Chapter 2.4, Problem 35E

To determine

Write a function that models the height of the water-skier over time. When is the water-skier 5 feet above the water? How long is the skier in the air.

Answer to Problem 35E

  $y(x)=-16x^2+6x+22.$

Water-skier 5 feet above the water $16x^2-6x-17=0.$

Skier in the air $x=1.235s$

Explanation of Solution

Given information: The table shows the heights y of a competitive water-skier x seconds after jumping off a ramp

  

Calculation :

$y(0)$		$y(0.25)$		$y(0.75)$		$y(1)$		$y(1.1)$
22		22.5		17.5		12		9.24
	0.5		-1.5		-3.5		-5.5
		-2		-2		-2

Because second differences are constant so we can modal the data with a quadratic function .

  $y(x)=a{x}^2+bx+c.$.......(1)

  $\begin{array}{l}$ y(0)=a\times {(0)}^2+b\times 0+c.$22=0+0+c.\\ c=22.\\ $ y(1)=a\times {(1)}^2+b\times 1+c.$12=a+b+22.\\ a+b=-\mathrm{10.........}(2)\\ $ y(0.25)=a\times {(0.25)}^2+b\times (0.25)+c.$22.5=(1/16)a+(1/4)b+22.\\ (1/16)a+(1/4)b=\mathrm{0.5.......}(3)\\ \text{solving}\text{(2)}\text{and}\text{(3)}.\\ a=-16,b=6,c=22.\\ \text{Putting}a=-16,b=6,c=22\text{in}\text{equation}(1)\\ y(x)=-16x^2+6x+{22.}^{}\end{array}$

Water-skier 5 feet above the water,

  $\begin{array}{l}5=-16x^2+6x+22.\\ 16x^2-6x-17=0.\\ x=1.235s.\end{array}$