Chapter 9.5, Problem 47E

To determine

To determine if the height of water can reach 50feet, if so, then find then distance.

Answer to Problem 47E

Water arc can be 50feet high.

Height of water ach will be 50feet at 157.5feet and 42.5feet of the river.

Explanation of Solution

Given:

Equation of water arc

  $y=-0.006x^2+1.2x+10$

Where $y$ is height and $x$ is distance.

Calculation:

According to question:

Maximum height of water arc will be,

  $x=\frac{-b}{2a}$

So,

  $x=\frac{-1.2}{2(-0.006)}$

So,

  $x=\frac{-1.2}{-0.012}$

Therefore,

  $x=100$ feet.

So, water arc can be 50feet high.

According to question:

  $y=50$

  $50=-0.006x^2+1.2x+10$

Converting the equation in standard form:

  $0.006x^2-1.2x-10+50=0$

So,

  $0.006x^2-1.2x+40=0$

On comparing the equation with $a{x}^2+bx+c$

  $a=0.006,b=-1.2,c=40$

Finding the value of $x$

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substituting the values,

  $x=\frac{-(-1.2)\pm \sqrt{{(-1.2)}^2-4(0.006)(40)}}{2(0.006)}$

Opening brackets,

  $x=\frac{1.2\pm \sqrt{1.44-0.96}}{0.012}$

So,

  $x=\frac{1.2\pm \sqrt{0.48}}{0.012}$

Solving root,

  $x=\frac{1.2\pm 0.69}{0.012}$

Therefore,

  $x=\frac{1.2+0.69}{0.012}$ or $x=\frac{1.2-0.69}{0.012}$

So,

  $x=\frac{1.89}{0.012}$ or $x=\frac{0.51}{0.012}$

Therefore,

  $x=157.5$ feet or $x=42.5$ feet

So, height of water ach will be 50feet at 157.5feet and 42.5feet of the river.