Dot Physics

Snow Board Jump Help

I really shouldn’t do this. I might be helping someone to set up something dangerous. But, I am going to anyway. Here is a question posted on some forum. (actually, it is from math help forum)

“I’m anticipating a good winter this year, one with lots of snow. My yard is sloped quite a bit and it would be the ideal place for a huge snowboard jump, only problem is I need to calculate how fast I will be traveling when I hit the jump, how high and what angle the jump should be, and the distance and angle of the landing ramp to optimize my range.”

So, what am I going to do? I am going to give an example of how to solve such a problem and then I am going to make the solution as a spreadsheet. This way, you can enter your own dangerous setup and make your own ramp. NOTE: if you hurt yourself, really it is your fault and not mine, right? Actually, I am going to show you how to do this so that you won’t do it. DON’T build a ramp and jump. Don’t.

I have actually done this problem before (most notably in the infamous giant water slide jump). But, I will go ahead and start over. Mainly because I want to include small calculations that would have a frictional force and to see if air resistance needs to be included (I am pretty sure that it doesn’t need to be included).

The Setup

In this calculation, I am going to start with:

  • Person of mass m
  • Starting on a slope of incline theta
  • Starting a distance of a up the slope
  • A coefficient of kinetic friction mu between the board and the snow
  • A ramp at an angle of alpha above the horizontal and of length b

Here is a diagram:

i-11d9b50549630c6c1ff8adfec2e65ae1-page-0-blog-entry-2-11.jpg

The first thing to calculate is the speed of the snow board as it goes down and then up the ramp. To do this, I will use the work-energy principle. This says:

i-cff3752fe012df61ddb1edb46df0f458-work_1.jpg

Basically, work on a system changes its energy. Then I have the definition of work and the energy. Simple. To use this, I first need to determine my system. In this case, my system will be the snow boarder and the Earth. This means that there will NOT be work done by the gravitational force on the snow boarder, but there WILL be a gravitational potential energy of the boarder-Earth system. Next, I need to determine what force will do work on the boarder. Here is a free body diagram of the snow boarder.

i-cd663b901ecffe5f9e79c29375c7362f-fbd_1.jpg

This is a force diagram for the boarder going down the slope (it would look a little different going up the slope). But, the key idea is that there is only one force that can do work. The normal force (FN) doesn’t do any work because it is perpendicular to the displacement. That leaves the frictional force. To find this force, I will use the normal model for friction:

i-03c7ba1dce3f2b16cecae1458c913363-friciton_1.jpg

I am using N as the normal force. From the diagram above, and the idea that the snow boarder does not accelerate perpendicular to the ground, I can find the normal force as:

i-97d931fb6d8279d8b328d0e57f19accb-nup_and_down_1.jpg

Since that is the only force that does work, I can write the work-energy principle as: (I figure you can see the skipped step of solving for the frictional forces)

i-4d70230929e6981af3e8230845afe95e-work_with_friciton_.jpg

Now, for energy, I need to consider the beginning and end of my interval. Of course the beginning is at the top of the slope. The end will be at the top of the ramp. To make things as easy as possible, I will call the top of the ramp y = 0 meters. This means that at the beginning, there is no kinetic energy, but there is gravitational potential energy. At the end, there is only kinetic energy. Thus my work-energy equation becomes:

i-4bb8c915b31ebe1a12ca44b1a5d927b4-plug_in_work_2.jpg

Solving this for the final velocity

i-b6bafdf1d52902bede6d0afec5f8dace-la_te_xi_t_1_1.jpg

Does everything look ok?

  • a*sin(theta) – b*sin(theta) is the change in height. If this is negative, then there will be no velocity at the end because it won’t make it that high
  • This expression has the correct unit (sqrt(m^2/s^2))
  • If the coefficient of friction is zero, the velocity should be the same as if you drop it – this does check out. Also, the greater the friction coefficient, the lower the final speed (because of the negative sign).

