Applications of Quadratic Functions in Kinematic Modeling
Lesson Plan
Chris Joy
HHMI-FT Internship
Summer 2012
Introduction and Motivation
Objectives: NCSCOS 2.02: The students will be able to use quadratic functions and inequalities to model
and solve problems.
a) Solve using tables, graphs, and algebraic properties
b) Interpret constants and coefficients in the context of the problem
NCTM: The students will be able to analyze functions of one variable by investigating rates of change,
intercepts, zeroes, asymptotes, and local and global behavior.
NOM: Much of the work of mathematicians involves a modeling cycle, consisting of three steps: (1)
using abstractions to represent things or ideas, (2) manipulating the abstractions according to some
logical rules, and (3) checking how well the results match the original things or ideas. The actual thinking
need not follow this order. 2C/H2*
NOM Webste: http://www.project2061.org/publications/bsl/online/index.php?chapter=2#chaptoc
The main focus of this lesson will be on using the kinematic equation h(t)=1/2gt2+v0t+h0 to
introduce students to graphing quadratic functions and how each parameter affects the graph. Using
kinematics to introduce the topic gives students a concrete example so that they can better visualize
how a, b, and c affect a quadratic graph. Instead of simply discussing the general parameters or features
of the graph, students can use their intuition about something like a ball flying through the air to
visualize how changing parameters would change a graph. So, instead of thinking how b would affect
the graph of a quadratic equation, students could think, “what would happen if I threw two baseballs
into the air, but threw one harder than the other?” the students would think in terms of the context of
the problem, “would the ball go higher or lower? Spend more or less time in the air?” and use their
intuition about the world around them to understand the idea first with the kinematics example and
later in a more general and abstract case.
Learning Activities
There are three learning activities in this lesson. The first is a discovery learning section on
kinematics, which works with knowledge the students already have about the world around them to
understand the kinematic equation h(t)=1/2gt2+v0t+h0. The next section generalizes the knowledge
highlighted in the first section to help students understand how each parameter of a quadratic function
causes changes in the graph. And the third and final activity is an in class game where students will be
asked to solve for different parameters by launching erasers into the air, timing them, and marking the
maximum height. All three activities are designed to focus on the mathematical content of quadratic
functions, but also on the greater implication of the applications of mathematical modeling in science
and using the mathematical though process highlighted in the Nature of Mathematics.
Kinematics Section
The warm-up section of this lesson will introduce the topic of kinematics, explaining what it is
and how it is used. From there the students will work through a discovery learning activity which
focuses on the kinematic equation h(t)=1/2gt2+v0t+h0. There will be four sections in the warm up. The
first will focus on the role of g in the equation. After the first section students should understand, within
the context of the problem, that positive acceleration causes objects to rise, negative acceleration
causes objects to fall and greater magnitudes (absolute values) of acceleration cause the objects to
either rise or fall at a faster rate. The second section will focus on the role of v0 in the equation. After
the second section the students should understand that increasing v0 will make objects stay in the air
longer and go higher and decreasing v0 will make objects stay in air for a shorter amount of time and go
lower. The third section will focus on the role of h0. This should be the easiest for students to
understand because it is the constant term in the equation, but students should understand that h0
essentially shifts the motion up or down (i.e. starting on a hill or from a ditch). The final section will
introduce the concept of the vertex by focusing on the maximum or minimum height that the object
reached, and presenting the vertex formula as t =
−v0
.
g
It will also explain the process to find the
maximum (or minimum) height itself by substituting the vertex formula back into the function for t and
evaluating.
The kinematics section will pose questions in each subsection to help the students find and
make sense of the answer for themselves. Overall for this section students will be asked to imagine the
flight of a baseball that they throw up into the air, somewhat hard, but not as hard as they can. From
there they have to answer questions about how changing parameters will affect the flight of the ball.
However, you should start with an example graph so the students can see generally how this path looks
on a graph.
