Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
3. Section 9.3
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Exercise 85 Page 513

Practice makes perfect
a The ball hit by the golfer will land on the ground when h=0. Therefore, we want to find the x-intercept of the equation that represents the ball's path. By equating it with 0, we can solve for the x-intercepts with the Zero Product Property.
h=- 16t^2+80t
0=- 16t^2+80t
0=- 16t(t-5)
0=t(t-5)
lct=0 & (I) t-5=0 & (II)
lt_1=0 t_2=5
At t=0 the ball leaves the ground, and when t=5 the balls land on the ground.
b A parabola is symmetrical. Therefore, if we can find two points on the graph that are on the same h-coordinate, they are going to be halfway from the parabola's vertex, which in this case describes the highest point of the golf ball. Two such points are the t-intercepts, which both lie on t=0.
Average: 0+5/2=2.5The maximum height occurs when t=2.5. Therefore, by substituting these values in the formula, we can find the maximum height of the golf ball.
h=- 16t^2+80t
h=- 16( 2.5)^2+80( 2.5)
h=- 16(6.25)+80(2.5)
h=- 100+200
h=100
The maximum height of the golf ball is 100 feet. We can see this situation on the following diagram.
c The ball follows a parabolic path from the moment it is hit to the moment it touches the ground. In Part A we have shown that the ball leaves the ground at t=0 to land back on it at t=5. Therefore, only times t between these two moments should be in our domain.

Domain : 0 ≤ t ≤ 5