Question

A ball is thrown across a playing field from a height of h = 3 ft above the ground at an angle of 45° to the horizontal at the speed of 20 ft/s. It can be deduced from physical principles that the path of the ball is modeled by the function
y = − 32/(20)^2x^2 + x + 3
where x is the distance in feet that the ball has traveled horizontally.

(a) Find the maximum height attained by the ball.

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

Answer 1

\[y=-\frac{ 32 }{ (20)^2 }x^2+x+3\]

Answer 2

I got 6.125 for A not sure how to find b?

Answer 3

@jdoe0001

Answer 4

so is a parabola

6.25 isn't the y-coordinate of its vertex btw
but you can find the vertex of any parabola at \(\bf \textit{vertex of a parabola}\\ \quad \\
y = {\color{red}{ -\frac{32}{20}}}x^2{\color{blue}{ +1}}x{\color{green}{ +3}}\qquad


\left(-\cfrac{{\color{blue}{ b}}}{2{\color{red}{ a}}}\quad ,\quad {\color{green}{ c}}-\cfrac{{\color{blue}{ b}}^2}{4{\color{red}{ a}}}\right)\\
\qquad \qquad \qquad\qquad \qquad \qquad \qquad \qquad\qquad \uparrow \\
\qquad \qquad \qquad\qquad \qquad \qquad\textit{maximum height}\)

Answer 5

hmmm missed the square there anyhow \(\bf \textit{vertex of a parabola}\\ \quad \\
y = {\color{red}{ -\frac{32}{20^2}}}x^2{\color{blue}{ +1}}x{\color{green}{ +3}}\qquad


\left(-\cfrac{{\color{blue}{ b}}}{2{\color{red}{ a}}}\quad ,\quad {\color{green}{ c}}-\cfrac{{\color{blue}{ b}}^2}{4{\color{red}{ a}}}\right)\\
\qquad \qquad \qquad\qquad \qquad \qquad \qquad \qquad \uparrow \\
\qquad \qquad \qquad\qquad \qquad \qquad\textit{maximum height}\)

Answer 6

hmmm actually "x" is meant the distance traveled... so
so to find "x" when it hits the ground, set y=0

that is \(\bf 0 = -\cfrac{32}{20^2}x^2+x+3\) and solve for "x"