The first section focuses on the role of gravity in the function. To see how gravity affects the
flight of the ball, use examples of other planets’ gravitational force. For example, “on Jupiter the
acceleration due to gravity is almost three times as large as that on Earth so the baseball would be three
times as heavy, how would you expect the flight of the ball to differ if you threw it on Jupiter instead of
on earth?” Students should come up with an answer that includes both the fact that the ball would not
go up as high and also that the ball would not stay in the air as long. Below are the graphs, (Earth – h(t)
= - 4.9t2 + 10t is on the left and Jupiter – h(t) = - 12.97t2 + 10t is on the right).
Next you could ask “On the moon the acceleration due to gravity is much less than that on
Earth, how would the flight of ball be different on the moon?” Again, students should come up with an
answer that includes both the fact that the ball will go higher and also that the ball will stay in the air
longer. Below again are the graphs, (Earth – h(t) = - 4.9t2 + 10t is on the left and the moon – h(t) = 0.81t2 + 10t is on the right).
Finally you can use the example of a sci-fi tractor beam to explain the difference between
positive and negative values for g. Students should understand that positive g causes the ball to rise and
negative g causes the ball to fall back down to Earth. Graphs are below (Earth – h(t) = - 4.9t2 + 10t is on
the left and the tractor beam – h(t) = 9.8t2 + 10t is on the right).
Have the students start with an example equation (on Earth) like h(t) = - 4.9t2 + 10t . As you
change the acceleration due to gravity have them first answer the question based on their intuition,
then have them plug in different values for g into a graphing calculator to see how these changes are
interpreted graphically. For Jupiter, you would use h(t) = - 12.97t2 + 10t and for the moon you would use
h(t) = - 0.81t2 + 10t.
For the section on v0 you can ask the students question about what would happen if they threw
the ball harder or softer. How would they expect the flight to change? Again students should come up
with an answer that includes both the fact that the ball will go higher and stay in the air longer when
they throw the ball harder and the ball will go lower and stay in the air for a shorter amount of time
when they throw the ball softer. Again, they should answer the question using their intuition first, then
try and confirm that intuition with changes on the graph. Again you can start with h(t) = - 4.9t2 + 10t and
for the ball thrown very hard you can use h(t) = - 4.9t2 + 50t and for the ball throw very soft you can use
h(t) = - 4.9t2 + 3t.
Here are the graphs again, (The ball thown somewhat hard – h(t) = - 4.9t2 + 10t is on the left and
the ball thrown very hard– h(t) = 9.8t2 + 50t is on the right and the ball thrown softly – h(t) = - 4.9t2 + 3t
is below).
However, the students also should see a couple examples of positive g values (i.e. “inside the
tractor beam”) to see how different values for initial velocity change the path under “anti-gravity.” Have
an example where they throw the ball the same as before and an example where they throw the ball
downward with the same speed; h(t) = 4.9t2 + 10t for the first and h(t) = 4.9t2 – 10t for the second. Note
that the students should only be focusing on values from t = 0 onward (no negative times). The ball
thrown upwards in the tractor beam is on the left and the ball thrown downwards inside the tractor
beam is on the right.
The next section on h0 will work in the same way except now the students will be asked to think
about the changes in initial height. What would happen if you threw the ball off of a hill? What would
happen if you threw the ball from down in a ditch? For the graphs you can start again with h(t) = - 4.9t2 +
10t. Then for the hill example you can use h(t) = - 4.9t2 + 10t + 10 and for the ditch example you can use
h(t) = - 4.9t2 + 10t – 10. Here they should again note the changes in hangtime and maximum height,
however an important note they need to take away from this section is that changing the initial height
only changes the maximum height, not the time at which it occurs.
Here are the corresponding graphs (hill on left, ditch on right).
