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 s
peed 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.)
Your function is a bit garbled, but I assume it's supposed to be y = -32/(20)2 * x2 + x + 4 This is plainly a parabolic function - it fits the form y = ax2 + bx + c. Since a is negative, it is an inverted parabola, one whose opening points downwards. Such a parabola has a maximum value, which in this case corresponds to the maximum height reached by the ball. The maximum or minimum value of a parabola occurs at the vertex, which itself is located at the x coordinate of -b/2a, based on our standard form. With a b of 1 and an a of -32/400=-2/25=-.08, this corresponds to x = 1/.16 = 6.25. We find the height the ball reaches at this point by plugging the x into our equation. y = -.08*(6.25)2 + 6.25 + 4 = -3.125 + 6.25 + 4 = 7.125 feet. The horizontal distance covered corresponds to the x value where the ball reaches a height of 0. I.e., it's one of the roots of the function, which can be found with the quadratic equation, x = [-b ± √(b2-4ac)]/2a Plugging in our coefficients, we have [-1 ± √(1 + 1.28)]/-.16 = [-1 ± √(2.28)]/-.16 = [-1 ± 1.51]/-.16 = 6.25 ± 9.44 There is one positive solution and one negative solution to this - for physical reasons, the positive one is the one we prefer, since we're looking for where the ball hits the ground ahead of the thrower. Thus, our answer is x = 15.7 feet.
