Ski jump landing and acceleration

My last Olympics post may have been a little complicated. I am going to try to make this one a little easier. In this post, I want to look at the landing portion of a ski jump. This could apply to THE ski jump, but there are some things in that even that make it a little more complicated (but I might come back to that in another post). For this case, I will consider the freestyle event - aerials. I didn't search too long, but here is a nice short video.

First, a quick estimation of how high they are "falling" from on the way down. In that video, the jumper takes about 1.5 seconds to get from the highest point to the landing. Using some kinematic equations, (or the work-energy principle) I can find a couple of useful things. How high? How fast?

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And how fast?

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These two numbers are really just for reference. And here is the big deal. What if you jumped off an 11 meter building? That would be bad, right? (although some people can jump off things like this - here is my dangerous, parkour, jumping calculator) So, this really has to do with acceleration. Acceleration is the change in velocity - let me write it like this:

i-caa83d68351f125689da77f0f9423e2f-2010-02-21_la_te_xi_t_1_17.jpg

Note that acceleration is a vector and so is the velocity. In short, a vector has both magnitude and direction (here is the long version of vectors).

Now I am going to draw a diagram for a person jumping something like an aerial, but on flat ground.

i-00a16492dd8ab9794bd904b754c7d3cb-2010-02-22_untitled.jpg

What is the acceleration during the landing? Well, this would be the change in velocity divided by how long this took. Let me draw this graphically.

i-ae32c080a1ce7b8d94a5648ae7287629-2010-02-22_untitled_2.jpg

Here I assume a time interval of 1 second so that the acceleration vector is the same length as the change in velocity vector. I labeled this aflat so that I could compare it to the next case. Needless to say, this acceleration can be quite large. You can make it survivable if you increase the time over which this change in velocity takes place - as they do in parkour by rolling or something.

Now I will look at landing on a sloped surface. Here is my new picture.

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Really the only difference in this case is that the final velocity is 'downhill' rather than flat. To make it realistic, I made the magnitude of the final velocity a little bit slower than the initial velocity. Now, let me draw the acceleration like I did before and compare to the acceleration when landing on the flat ground.

i-8a6725fc6d9163536cc3a3a917714c03-2010-02-22_untitled_4.jpg

So, the acceleration when landing on the sloped area is smaller in magnitude than on flat ground. In short this is because the velocity (vector) does not change as much. In an ideal landing, the slope would be almost in the same direction as the skier for a very small acceleration.

But wait! I know what you are thinking, the skier still has to stop - right? Of course you are correct. But in the case of the sloped landing, how does the skier stop? The slope gradually changes from going down to horizontal. This gradual change means that the change in velocity to zero can take a long time making the acceleration smaller.

A similar thing happens on the traditional ski jump - notice how they are landing on a downward slope.

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"A similar thing happens on the traditional ski jump - notice how they are landing on a downward slope."

On a traditional ski jump, the jumper's velocity vector is also much closer to horizontal than it is for a freestyler.

But on a traditional ski jump, you are falling much longer so you really need that 55 mph horizontal velocity at takeoff, even if you take flying into account.

In any case, my recollection is that the landing slope is less steep for a traditional ski jump (even ski flying) than it is for freestyle aerials. The slope at the inflection point takes those vectors into account.

A bigger deal is that the snow is kept loose in the landing area for aerials, but there is nothing but hard packed snow to land on if you get too far past the transition point and onto the flat of a jumping hill.

By CCPhysicist (not verified) on 27 Feb 2010 #permalink

@CCPhysicist,

I think you are right about the traditional jump. If you have a greater horizontal speed, after landing you will effectively still be moving down a lot on a shallow slop. Good point.