The final section will focus on finding the maximum height. For this section the students will be
asked which two parameters affect the time at which the maximum height occurs (but they should note
that all three affect the maximum height itself). They should come up with the fact that g and v0 affect
when the maximum height happens. After they come up with that answer they should be given the
vertex formula in the context of the problem, t =
−v0
.
g
From there you can ask students how they would
use that and the kinematic equation to find the maximum height. By the end of this section they should
understand how to substitute the vertex formula into the kinematic equation to get the maximum (or
minimum) height.
To help guide students through the kinematics section, tell them to trust their intuition. The
idea of this section is that students can use their knowledge of the world around them to understand
the effect of each parameter in the real-world context and then, in the content section, generalize and
abstract that knowledge to understand, in a general sense, graphing quadratic equations.
Content Section
The content for this lesson is designed to help the students relate what they learned about
kinematics back to the more abstract case of general quadratic equations. This section briefly goes over
what a quadratic function is, what the graph looks like, what the vertex is and how to find it, and how
each of the parameters, a, b, and c, affect the graph of the function (related back to the warm-up
section).
This section has the same breakdown as the previous section on kinematics, except now the
students are asked to generalize their knowledge from the first section to the case of general quadratic
functions. There will be a section on “a” (relating back to g), a section on “b” (relating back to v0), a
section on “c” (relating back to h0), and finally a section on finding the vertex. Once again the students
can work through a series of examples with corresponding graphs and use their knowledge from the
previous section to help them understand the general case.
Most students will pick up reasonably well on the effect that the parameters “a” and “c” have
on the graph (i.e. that the sign of “a” determines which way the parabola opens and larger absolute
values for “a” make the parabola thinner; that “c” simply shifts the graph up or down) but to some
students, the role of “b” is unclear. By beginning with the application in kinematics, students should be
able to more easily understand this effect. Students can think in terms of throwing something up into
the air. Ask the students guiding questions such as “what do you expect would happen if you threw
something into the air very hard? Very soft? Would the object stay in the air longer or shorter? Go
higher or lower?” Relating back to the warm-up section will help students make sense of the general
case of quadratics for themselves, by using the application in kinematics.
The content section flows directly into the activity, where the students will have to apply their
new knowledge of kinematics and quadratic functions to find the initial velocity for an eraser being shot
up into the air and also to calculate the height of desks in the classroom by timing the eraser’s fall.
Activity
The activity for this lesson is an in class game called Eraser Chaser. For this game you will need:
At least 3 chalkboard or white board erasers
2 Rubber bands (relatively thick ones)
Duct Tape
4 -5 Yard Sticks (a measuring tape can be used in place of the fifth yardstick)
4 Pens/Pencils
At least 3 Stopwatches or time keeping devices
The game basically works by launching erasers upwards, using the yardsticks and rubber bands as a
catapult, checking the maximum height and having students calculate the initial velocity. To set up the
“catapults” you will need to tape the pens/pencils perpendicular to the yardsticks towards the bottom
(experiment so that you can find a good height that allows the students to launch the erasers up, but
not so much that they hit the ceiling) and tie the rubber bands around the pencils. Separate the
students into groups (~5 or 6students/ group more groups can be formed if there are more students).
Two groups will start on the “catapults” which should be located near a chalkboard or whiteboard. One
student will launch the eraser upwards, one student will time how long it takes to reach its maximum
height, and one student will mark the maximum height on the board and subtract the height that the
eraser started from (which can be measured in meters from the yardstick). Have the students do about
5 runs at each experiment and take an average of the heights and times. Given the equation
h(t)=1/2at2+v0t+h0 have the students calculate the initial velocity by using the vertex formula (students
should note that to find the time at which the maximum height occurs, the initial height does not come
into play, which should reinforce the concept of a “vertical shift” if it was unclear from the previous
sections. The shape of the graph doesn’t change, it is simply “shifted” by the c value or initial height.
Basically, h0 affects the height of the vertex, but not when it occurs.). Another station(s) can be set up
in which students gradually tip an eraser off of a desk or chair and time how long it takes the eraser to
hit the ground and from that information calculate the height of the chair or desk (i.e. start with
h(t)=1/2at2+v0t+h0 , and students should note that v0=0 since it is starting from rest so the equation
becomes h(t)=-4.9t2+h0, and since they are timing the eraser falling to the ground, or h=0, the equation
becomes 0=-4.9t2+h0. From there the students only need to substitute the time they calculated into the
equation to calculate the initial height and compare that to the measured height of the desk or chair).
Like in the other station, the students should do about 5 runs and take an average of the times.
Common errors in running these types of experiments can include mistakes in units (i.e. lengths need to
be in meters, not inches or centimeters), also many students forget the negative sign before the “a”
coefficient, which can yield some strange answers. At the end of the activity, have the students all put
their data up on the board and compare answers.
Quiz
The quiz at the end will check that the students met the learning objectives for this lesson. There will be
4 conceptual questions about generally graphing quadratic equations, 4 computational questions with
real-world applications of quadratic equations, and 2 “challenge” problems which also have real-world
applications, but push the students a step further. For these problems they will be given the kinematic
equation h(t)=1/2at2+v0t+h0, and the vertex formula 𝑥 =
−𝑏
2𝑎
. Students will be instructed to round, only
the final answer, to one decimal place. The quiz is multiple choice and is included in a printer-friendly
version below the answer key.
Quiz Answer Key
1 “What are the x and y coordinates of the vertex of y = 2x2 + 4x + 12? Is it a maximum or a minimum?”
The correct answer is (-1, 10) minimum. This question checks that students can work with a
quadratic in a generic form and find the vertex and also check that they understand when the vertex is a
minimum and when it is a maximum based on the sign of “a.” Students should recognize that “a” is
positive and from there understand that the vertex will be the minimum.
2 “In his 2nd round at the 2012 British Open, Tiger Woods chipped in for birdie on the 18th hole from a
treacherously deep bunker. The sand trap was close to 3 meters deep in some places! Assuming Tiger’s
ball left his club face at around 25m/s, which equation would model the path of his ball?”
The correct answer is h(t) = - 4.9t2 + 25t - 3. This question is designed to check that students can
interpret the initial height from a word problem and then represent it in the equation.
3 “Consider the graph of y = x2. What happens to the graph if you change a to be -5?”
The correct answer is the parabola gets thinner and opens downward. This is a conceptual
question to see whether the students can work back and forth between the applications and the general
case of quadratic equations and graphs. In this case, the question checks that the students understand
in a general sense the role of a in quadratic equations.
4 “In his rookie season, Panthers’ quarterback Cam Newton broke the NFL’s rookie season passing
record previously held by Peyton Manning from his 1998 rookie season. Part of Newton’s passing
effectiveness comes from his height; at 6’ 5” (about 2m) Newton is easily able to keep his passes from
being swatted down by the defensive line. Assuming that a defensive linemen’s hand can reach up to 10
ft (3.05m) in the air and that Cam Newton’s passes cross the defensive line about half a second after
release, which equation would model a pass from Cam Newton that clears the defensive line (doesn’t
get swatted down)? (Hint: Think about what the initial height of the football would be in this problem)”
The correct answer is h(t) = - 4.9t2 + 5t + 2. For this problem, students need to substitute .5s into
each equation for “t” and check that the height is greater than 3.05m. However, the students also need
to consider that the ball is leaving from the quarterback’s hand (i.e. not from the ground) so the initial
height should be about the height of the quarterback (i.e. h0 = 2m). From the choices given, two will be
high enough to clear the d-line, but only one will have considered the interpretation of h0 as Cam
Newton’s height.
5 “How would the graph of y = 5x2 + x change if you change b to be -3?”
The correct answer is that the graph moves to the right and the vertex moves down. This is
another conceptual question to gauge how well the students understand the role of b in quadratic
equations and the interaction between b and a.
6 “The equation y = 4x2 + 8x has its vertex at (-1, -4). What are the coordinates of the vertex of the
equation y = 4x2 + 8x + 8?”
