Question

0 - 91 college algebra
question 25, 3.2.35 - qc
part 9 of 10
hw score: 70.99%, 24.14 of 34 points
points: 0 of 4
hw 7(3.1 - 3.2)
a golf ball is hit with an initial velocity of 130 feet per second at an inclination of 45° to the horizontal. in physics, it is established that the height h of the golf ball is given by the function
h(x)=\frac{-32x^{2}}{130^{2}}+x
where x is the horizontal distance that the golf ball has traveled. complete parts (a) through (g).
(f) use a graphing utility to determine the distance that the ball has traveled when the height of the ball is 90 feet. choose the correct answer below and, if necessary, fill in the answer box to complete your choice.
a. the ball has traveled 115.07, 413.05 feet. (use a comma to separate answers as needed. round to two decimal places as needed.)
b. the ball has not traveled.
(g) create a table with tblstart = 0 and δtbl = 25. to the nearest 25 feet, how far does the ball travel before it reaches a maximum height? what is the maximum height?
to the nearest 25 feet, the ball travels 275 feet.
the maximum height is feet. (round to two decimal places as needed.)

Explanation:

Step1: Identify the height - distance function

The height function of the golf - ball is $h(x)=\frac{-32x^{2}}{130^{2}}+x$.

Step2: Set up the equation for part (f)

We want to find $x$ when $h(x) = 90$. So, we set up the equation $\frac{-32x^{2}}{130^{2}}+x = 90$. First, simplify the coefficient: $\frac{-32x^{2}}{16900}+x - 90=0$. Multiply through by 16900 to get $-32x^{2}+16900x - 1521000 = 0$. Divide by - 4 to simplify further: $8x^{2}-4225x + 380250=0$.
Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$, where $a = 8$, $b=-4225$, and $c = 380250$.
$x=\frac{4225\pm\sqrt{(-4225)^{2}-4\times8\times380250}}{2\times8}=\frac{4225\pm\sqrt{17855625 - 12168000}}{16}=\frac{4225\pm\sqrt{5687625}}{16}=\frac{4225\pm2384.88}{16}$.
We get $x_1=\frac{4225 + 2384.88}{16}\approx413.05$ and $x_2=\frac{4225 - 2384.88}{16}\approx115.07$.

Step3: For part (g)

The function $h(x)=\frac{-32x^{2}}{130^{2}}+x$ is a quadratic function of the form $y = ax^{2}+bx + c$ with $a=\frac{-32}{130^{2}}$ and $b = 1$.
The x - coordinate of the vertex of a quadratic function $y = ax^{2}+bx + c$ is $x=-\frac{b}{2a}$.
$x=-\frac{1}{2\times\frac{-32}{130^{2}}}=\frac{130^{2}}{64}=\frac{16900}{64}=264.0625\approx275$ (to the nearest 25 feet).
To find the maximum height, substitute $x = 264.0625$ into $h(x)$:
$h(264.0625)=\frac{-32\times(264.0625)^{2}}{130^{2}}+264.0625$.
$h(264.0625)=\frac{-32\times69737.515625}{16900}+264.0625$.
$h(264.0625)=\frac{-2231600.5}{16900}+264.0625$.
$h(264.0625)=-132.0474+264.0625 = 132.0151\approx132.02$ feet.

Answer:

(f) The ball has traveled 115.07, 413.05 feet.
(g) To the nearest 25 feet, the ball travels 275 feet. The maximum height is 132.02 feet.