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A ball is thrown across a playing field from a height of h = 6 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)^2] x^2 +x +6

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

AFind the maximum height attained by the ball. (Round your answer to three decimal places.)

BFind the horizontal distance the ball has traveled when it hits the ground. (Round your answer to one decimal place.)

Verified Answer :

Athe maximum height, occurs at the vertex of the quadratic, that is, the U-turn of the parabolic graph, and that occurs at \(\bf \qquad \qquad \textit{vertex of a parabola}\\ \quad \\ \begin{array}{lccclll} y=&\cfrac{-32}{20^2}x^2&+1x&+6\\ &\uparrow &\uparrow &\uparrow \\ &a&b&c \end{array} \qquad \qquad \left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)\)

so the y-coordinate is the height at that point, or \(\bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\)

b)
when the ball hits the ground the height is 0, because... well, the ball is all the way down, down to the x-axis, at that time y = 0
thus \(\bf y=\cfrac{-32}{20^2}x^2+1x+6\implies 0=\cfrac{-32}{20^2}x^2+1x+6\)

solve for "x"

boffeemadrid

The maximum height attained by a ball and the horizontal distance traveled at the end of the ball's path which is thrown is required.

The maximum height attained by the ball is 9.125 ft from the ground.

The horizontal distance traveled is 16.93 ft.

The given equation is

\(y=-\dfrac{32}{(20)^2}x^2+x+6\)

Since, the equation is of a parabola its vertex x axis point is

\(-\dfrac{b}{2a}=-\dfrac{1}{2\times-\dfrac{32}{(20)^2}}=6.25\)

At this point the in the x axis the ball will be at the highest point so,

\(y=-\dfrac{32}{(20)^2}\times 6.25^2+6.25+6\\\Rightarrow y=9.125\)

Hence, the maximum height attained by the ball is 9.125 ft from the ground.

At \(y=0\) the ball would be on the ground

\(0=-\dfrac{32}{(20)^2}x^2+x+6\\\Rightarrow -0.08x^2+x+6=0\\\Rightarrow -8x^2+100x+600=0\)

Solving the equation

\(x=\dfrac{-100\pm \sqrt{100^2-4\left(-8\right)\times 600}}{2\left(-8\right)}\\\Rightarrow x=-4.43,16.93\)

The horizontal distance traveled is 16.93 ft.

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