$	$0.5$	$1$	$1.5$	$2$	$2.5$
$\text{Height(feet)},x$	$150$	$146$	$134$	$114$	$86$	$50$

General form of the equation of a parabola is $y=a{x}^2+bx+c$ .

Equations of the parabola with the given points are

For the point $(0,150)$ is $150=a{(0)}^2+b(0)+c\to c=150$$\mathrm{...}(1)$

For the point $(1,134)$ is $134=a{(1)}^2+b(1)+c\to a+b+c=134$$\mathrm{...}(2)$

For the point $(2,86)$ is $86=a{(2)}^2+b(2)+c\to 4a+2b+c=86$$\mathrm{...}(3)$

By solving the above three questions we get,

  $a=-16,b=0,c=150$

Substitute $a=-16,b=0,c=150$ in $y=a{x}^2+bx+c$ ,

  $y=-16x^2+150$

Therefore, the above equation is the equation of the parabola with the given points.

To find the time taken by the object in the air we have to find the $x-$ intercepts,

Substitute $y=0$ in $y=-16x^2+150$ ,

  $\begin{array}{l}0=-16x^2+150\\ -150=-16x^2\\ x^2=\frac{-150}{-16}\\ x^2=9.375\\ x=\pm \sqrt{9.375}\\ x=\pm 3.06\end{array}$

Therefore the object is been in the air for $3.06$ seconds.