Aball 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.
(a) 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.)

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

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

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

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

At the ball would be on the ground

Solving the equation

The horizontal distance traveled is 16.93 ft.

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step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)

he sells x mount because thats eqivilqant to the maz and then divided into quarters

step-by-step explanation: