Question

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 functiony = −32/((20)^2)x^2 + x + 6 where x is the distance in feet that the ball has traveled horizontally Find the maximum height attained by the ball. (Round your answer to three decimal places.)

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

Answer 1

So the question ask base on your problem, the following are the answers to your question are the following:
A. the maximum height attained by the ball is 9.125 ft.
B. the horizontal distance the ball has traveled when it hits the ground is 25ft. 

I hope you are satisfied with my answer and feel free to ask for more if you have further clarification 

Answer 2

Answer:

maximum height attained by the ball: 9.125 ft

horizontal distance the ball has traveled when it hits the ground: 16.93 ft

Step-by-step explanation:

The function

y = −32/((20)^2)x^2 + x + 6

y = -0.08x^2 + x + 6

is a parabola. Its maximum is at its vertex. x-coordinate of the vertex:

x = -b/(2a) = -1/(2*-0.08) = 6.25

y-coordinate of the vertex:

y = -0.08*6.25^2 + 6.25 + 6 = 9.125 ft

When the ball hits the ground, y = 0, so we have to find the roots of the function with help of the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$

$x = \frac{-1 \pm \sqrt{1^2 - 4(-0.08)(6)} }{2(-0.08)}$

$x = \frac{-1 \pm 1.7088}{-0.16}$

$x_1 = \frac{-1 + 1.7088}{-0.16}$

$x_1 = -4.43$

$x_2 = \frac{-1 - 1.7088}{-0.16}$

$x_2 = 16.93$

The negative solution is not correct because it doesn't have a physical meaning