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Elevated Projectile Motion

  1. Dec 23, 2007 #1
    Hello,
    I hope I posted this question under the right forum. :blushing:

    I am working through a problem I thought of a few days ago, just for fun, and it's driving me insane. I've gone through numerous text books, none of which seem to have just the right equations.

    Here is the scenario: I have a ball which weighs .1kg that is being launched from "0" height, to an elevated target 20ft away, which is 10ft above the ground.

    I am trying to figure out what the required force is to launch the ball in a parabolic path to the target. This seems like there should be an equation that you can plug in the numbers and angles, but everyone I have come across has had 1 too many unknowns in it. Any of my attempts to eliminate a variable have been in vain. Any ideas? :confused:

    Thanks in advance for the help, - Peter
     
  2. jcsd
  3. Dec 23, 2007 #2
    If air resistance is negligible, the resulting equations of motion describe a parabola that passes through the points (0,0) and (20, 10). You can either hit the target when the ball is decelerating (on its way up) or when it is accelerating (on its way down). Let's suppose you want to hit it when the ball is at apex, so no energy is wasted. Then the equation of the parabola is of a parabola concave down with vertex at (20, 10) and a root at the origin. From your knowledge of curves, you can derive the Cartesian equation as y -10 = S*(x - 20)2 where S is a scale factor that places our roots. Solving for S by plugging in the point (0,0), we get y - 10 = -(x - 20)2/40 as the path of our ball. With what velocity do you want it to impact the target? Assuming that velocity is v_M, we note that this will only be the horizontal velocity of the projectile, as a projectile at its apex has a vertical velocity component of 0. We then replace x in our equation with the parametrization x(t) = v_M*t (standard equation of an object moving at uniform velocity) to get y(t) instead of y(x). We note that v(0, 0) = (x'(0), y'(0)). The magnitude of the velocity (Sqrt[x'(0)2 + y'(0)2) is all we need, as we're not interested in applying two separate forces in each Cartesian direction, just one at the angle we want. Then we note F = m*dv/dt. In this case, The force required is up to your firing mechanism, as you need to supply a dt to get a force back. dt is usually very small. As a better approach, you can use energy, in which case, you will have a definite result. The kinetic energy of the ball is 1/2*m*v2 (just remove the square root in the equation for the magnitude of the velocity). This is the minimum amount of energy you need to apply to the ball to get it to hit the target, as a lot of the energy will be wasted in heat, friction, and compressing the ball instead of actually moving it.
     
    Last edited: Dec 23, 2007
  4. Dec 23, 2007 #3
    The ball is a standard tennis ball. Air resistance does need to be taken into consideration.

    http://wings.avkids.com/Tennis/Book/ball-01.html

    The specs from that website should be appropriate. Thanks for the quick reply, - Peter
     
  5. Dec 23, 2007 #4
    In the case of air friction, the shape of the path is not a parabola, but a parabolic-like arc that is slightly squished in from the side away from the launch.
    We know that the force on the ball at any particular moment is [itex]F_net = F_g + F_f[/itex] where Fg is the force due to gravity and Ff is air friction. The standard expression for air friction of large objects moving at non-negligible speeds is [itex]F_f = 0.5\rho v^2 C A[/itex] where [itex]\rho[/itex] is air density, C is the drag coefficient that depends on the object's geometry, and A is the cross-sectional area of the object. You can get an idea of C from this page, where it gives typical coefficients for rough and smooth spheres, so yours is between 0.1 and 0.4. This is assuming you're launching the ball at a reasonable velocity, not slowly.
    The challenge in the case of friction is in solving the differential equation [itex]m \frac{dv}{dt} = (0, -mg) + (F_f, 0)[/itex]. I assumed that there would be negligible friction in y for simplicity. If the launch angle is great, you will have to account for friction in y, however. in the simpler case above, we have two equations m*dv1/dt = -k*v2 where k takes care of all the constant terms in the formula for air friction and m*dv2/dt = -mg, which easily integrates to v2(t) = g*t + v2(0). With regards to the horizontal equation, we get v1(t) = -1/[K(t - 1) - 1/v1(0)] by separation, where K = k/m. We can now integrate to get distance. The integration of the y direction is no problem and it gives us the equation y(t) = -(1/2)*g*t2 + v2(0)*t as we expect, allowing us to solve for v2(0) as we want y(T) = 10 for some time T of your choosing, as in the previous problem. Similarly, we get the expression for x(t) by integrating v1(t) with respect to t, giving x(t) = -ln(1+K*v(0)*(1-t))/K. We can now get x(T) = 20 and use it to solve for v1(0). The less time you give for T, the greater the force will be.
    We then have our initial velocity vector v = (v1(0), v2(0)) = ((e-20K-1)/(K*(1-T)), 10/T + (g*T/2)), which you can use as in the previous post to get the minimum force or energy required. If the ball is launched very quickly (ie., by impact), you can also work out how much energy is wasted by your ball when launched by dropping the ball so that it accelerates to near launch velocity and measuring the height of the bounce, then taking the change in potential energy mg(h2 - h1) where h is height that the ball is dropped from. Since you have to accelerate the ball from 0 to the magnitude of the velocity vector here, you have again F = m*v/t_I, where t_I is the total time your machine applies force to the ball, which should be pretty small.
     
