A football player kicks a ball at an angle of c above the ground with an initial speed of 20 meters/second. The height, h, as a function of the horizontal distance traveled, d, is given by h = 0.75d - 0.0192 d².

Draw a right triangle with one leg of length h, opposite the 37° angle, and

perpendicular to the ground and to the 2nd leg of length d and adjacent to the 37° angle. 

a) When the ball hits the ground, how far is it from the spot where the football player kicked it? 

You want to solve for d when h = 0. 

0 = 0.75d - 0.0192 d² 

d(0.75 - 0.0192d) = 0

0.75 - 0.0192d = 0 

d = 0.75/0.0192 = 38.82

b) What is the maximum height the ball reaches during its flight? Using completing the square.

h = 0.75d - 0.0192 d² is a parabola with its maximum h at the vertex.

In order to find the vertex by "completing the square", transform the equation from the y = Ax² + Bx + C form into the vertex form y = a(x-H)² + K where (H,K) is the vertex. 

y = A(x-H)² + K = Ax² - 2AHx + AH² + K.

So B = -2AH => H = -B / 2A.

C = AH² + K => K = C - AH² = C - A(-B / 2A)² = C - B²/(4A).

h = 0.75d - 0.0192 d² = - 0.0192 d² + 0.75d + 0 => A = -0.0192, B = 0.75 and C = 0.

H = - B / 2A = - 0.75 / 2(-0.0192) = 0.5 * 0.75 / 0.0192 = 19.53125.

K = C - B²/(4A) => K = (0.75)²/(4(0.0192)) = 7.32

So max. height h = K = (0.75)²/(4(0.0192)) = 7.324.

c) What is the horizontal distance the ball has traveled when it reaches its maximum height?

max. d = H = 0.5 * 0.75 / 0.0192 = 19.53125.