The correct answer is (-1, 4). This question checks the students’ understanding of “c” and also
checks that they understand graphically and algebraically the concept of a “vertical shift.” Students
should note that the x-coordinate of the vertex is only dependent on “b” and “a” but the y-coordinate is
dependent on “c”
7 “In game 1 of the 2012 NBA Finals, LeBron James scored 30 points! Ideally, LeBron releases his jump
shot at the top of his jump around 10ft (3.05m). For a three-pointer, the ball travels through the air for
about 1 second before reaching the basket which is 10ft high (3.05m). If LeBron shoots the ball at
4.9m/s will he make the shot?”
The correct answer is yes, he will make the shot. For this question the students should interpret
all the given information and put it appropriately into the kinematic equation (i.e. h(t) = - 4.9t2 +4.9t +
3.05). From there the students should substitute t=1 into the equation and see if the answer is 3.05m
(i.e. if the ball is at the right height to go in after 1 second).
8 “An irresponsible driver who was texting and driving missed a turn while driving on a hill that was 10m
high and flew for 5 seconds before hitting the ground below. How fast was the car travelling when it
went off the top of the hill?”
The correct answer is 22.5m/s. For this problem the students need to recognize that h0=10m
(the height of the hill) and h(5)=0. From there they should substitute t=5 into the equation (which
equals 0) and solve for v0). This problem is designed to be one of the “challenge” problems.
9 “The human cannonball is a circus act in which a person is placed inside of an air-powered cannon and
launched into the air, later landing safely in a well-placed net. Most of these cannons eject their human
cannonballs at around 25m/s, however, for more of a thrill, some circuses use cannons that reach
speeds up to 40m/s! What is the minimum height the circus tent could be to safely accommodate this
feat with the faster cannon?”
The correct answer is at least 82m. For this problem the students need to use the vertex
formula to find the maximum height that the human cannonball would reach, then find the lowest
height from the choices that is greater than the maximum height.
10 “As you saw in the first section of this module, the acceleration due to gravity on the moon is
1.62m/s2 compared to Earth’s of 9.8 m/s2. If you shot off a bottle rocket from ground level, with an
initial velocity of 30m/s how much higher or lower would the bother rocket go on the moon than on
Earth?”
The correct answer is 231.9m higher. For this problem, students should use their understanding
from the first section to note that the rocket will go higher on the moon than on Earth. From there,
students can find the x value for the vertex (time when the rocket reaches its maximum height) from the
vertex formula and substitute that back into the kinematic equation to find the maximum height of each
rocket (45.9m for Earth and 277.8m for the moon). Subtracting the two heights will give 231.9m.
Quiz
For this quiz, use the kinematic equation h(t)=1/2gt2+v0t+h0 (where g= - 9.8 m/s2 unless otherwise
specified in the problem) and the vertex formula 𝒙 =
−𝒃
.
𝟐𝒂
Round your final answer to one decimal
place. Remember the function uses height in meters. Good Luck!
1. What are the x and y coordinates of the vertex of y=2x2 + 4x + 12? Is it a maximum or a minimum?
A. (-2, 12) Minimum
B. (-2, 12) Maximum
C. (-1, 10) Minimum
D. (-1, 10) Maximum
2. In his 2nd round at the 2012 British Open, Tiger Woods chipped in for birdie on the 18th hole from a
treacherously deep bunker. The sand trap was close to 3 meters deep in some places! Assuming
Tiger's ball left his club face at around 25m/s, which equation would model the path of his ball?
