Question: After a football is kicked by a punter, it’s subjected to gravity, and follows a parabolic
path. The height of one particular punt is shown in the graph. The first coordinates
represent the horizontal distance from the punter, and the second coordinates represent
the height of the ball above the ground. Both measurements are in feet. Answer each
question with a full sentence.
1. About bow high above the ground was the ball when it was kicked?
2. How far away from where it was kicked did the ball land?
3. How high did the punt go?
4. Use the graph and some ingenuity to estimate the actual distance the punt
travelled. (This is different from how far away it landed! Think of the parabola as
a road, and estimate how far you’d travel on that road.) There’s no right answer
here: A big part of the question is describing how you got your estimate.
The next graph also shows the height of the ball on the 𝒚𝒚 axis, but this time the 𝒙𝒙
coordinates are the number of seconds since the ball was kicked.
5. The hang time of a punt is how long it’s in the air before returning to the ground.
What was the hang time for this punt?
6. After how many seconds was the ball 45 feet above the ground?
7. The equation of the parabola representing the ball’s height 𝒇𝒇(𝒙𝒙) = −𝟏𝟏𝟏𝟏(𝒙𝒙 −
𝟏𝟏. 𝟖𝟖)𝟐𝟐 + 𝟔𝟔𝟔𝟔, where 𝒇𝒇(𝒙𝒙) is the height in feet and 𝒙𝒙 is the number of seconds after
it was kicked. Find the exact height after the number of seconds you found in
Question 6. How accurate were your answers to Question 6? Edit
Answer: We consider the image 1, and get a few data points (approximately) from the given image.
It can be given as:
Distance (𝑥𝑥) 10 25 45 75 105 125 140
Height (𝑦𝑦) 15 35 50 60 50 35 15
We can claim that the graph is parabolic and hence the relation between 𝑦𝑦 and 𝑥𝑥 is parabolic.
Let us approximate the equation as:
𝑦𝑦𝑖𝑖 = 𝑎𝑎𝑥𝑥𝑖𝑖
2 + 𝑏𝑏𝑥𝑥𝑖𝑖 + 𝑐𝑐 + 𝜖𝜖𝑖𝑖 for 𝑖𝑖 = 1,2, … ,7
Given the data, we would try to estimate the unknown constants 𝑎𝑎, 𝑏𝑏, 𝑐𝑐 by the method of least
squares. Define, 𝑒𝑒𝑖𝑖 = 𝑦𝑦𝑖𝑖 − 𝑦𝑦�𝚤𝚤 = 𝑦𝑦𝑖𝑖 − 𝑎𝑎𝑥𝑥𝑖𝑖
2 − 𝑏𝑏𝑥𝑥𝑖𝑖 − 𝑐𝑐; and define the sum of squares
𝑆𝑆 = �𝑒𝑒𝑖𝑖
= �(𝑦𝑦𝑖𝑖 − 𝑎𝑎𝑥𝑥𝑖𝑖
2 − 𝑏𝑏𝑥𝑥𝑖𝑖 − 𝑐𝑐)2
The unknown constants are calculated by minimizing 𝑆𝑆; i.e. solving the equations
𝜕𝜕𝜕𝜕 = 0;
𝜕𝜕𝜕𝜕 = 0 and 𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕 = 0
Or, there is another way to solve for the given model. We put the obtained data in the
model: 𝑦𝑦𝑖𝑖 = 𝑎𝑎𝑥𝑥𝑖𝑖
2 + 𝑏𝑏𝑥𝑥𝑖𝑖 + 𝑐𝑐 + 𝜖𝜖𝑖𝑖, and obtain the linear model:
𝒚𝒚 = 𝑋𝑋𝜷𝜷 + 𝜖𝜖
Where 𝒚𝒚 = (15 35 50 60 50 35 15)′ and 𝜷𝜷 = (𝑎𝑎 𝑏𝑏 𝑐𝑐)′ with matrix 𝑋𝑋, whose rows are
(1 𝑥𝑥𝑖𝑖 𝑥𝑥𝑖𝑖
2); 𝑖𝑖 = 1(1)7. The general solution of the linear model is given as:
𝜷𝜷� = (𝑋𝑋′
We obtain this general solution of 𝜷𝜷�, using R statistical software. The code is given as:
The obtained solution is: 𝑎𝑎 = −0.011, 𝑏𝑏 = 1.603 and 𝑐𝑐 = 0.047. Thus, the given model is:
𝑦𝑦 = 0.047 + 1.603𝑥𝑥 − 0.011𝑥𝑥2
1. Now given our measuring standards, the distance (𝑥𝑥) when the ball is kicked is 0, and hence
the height of the ball is 0.047. However, this cannot be true and corresponds to an error. So,
we observe the graph, and find the height to be approximately 2.5 ft when the ball is kicked.
2. Again, if we look at the graph, we find that the ball lands at 150 ft away.
3. To check the maximum height attained by the ball, we first need to find at what distance it
attains maximum height. It is obtained by solving: 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑 = 0, we find 𝑥𝑥 = 72.86 ft. The maximum
height is then 58.45 ft.
4. The distance covered by the ball is equivalent to the length of the arc. Let 𝑥𝑥0 be the horizontal
distance from the origin at any time and 𝑦𝑦0 be the height if the ball then. Then by Pythagoras,
2 + 𝑦𝑦0
2 = the square of the linear distance of the ball from the origin = 𝑙𝑙0
2 (say). So, the
instantaneous distance covered is hence
𝑑𝑑𝑑𝑑 = �(𝑑𝑑𝑑𝑑)2 + (𝑑𝑑𝑑𝑑)2 = �1 + �
Thus, we get the formula:
𝑙𝑙 = � 𝑑𝑑𝑑𝑑
= � �1 + [𝑦𝑦′(𝑥𝑥)]2
𝑑𝑑𝑑𝑑 = � �1 + (1.603 − 0.022𝑥𝑥)2
We again compute this integral using R statistical software and find the answer to be 202.75 ft.
The code is given as:
integrate(f, lower = 0, upper = 150)
5. If we consider the second image, the hang time is nearly equal to 3.5 seconds.
6. Judging by the graph, the ball was 45 ft above the ground approximately after 0.8 seconds
and 2.8 seconds.
7. The equation given to us is: 𝑓𝑓(𝑥𝑥) = −18(𝑥𝑥 − 1.8)2 + 60, where 𝑓𝑓(𝑥𝑥) is the height in feet
and 𝑥𝑥 is the number of seconds after it was kicked. To find the correct solution of the question
presented in 6, we put 𝑓𝑓(𝑥𝑥) = 45 and solve for it. We get 0.88 seconds and 2.71 seconds. The
answer provided in 6 are fairly accurate. Edit
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