Math abounds in the Olympics, whether it’s
data around Olympic records, medals won, or
other statistics. In this issue we will look at the
dramatic sport of ski jumping, and its
application to quadratics and other non-linear
models. Take a look at this interactive to learn
more about ski jumping (Source: NY Times).
The path of the ski
jumper can be
approximated with a
parabola. From the
NY Times interactive
we learn that the skier
starts her jump 15 ft
off the ground and
can cover the length
of a football field. We
can sketch out a
model for this.
We get three
points: (0, 15), the
(h, k), the vertex of
the parabola, and
(x, y) where the
skier lands. The
covered is 300 ft.
and is the
hypotenuse of a
Use the Pythagorean Theorem to find the
You can now find the coordinates of the point
where the skier lands.
We now have two coordinates and can
conceive of the vertex of the parabola, (h, k).
From the video we see that the skier elevates
by about 3 ft. We can then assign a value to k.
The equation for a parabola in vertex form has
two variables, a and k, that are unknown. But
we do have two coordinates and can solve for
these two values.
Here is the solution using the first set of
coordinates. We end up with an equation that
shows a as a function of k.
Here is the solution using the second set of
coordinates. We end up with a second equation
that shows a as a function of k.
Substitute one equation into the other to solve
Substitute the value of h into one of the
equations to find a. We now have all the
parameters for the parabola in vertex form.
Here is the graph of the parabola, with two key
Now that you’ve seen how to construct a
quadratic model using curve-fitting techniques,
expand on your work:
• Try different values for k to see what impact
it has on how far the ski jumper jumps.
• Try a longer slant height to see what the
equation is for different values of k.