A. h(t)= - 4.9t2 + 25t - 3
B. h(t)= - 4.9t2 + 25t + 3
C. h(t)= - 4.9t2 - 25t - 3
D. h(t)= 4.9t2 + 25t - 3
3. Consider the graph of y=x2. What happens to the graph if you change the "a" value to be -5?
A. The parabola gets wider and opens downward.
B. The parabola gets wider, but still opens upwards.
C. The parabola gets thinner and opens downward.
D. The parabola gets thinner, but still opens upwards.
4. In his rookie season, Panthers' quarterback Cam Newton broke the NFL's rookie season passing
record previously held by Peyton Manning. Part of Newton's passing effectiveness comes from his
height; at 6' 5" (about 2m) Newton is easily able to keep his passes from being swatted down by the
defensive line. Assuming that a defensive linemen's hand can reach up to 10 ft (3.05m) in the air and
that Cam Newton's passes cross the defensive line about half a second after release, which equation
would model a pass from Cam Newton that clears the defensive line (doesn't get swatted down)?
(Hint: Think about what the initial height of the football would be in this problem and what it means
in the equation for the ball to clear the D-line)
A. h(t)= - 4.9t2 + 5t + 2
B. h(t)= - 4.9t2 +10t
C. h(t)= - 4.9t2 + 2t + 2
D. h(t)= - 4.9t2 - 5t + 2
5. How would the graph of y = 5x2 + x change if you change "b" to be -3?
A. The parabola moves to the right and the vertex moves up.
B. The parabola moves to the right and the vertex moves down.
C. The parabola moves to the left and the vertex moves down.
D. The parabola moves to the right and the vertex moves up.
6. The equation y = 4x2 + 8x has its vertex at (-1, -4). What are the coordinates of the vertex of the
equation y = 4x2 + 8x + 8?
A. (-1, -4)
B. (1, 4)
C. (1, -4)
D. (-1, 4)
7. In game 1 of the 2012 NBA Finals, LeBron James scored 30 points! Ideally, LeBron releases his jump
shot at the top of his jump around 10ft (3.05m). For a three-pointer, the ball travels through the air
for about 1 second before reaching the basket which is 10ft high (3.05m). If LeBron shoots the ball at
4.9m/s will he make the shot?(Hint: For LeBron to make the shot, h(1) needs to equal the height of
the basket.)
A. Yes, he will make the shot.
B. No, he will miss the shot.
8. An irresponsible driver who was texting and driving missed a turn while driving up a hill that was
10m high and flew for 5 seconds before hitting the ground below. How fast was the car travelling
when it flew off the hill? (Hint: Think about which parameters you have, which one you're looking for,
and what h(5) is if the car reaches the ground after 5s.)
A. 50m/s.
B. 22.5m/s.
C. 38.7m/s.
D. 2m/s.
9. The human cannonball is a circus act in which a person is placed inside of an air-powered cannon
and launched into the air, later landing safely in a well-placed net. Most of these cannons eject their
human cannonballs at around 25m/s, however, for more of a thrill, some circuses use cannons that
reach speeds up to 40m/s! What is the minimum height the circus tent could be to safely
accommodate this feat with the faster cannon?
A. 40 m
B. 56.2 m
C. 121.6 m
D. 82 m
10. As you saw in the first section of this module, the acceleration due to gravity on the moon is 1.62m/s2 compared to Earth's of -9.8 m/s2. If you shot off a bottle rocket from ground level, with an
initial velocity of 30m/s how much higher or lower would the bottle rocket go on the moon than on
Earth?
A. 231.9 m Higher
B. 231.9 m Lower
C. 8.2 m Higher
D. Same Height
Eraser Chaser
Student Activity Sheet
Name:
Station 1:
h0 =
hmax1 =
hmax2 =
hmax3 =
hmax4 =
hmax5 =
t1 =
t2 =
t3 =
t4 =
t5 =
hmax(avg) =
t (avg) =
t1 =
t2 =
t3 =
t4 =
t5 =
hmax(avg) =
t (avg) =
v0 =
Station 2:
h0 =
hmax1 =
hmax2 =
hmax3 =
hmax4 =
hmax5 =
v0 =
Station 3:
v0 = 0 m/s
t1 =
t2 =
t3 =
t4 =
t5 =
t (avg) =
h0 =
Station 4:
v0 = 0 m/s
t1 =
t2 =
t3 =
t4 =
t5 =
t (avg) =
h0 =