y = [-32/(20)^2]x^2 + x + 5
a.) For maximum height, y' = 0
y'= (-32/200)x + 1 = 0
32/200x = 1
x = 200/32 = 6.25
The maximum height attained by the ball is [-32/(20)^2](6.25)^2 + 6.25 + 5 = 8.125 feet.
b.) At the point the ball hits the ground, y = 0
y = [-32/(20)^2]x^2 + x + 5 = 0
x = 16.3 ft