Ok, now what about the after it leaves the ramp? Of course, I have done projectile motion before, so I will try to be brief. The key idea in projectile motion (assuming air resistance is small enough to be ignored – and I will look at that in later) is that the x- and y-motions are independent. This means that the following can be written:

i-79b1ebccaa1be1232eed0704eaa77207-projectile.jpg

The initial x- and y-velocities are:

i-8b5826ca5e92d0b51d6363ad367845b9-la_te_xi_t_1_2.jpg

In order to solve these two equations, I need to know how high (compared to the end of the ramp) the landing point will be. How about I call this s – the y value of the landing point (remember that the end of the ramp is at y = 0 meters). This means that s = positive is a landing point higher than the ramp, and s = negative would be lower.

Plugging in stuff, you will see that a quadratic equation needs to be solved. I am not going to write that out (but it isn’t too bad). If I call x1 = 0 meters (at the end of the ramp), then the landing location will be:

i-baef1fccae7219b043a39260a2934dbf-la_te_xi_t_1_4.jpg

I could combine this with the velocity above, but I am not going to write that out. I will put it in a spreadsheet for you though.

I put in some initial values. I found a site that said that coefficient of static friction between waxed skis and snow was 0.05 (http://www.newi.ac.uk/buckleyc/forces2.htm). REMEMBER – this is only for educational purposes. There totally could be an error in here. I played with it in the limiting cases and it seems ok, but you just never know. I have made mistakes in the past, I am sure I will makes mistakes again. Oh! Also, don’t forget about units. I put down my units, if you want to do it in feet, convert.

Well, what about the air resistance?

I said I would address this, and now I will. I will not model motion with air resistance but instead do a quick calculation to see if it even needs to be included. Let me look at the horizontal motion (since it is constant without air resistance). If the horizontal velocity is vx, then the magnitude of the air resistance can be modeled as:

i-44f2c222f64d0fa8dcfe01faf546be17-la_te_xi_t_1_51.jpg

Or basically, some constant times the magnitude of the velocity squared. I don’t want to find all these instead I will use the idea that the terminal velocity of a sky diver is about 120 mph (54 m/s). In terminal velocity, the air resistance is equal to the weight. So, I call call the air resistance force as Kv2, then:

i-26b48f1e2aeb1eb175fd1b6c24b197e1-la_te_xi_t_1_61.jpg

Where vt is the terminal velocity. If I put in values of m = 65 kg, then K = 0.22 Ns2/m2. Now I can calculate the horizontal air resistance force on the jumper. (yes, I know I made some assumptions here). If the initial horizontal velocity is 5 m/s, then the air resistance would be Fair = 5.5 Newtons. Over the course of the jump, this would only change the velocity a very small amount. I think it is ok to leave it off.

Comments

  1. #1 monika hardy
    September 14, 2009

    nice.
    any chance you could have a couple conversations with my kids about this? we have ning site and some have created a snomath group.. working on this exact idea. would love to empower them with online collaborations from experts…

  2. #2 Anonymous Coward
    September 15, 2009

    For a relatively short ramp, I would guess the dominant “non-ideal” effect would be the inelastic nature of the skier leaving the jump, rather than air resistance or friction.

    By “jumping” up at takeoff (or at the transition from a to b), the skier can increase her translational kinetic energy, in a way that would favor shallower ramp angles (for maximum distance travelled). Conversely, if the transition from section a to section b is too abrupt and the skier’s legs buckle, this could reduce the distance travelled.

  3. #3 Rhett
    September 16, 2009

    @AC,

    You are totally correct. I left out the effect of the human jumping. Of course, this is something I can’t really take into account since everyone would jump differently. Good point though.

  4. #4 name
    November 18, 2009

    I just Told Everyonne I know About Your site,

  5. #5 csa
    February 4, 2010

    Just happened to stumble across this post, which I plan to steal to use with my senior physics class. If this theft isn’t alright with you, please don’t hesitate to let me know via email. I’m looking forward to reading more of your work!

  6. #6 Rhett Allain
    February 4, 2010

    @csa,

    Steal away. Take what you like and use it for good, not evil. This is what makes the internet so great.

  7. #7 johnny
    February 23, 2010

    how about this . start small with your jump and slowly build it up including the landing, so you dont make it too big and over shoot your landing. at least if you do over shoot your landing at first you wont be very high in the air. a flat top jump is more forgiving to coming up short than a double jump.

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