Question

algebra
faduma musse 10/09/25 7:09 am
question 25, 3.2.35 - gc
part 7 of 10
hw score: 70.99%, 24.14 of 34 points
points: 0 of 4
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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).
(e) use a graphing utility to graph the function h = h(x).
choose the correct graph below.
0,600 by 0,140, xscl = 100, yscl = 20
(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 feet.
(use a comma to separate answers as needed. round to two decimal places as needed.)
b. the ball has not traveled.
view an example get more help

Explanation:

Step1: Analyze the function form

The function $h(x)=\frac{-32x^{2}}{130^{2}}+x$ is a quadratic function of the form $y = ax^{2}+bx + c$ where $a=\frac{-32}{130^{2}}<0$, $b = 1$ and $c = 0$. Since $a<0$, the graph of the function is a parabola opening down - ward.

Step2: Consider the domain and range

The domain $x\geq0$ (distance cannot be negative) and we are given the viewing window $[0,600]$ for $x$ and $[0,140]$ for $y$. The vertex of the parabola $x=-\frac{b}{2a}=-\frac{1}{2\times\frac{-32}{130^{2}}}=\frac{130^{2}}{64}\approx264.06$. The maximum value of the function occurs at the vertex.

Step3: Solve for part (f)

Set $h(x)=90$, so $\frac{-32x^{2}}{130^{2}}+x = 90$. Multiply through by $130^{2}$ to get $-32x^{2}+130^{2}x=90\times130^{2}$. Rearrange to $32x^{2}-130^{2}x + 90\times130^{2}=0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 32$, $b=-130^{2}$ and $c = 90\times130^{2}$.
$x=\frac{130^{2}\pm\sqrt{130^{4}-4\times32\times90\times130^{2}}}{2\times32}=\frac{130^{2}\pm130\sqrt{130^{2}-4\times32\times90}}{64}$.
$x=\frac{16900\pm130\sqrt{16900 - 11520}}{64}=\frac{16900\pm130\sqrt{5380}}{64}=\frac{16900\pm130\times73.35}{64}$.
$x_1=\frac{16900 + 130\times73.35}{64}\approx\frac{16900+9535.5}{64}=\frac{26435.5}{64}\approx413.06$
$x_2=\frac{16900 - 130\times73.35}{64}\approx\frac{16900 - 9535.5}{64}=\frac{7364.5}{64}\approx115.07$

Answer:

(e) The graph of a quadratic function $y = ax^{2}+bx + c$ with $a<0$ is a parabola opening down - ward. Among the options, the correct graph is the one that is a downward - opening parabola within the given viewing window. Without seeing the exact details of the options in a non - visual way, we know it should be a downward - opening parabola.
(f) A. The ball has traveled 115.07,413.06 feet.