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Initial Velocity for X displacement with air resistance

  1. Feb 1, 2006 #1
    I'm in a high school AP physics class but I suspect this question is considerably above the course level because I don't think it was intended to be a part of the project, so I'm not sure which forum to put it in.

    I would like to find the initial velocity required to launch a ping pong ball 25 meters at x angle (whatever the optimum angle for this would be), accounting for air resistance but assuming no wind.

    I measured the radius of the sphere (ping pong ball) to be 18.8 mm (.018 m) giving it a cross sectional area of .0111 m^2. The mass is 2.6g. The outside air density is about 1.22 kg/m^3.

    I came across the equation: F = 1/2*cd*p*a*v^2 where F is the force of air resistance on the object, cd is a coefficient of drag (I guess .5 for sphere?), p is air density, a is the cross section area of the object, and v is velocity.

    I can see how this is useful in determining the force of air resistance at a specific velocity, but I have no idea where to go from here to calculate initial velocity required to get it to fly a specific distance since the velocity, and consequently force of the air resistance, are dependant on each other and constantly changing.

    The reason I would like to know is because my physics teacher has designed a competition in which we must build a catapult that launches a ping pong ball to hit a target. Points (translated later into the grade for the project) are based on how accurately you can hit the target. Unfortunately, the target distance is random for each person (lol?.. it gets worse) from 5 meters to 25 meters. Just at a glance I have not the slightest clue how someone aiming for a target 25 meters away is going to compete with someone aiming for a target 5 meters away. However, factor in air resistance on PING PONG balls and the distance disadvantage/advantage is exponential, thus giving someone with a shorter distance a ginormous advantage (much more than 5x in 5 m vs 25 m because it's exponential). Since I'de rather my grade not be decided by lottery, I'de like to have some mathematical evidence to show how ridiculous this competition is. It is worth noting that the catapult cannot exceed w/l/h dimensions of 2'x2'x3' respectively. Assuming putting energy into this mechanism wasn't a problem, it seems even getting to the initial velocity required in 2 feet would require an acceleration so large that it would be near impossible to build something capable of withstanding that force.
     
    Last edited: Feb 1, 2006
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  3. Feb 2, 2006 #2

    andrevdh

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    The problem is compounded if the launching mechanism put some spin on the ball - it will drift sideways due to the Magnus force.
     
  4. Feb 2, 2006 #3

    Galileo

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    If we assume the resisting force varies linearly with the velocity an exact solution can be found. But a quadratic velocity dependence is probably closer to reality.
    I don't think they expect you to solve the trajectory of the particle analytically, but you could try numerical methods with the computer.
     
  5. Feb 2, 2006 #4
    As far as how fair the contest is if the target gets smaller as the distance gets closer you really have no advantage considering the distance.

    Anyhow, unless your class is doing some differential equations you really can't calculate any kind of basic air resistance. I think the best method to maximize your grade for this class is stick with kinematics and experimentation. Build your catapult, find out how it shoots, use some kinematics to come up with an Intial launch velocity and just try to get a ball park figure. Your physics teacher doesn't expect you to land a rover on mars, he just wants you to think about how to use some of the stuff you have been covering.
     
  6. Feb 2, 2006 #5
    I doubt he had any plans for us to solve the trajectory at all, it's a fairly basic physics class and this type of air resistance calculation is way beyond what we've ever done. I've looked for computer programs to help solve this (I'm sure they exist) but haven't had any luck finding one.

    The target would have to get exponentially bigger as the distance becomes larger because the margin of error is increased exponentially with larger distances I believe. Regardless, the target doesn't change - it's a plastic tub with a 8"x9" opening 5" or so tall.

    Yeah we're really not meant to calculate this, but I consider this competition almost entirely based on luck and just about impossible if you are unlucky enough to get a far target. The idea behind the random target distance is just so we have to build a launcher with variable distance, unfortunately I don't think he realized the problems with doing that.

    The reason for calculating this not to get the perfect launch angles/velocities/spin/etc, I'm interested in proving that it is not possible to hit a 25m target with the restrictions he gave us.

    I found a vacume powered ping pong ball cannon that barely launched the ball 25m. This cannon was 1m long and launched the ball with a velocity of over 150 m/s giving it an average acceleration of over 1,000 times that of gravity while in the cannon. There is simply no way anyone is going to be able to build a catapult with that sort of acceleration. Even if it were possible, no one could be accurate with numbers that high trying to get into a basket so small.
     
    Last edited: Feb 2, 2006
  7. Feb 2, 2006 #6
    well if you use the range equation [tex] R=\frac{V_o^2sin2\theta}{g} [/tex] and allow [tex] \theta [/tex] to be equal to 45 degrees then you get

    [tex] \sqrt{Rg}=V_o [/tex] now you know roughly how fast something has to be going initially to go the distance you want.

    I think this project is completely reasonable and can be done. You just need to TRY AND THINK OF A WAY TO COMPLETE IT. Try something, research designs, make theoretical calculations...anything!

    Here is something that might help you design something

    http://www.fsea.org/pdf/CT2 Advanced Catapultpdf.PDF


    As far as the error, what you stated is exactly what I stated. If the target is smaller up close then it would have to be larger far away....THE EXACT SAME THING I STATED.
     
  8. Feb 2, 2006 #7
    That works great if air resistance was negligible, but this is a ping pong ball. It flies slightly better than a balloon. Air resistance is the real problem here, not gravity.

    I'm very competitive with this sort of thing and have thought of at least 20 different designs from gravity powered trebuchets to elastic powered spinning arms to high pressured air shot behind the ball. Unfortuantely, because of the massive role of air resistance on the trajectory of a ping pong ball, I find it highly impractical to get it 25 meters, let alone with a shred of accuracy. If this was a golf ball, no problem, but it's not - that's why I'm here in the first place.

    Sorry, perhaps I worded it badly. I was not disagreeing with you, just stressing the point that the difference would be very large to make it fair (exponential rather than say twice the size at twice the distance - I know this doesn't go against what you said at all) yet the target does not change in size at all.
     
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