Activity
1. The function defined by gives the height in feet of a cannonball seconds after the ball leaves the cannon.
a. What does the term 50 tell us about the cannonball?
Compare your answer:
The 50 tells us the cannonball is fired from 50 feet off of the ground.
b. What does the term tell us about the cannonball?
Compare your answer:
The tells us that the initial upward velocity is 312 feet per second.
c. What does tell us about the cannonball?
Compare your answer:
The represents the effect of gravity pulling the cannonball back to the ground.
2. Use the graphing tool or technology outside the course. Graph the function. Adjust the graphing window to the following boundaries: and .
Compare your answer:
3. Observe the graph and:
a. Describe the shape of the graph. What does it tell us about the movement of the cannonball?
Compare your answer:
The distance from the ground increases, reaches a peak, and then decreases, which makes sense in the situation. It also looks like the distance from the ground decreases in the same way that it increased on the way up. The graph tells us the cannonball went up in the air, and then fell back down to the ground.
The shape of the quadratic function is called a parabola. It has a “U” shape.
b. Estimate the maximum height the ball reaches. When does this happen?
Compare your answer:
The cannonball gets up to a little over 1500 feet, maybe to 1600 feet, at around 10 seconds.
The peak of the parabola, where the maximum occurs, is also called the vertex of the graph.
c. Estimate when the ball hits the ground.
Compare your answer:
The cannonball hits the ground a little bit before 20 seconds.
The zero of a function is where the function equals 0, or where it crosses the -axis. In this case, there is only one zero where the cannonball hits the ground. The parabola does cross the -axis in the negative -values, but that does not make sense in this problem.
4. What domain is appropriate for this function? Be prepared to show your reasoning.
Compare your answer:
The equation is a good model for the height of the cannonball from the time it is launched, , until the time it lands, just before . For times outside of these values, the function values do not mean anything in the context (for example, for the function would place the cannonball underground).
Are you ready for more?
Extending Your Thinking
If the cannonball were fired at 800 feet per second, would it reach a mile in height? Remember there are 5280 feet in a mile. Be prepared to show your reasoning.
Compare your answer:
The equation here would be . Since and there are 5280 feet in a mile, this cannonball would reach over a mile in height. This could also be seen by graphing the function.
Video: Using Quadratic Functions to Describe Height
Watch the following video to learn more about quadratic functions.
Self Check
A toy rocket is launched from a platform 4 feet above the ground at a velocity of 128 feet per second. Its height is graphed below. Approximately what is the rocket’s maximum height?
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128
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110
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260
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8
Additional Resources
Reading Graphs of Quadratics
An object is thrown upward from a height of 5 feet with a velocity of 60 feet per second. Its height, , in feet after seconds is modeled by the function .
Step 1 -Identify the starting height.
5 feet
Step 2 -Identify the maximum height by looking at the vertex or peak of the parabola.
approximately 62 feet
Step 3 -Identify when the object reaches its maximum by matching the maximum with its matching time from the -axis.
approximately 1.75 seconds
Step 4 -When does the object hit the ground?
Look at where the object has a height of 0 feet, or the zero of the function. This happens at approximately 3.75 seconds. Notice there would be another zero on the graph of the function to the left of , but that does not make sense in the problem because the object was thrown from 5 feet.
Try it
Try It: Reading Graphs of Quadratics
A rock is launched into the air, and its height is represented by the graph below.
What is the maximum height of the rock?
Here is how to find the maximum height:
Look at the peak, or vertex of the parabola.
The maximum height is 25 feet after 2 seconds.