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)^{2}

x^{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. (Round your answer to three decimal places.) ft

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

Thank you for the opportunity to help you with your question! the solution is as such:

for horizontal distance, y=0

So, y = − 32/(20)^2x^2 + x + 3 =>0 = − 32/(20)^2x^2 + x + 3 ,

Solving the equation we get , x=13.9 ft , is the horizontal distance.

a) To obtain the maximum height, differentiate y w.r.t x and equate it to zero.

-32/200 x +1 =>x=200/32 = 25/4

Put x=25/4 in y equation , to get the maximum height =>

y= 13.2 ft, is the maximum height attain.

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