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Terminal velocity of a lacrosse ball?

  1. Mar 27, 2007 #1
    Can anyone help me with the terminal velocity of a lacrosse ball?
    Approximations would help too.
    Thanks.
     
  2. jcsd
  3. Mar 27, 2007 #2

    berkeman

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    Thread moved to Homework Help, Intro Physics.

    bolini, you must show us some of your own work in order for us to help you. What are the relevant equations that determine the terminal velocity of an object? Tell us what you know.
     
  4. Mar 28, 2007 #3
    still need help

    The equation for terminal velocity is the square root of 2mg/CpA. I used 9.8 m/s for gravity, .5 or the drag co on a ball and 1.29 kg/m3 for air density. So I come up with approx 33 m/s.

    My question now is how do I come up with the drop height necessary to reach terminal velocity. Is there an acceleration formula that must be used?
     
  5. Mar 28, 2007 #4

    berkeman

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    I'd start with the kinematic equations of motion, and add in the air resistance term. Do you know how the forumula that you cite is derived?
     
  6. Mar 28, 2007 #5
    And the answer - pretty darn high. Without drag I get 55 meters using d=Vf^2 / Vi^2 +2*a with Vi being 0 and Vf being 33m/s and 'a' being 9.8

    Not sure how to include resistance in the equation because it varies depending on the velocity which is changing until 33m/s is achieved.
     
  7. Mar 28, 2007 #6

    berkeman

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    Yeah, there are different models at different levels of complexity for air resistance. At the simple end is having it vary as the square of the velocity, and at the complex end (like what the battle tank computers use to compute trajectories) I'm sure there are many, many terms.

    That's why I asked if you knew how they computed the simple terminal velocity formula. If you saw how they derived it, that would give you a clue as to which version of the air resistance force equation to use.
     
  8. Mar 28, 2007 #7
    I'm attempting to help my son with a elementary school project that measures the impact crater of a ball at different heights. I thought it would be cool to see if we could achieve terminal velocity, but at 180 feet without considering air resistance, I think we are already too high to practically consider.

    I believe the first equation basically used air resistance as the square of velocity. I am curious now, even though we probably won't look for a building or bridge high enough, what the height actually would be. If you could help me incorporate air resistance into the kinematic equation I used, that would be helpful.

    Thanks for your help up until this point. I forgot how interesting physics used to be for me.
     
  9. Mar 28, 2007 #8

    PhanthomJay

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    Actually, an object never reaches terminal velocity, it only approaches it. The solution to the time and distance it takes to 'approach' terminal velocity (say reach .99V_t) involves a non linear differential equation (mg - cpAv^2 = mdv/dt), the solution of which is beyond me at this point, but it involves a natural log term I think and probably 'e' raised to some power. But on a practical level, terminal velocity is ordinarily reached (approached) in less than 9 seconds (but much less for lighter objects like a feather or penny). So you are correct in that it would take a height much greater than 180 feet for the ball to attain a speed of 33m/s or so when you consider air resistance. I'm guessing that it might be around 500 feet or so.
    I believe there is a site I found awhile back that calculates the speed of an object in air at various points in its fall, and gives the speed at impact based on the height it is dropped from (and the objects weight and projected area, etc.). I'll see if I can find it.
     
  10. Mar 28, 2007 #9
    this one is for rockets and only computes Vterm: http://exploration.grc.nasa.gov/education/rocket/termvr.html

    I've thankfully only had one come in at terminal velocity, dug a hole and left a debris field of about 10 yeards radius. The circuit boards had all their components stripped off--not a pretty sight.

    PS: the full eqn for terminal velocity=

    sqrt(2mg/pACd)*{tanh(t)*sqrt[(g*p*Cd*a)/(2*m)]}
     
    Last edited: Mar 28, 2007
  11. Mar 28, 2007 #10

    PhanthomJay

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  12. Mar 28, 2007 #11
    Cool! Very helpful, thanks for the link.
     
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