The path of the ball for many golf shots can be modeled by a quadratic function. The path of a
golf ball hit at an angle of about 10 degrees to the horizontal can be modeled by the function,
h=-0.002d^2 + 0.4d, where h is the height of the ball, in meters, and d is the horizontal distance
the ball travels, in meters, until it first hits the ground.
a) What is the maximum height reached by the ball?
b) What is the horizontal distance of the ball from the golfer when the ball reaches its
maximum height?
c) What distance does the ball travel horizontally until it first hits the ground?

This is a parabola that curves up and then turns down. The maximum height of this curve occurs when the slope changes from positive (going up) to negative (coming down). The slope is zero at this pint.

I answer in the order of b., a., and then c.

b. The first derivative of the equation gives us the slope for any point d, the distance. Take the first derivative and set the result equal to zero, then solve for d:

h'(d)= -0.004d + 0.4

0 = -0.004d + 0.4

-0.4 = -0.004d

d = 100 meters (the point at which the ball is maximum height and starts to turn down. [I.e., it's slope is zero]).

a. Now solve for h in the original equation, using d = 100

h(d) = -0.002d^2 + 0.4d

h(d) = -0.002(100)^2 + 0.4(100)

h(d) = 20 meters, the ball's maximum height

c. h(d) = -0.002d^2 + 0.4d

0 = -0.002d^2 + 0.4d [What values of d will result in a height, h(d), of 0?]

0 = -0.002d^2 + 0.4d

0.002d^2 = 0.4d

d^2 = 200d

d^2 - 200d = 0

d(d-200) = 0

The two solutions are d = 0 and d = 200. Both values will result in the equation returning h(d) of 0. The ball will be at ground level at both 0 seconds (the start) and 200 seconds (returns to ground).

We can graph the equation to confirm these results. See attached.