c Look for the horizontal axis intercept of the parabola.
A
a After 1.38 and 5.62 seconds.
B
b No, see solution.
C
c 7.05 seconds.
Practice makes perfect
a We are given an equation that models the position of a ball. The graph below illustrates the movement. We will also a horizontal line on the graph that represents the height of 130 feet. We can see that the ball will reach 130 feet once on the way up and again on the way down.
To find the time when the ball reaches 130 feet, we need the first coordinate of the intersection points of the parabola and the line. Let's use a calculator to find these points. Start by pushing the Y= button and typing the equations in the first two rows.
To see the graphs, you may need to adjust the window. Push WINDOW, change the settings, and push GRAPH to see the graphs.
Next, to find the intersection, push 2nd and TRACE. From this menu, choose intersect. The calculator will prompt us to choose the first and second curve and to provide the calculator with a best guess of where the intersection might be. We will need to repeat the process to find both intersection points.
Looking at the results on the calculator screen, we can see that the ball reaches 130 feet on the way up after 1.38 seconds and again on the way down after 5.62 seconds.
b Let's draw the graph again, but this time with a line at 250 feet. To find the maximum height the ball can reach, push 2ND and TRACE again and choose maximum. The calculator will prompt you to choose a left and right bound and to provide the calculator with a guess of where the maximum might be.
The throw is not strong enough, so the ball will not reach 250 feet. The maximum height the ball can reach is 202 feet.
c Let's review the graph one last time. To find the time needed for the ball to hit the ground, push 2ND and TRACE again and choose zero. The calculator will prompt us to choose a left and right bound and to provide the calculator with a best guess of where the horizontal axis intercept might be.
Notice that the calculator uses numerical calculations, so it only found an approximately 0 second coordinate. However, this approximation is very close. We can see that the height after 7.05 seconds is 0. The ball will hit the ground 7.05 seconds after its release.
