Chapter 1.9, Problem 75PS

Solution

(a)

To Write and Solve: The inequality to find the values of $x$ when the ball is low enough to touch the goal.

The values of $x$ for which the ball is low enough to go into goal are $x\le 8$ or $x\ge 18.5$ .

Given information:

Given, the model is $y=-0.0540x^2+1.43x$

Football goal height $=8ft$

  $x=$ Horizontal distance

  $y=$ Corresponding height

Calculation:

Given that, football goal height is $=8ft$ , so the height of the ball must be less than $8$ feet to go into goal since the soccer goal is $8$ feet high.

Substitute $0.0540x^2+1.43x$ for $y$:

  $0.0540x^2+1.43x<8$

Replace the inequality sign with $=$

sign:

  $0.0540x^2+1.43x=8$

Subtract $8$

from both the sides:

  $0.0540x^2+1.43x-8=8-8$

  $0.0540x^2+1.43x-8=0$

Use, the quadratic form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a},$

to solve the equation:

where $a,b,c$ are real numbers and $a\ne 0$ .

  $\begin{array}{l}x=\frac{-1.43\pm \sqrt{{\left(1.43\right)}^2-4\left(-0.0540\right)\left(-8\right)}}{2\left(-0.0540\right)}\\ x\approx \frac{-1.43\pm 0.5629}{2\left(-0.0540\right)}\end{array}$

Simplify the above expression:

  $\begin{array}{c}x\approx \frac{-1.43+0.5629}{-0.108}\\ x\approx 8.02\\ x\approx \frac{-1.43-0.5629}{-0.108}\\ x\approx 18.45\end{array}$

The critical $x-$ values of $0.0540x^2+1.43x<8$ is $8.02$ and $18.45$ . Since, the values do not satisfy the inequality, plot these points on a number line using open dots.

  

There are three intervals on the number line based on the critical $x-$ values. In each interval, check whether the $x-$ value satisfies the inequality.

$x$	$0.0540x^2+1.43x<8$
$6$	$6.636<8$
$12$	$9.384<8$
$20$	$7<8$

It can be seen that the inequality is satisfied in the interval $x=8.02$ or $x=18.45$ .

  

So, the solution is $x<8.02$ or $x>18.45$ . The values of $x$ for which the ball is low

enough to go into goal are $x\le 8$ or $x\ge 18.5$ .

(b)

To Find: Whether the player score the goal if no one was blocking or not.

Yes, the player scores the goal and ball will go $1.3$ feet over the goal.

Given information:

Given, the model is $y=-0.0540x^2+1.43x$

Football goal height $=8ft$

The distance from the ball was kicked towards the goal $=15ft$

  $x=$ Horizontal distance

  $y=$ Corresponding height

Calculation:

From part $\left(a\right)$ , its been found that the ball is low enough to go into the goal is $x\le 8$ or $x\ge 18.5$ .

In this scenario, a soccer player kicks the ball $15$ feet toward the goal.

When the horizontal distance is less than $8$ or greater than $18.5$ only a goal will be scored.

Hence, kicking the ball from $15$ feet will not result in a goal.

To find the height corresponding to a horizontal distance of 15 feet, substitute $x=15$ in the equation $y=-0.0540x^2+1.43x$:

  $\begin{array}{l}y=-0.0540{\left(15\right)}^2+1.43\left(15\right)\\ y=-0.0540\left(225\right)+21.45\\ y=9.3\end{array}$

The distance the ball will cross the goal can be calculated by subtracting the height of the soccer goal from $9.3$ feet:

  $9.3-8=1.3$

As a result, the ball will go $1.3$ feet over the goal.