College Algebra, Chapter 4, 4.1, Section 4.1, Problem 64

Suppose a ball is thrown across a playing field from a height of 5 ft above the ground at an angle of $45^{\circ}$ to the horizontal at a speed of 20 ft/s. It can be deduced from physical principles that the path of the ball is modeled by the function

$\displaystyle y = - \frac{32}{(20)^2} x^2 + x + 5$

where $x$ is the distance in the feet that the ball has traveled horizontally.

a.) Determine the maximum height attained by the ball.

The function is a quadratic function with $\displaystyle a = \frac{-32}{(20)^2}$ and $b = 1$. Thus, the maximum value occurs when

$\displaystyle x = - \frac{b}{2a} = - \frac{1}{\displaystyle 2 \left( \frac{-32}{20^2} \right)} = \frac{25}{4}$

Since $a < 0$, the maximum value is

$\displaystyle f\left( \frac{25}{4} \right) = - \frac{32}{(20)^2} \left( \frac{25}{4} \right)^2 + \frac{25}{4} + 5 = \frac{65}{8}$ ft or $= 8.125$ ft

The maximum height attained by the ball is $8.125$ ft

b.) Determine the horizontal distance the ball has traveled when it hits the ground.

To find the horizontal distance, we set $y = 0$. So


$\begin{equation} \begin{aligned} 0 =& - \frac{32}{(20)^2} x^2 + x + 5 && \\ \\ 0 =& x^2 - \frac{(20)^2}{32} x - \frac{5(20)^2}{32} && \text{Add } \frac{5 (20)^2}{32} \\ \\ \frac{20^4}{64^2} + \frac{5(20)^2}{32} =& x^2 - \frac{(20)^2}{32} x + \frac{20^4}{64^2} && \text{Complete the square: add } \left( \frac{\displaystyle - \frac{(20)^2}{32}}{2} \right)^2 = \frac{(20)^4}{(64)^2} \\ \\ \frac{20^4 + 5 (20)^2 (128)}{64^2} =& \left( x - \frac{20^2}{64} \right)^2 && \text{Take the square root} \\ \\ \pm \frac{5 \sqrt{65}}{4} =& x - \frac{20^2}{64} && \text{Add } \frac{20^2}{64} \\ \\ x =& \pm \frac{5 \sqrt{65}}{4} + \frac{20^2}{64} && \text{Simplify} \\ \\ x =& \frac{5 \sqrt{65}}{4} + \frac{20^2}{64} \text{ and } x = \frac{-5 \sqrt{65}}{4} + \frac{20^2}{64} && \\ \\ x =& 16.33 \text{ and } x = -3.83 && \end{aligned} \end{equation}$


Choose $x > 0$, so the distance the ball has traveled when it hits the ground is $16.33$ ft.