    Last edited: Dec 23, 2007
  6. Dec 23, 2007 #5
    Sorry I didn't mention that earlier. Here is a really simple diagram to help illustrate the scenario: http://img515.imageshack.us/img515/2257/ballsb9.jpg

    Simple, but I hope it helps convey the idea. Thanks for the help so far! - Peter
     
  7. Dec 24, 2007 #6

    arildno

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    Of course you have too many unknowns.
    You have two important initial parameters, namely initial speed and launch angle.
    Various combinations of these parameters will yield different solutions.

    Let us examine the general case in which from initial position is (0,0), and "final" position is (a,b) (in your case, measured in feet, a=20, b=10).

    Now, assume launch speed [tex]v_{0}\neq{0}[/tex] and launch angle [tex]\theta\neq\frac{\pi}{2}[/tex] measured in radians (i.e, we do not shoot straight up!)

    We therefore have, as functions of time t:
    Horizontal displacement: [tex]x(t)=v_{0}\cos\theta{t}[/tex]
    Vertical displacement: [tex]y(t)=v_{0}\sin\theta{t}-\frac{gt^{2}}{2}[/tex]
    where g is the acceleration due to gravity.
    Therefore, since [tex]t=\frac{x}{v_{0}\cos\theta}[/tex], we have the parabolic shape:
    [tex]y(x)=\tan\theta{x}-\frac{gx^{2}}{2v_{0}^{2}\cos^{2}\theta}[/tex]
    Thus, since we are to have y(a)=b, the following equation must hold:
    [tex]b=\tan\theta{a}-\frac{ga^{2}}{2v_{0}^{2}\cos^{2}\theta}\to{v}_{0}^{2}=\frac{ga^{2}}{2\cos^{2}\theta(a\tan\theta-b)}[/tex]
    Note therefore, that as long as tan exceeds b/a, you can always adjust your initial speed to get to your desired point.

    And b/a is, of course, the slope of the straight line going through your initial point (0,0) and your endpoint (a,b).
     
    Last edited: Dec 24, 2007
  8. Dec 24, 2007 #7

    HallsofIvy

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    A TENNIS ball that weighs 0.1 kg? I don't believe it!
     
  9. Dec 25, 2007 #8
    HallsofIvy: I think tennis balls are supposed to have a mass of 57 grams, close to .1kg just for mathematical simplicity, at least while I try to understand the math. :smile:

    Thanks a lot for the help! I knew there was a "simple" way to figure this problem out. Special thanks to Arildno for putting the solution in LaTeX format. :smile:
     
  10. Dec 25, 2007 #9
    Sorry for the double post, but, something about this is not working out correctly.

    When I do my work, assuming: g = 9.81, a = 20, b = 10
    I get 73.533m/s for the initial velocity, which is close to 165mph! Now, for a quick reality check, I know for a fact that I don't have to chuck a tennis ball at 165mph to get it from (0,0) to (20,10). Am I simply doing the math wrong, or is there something else?

    Thanks in advance, - Peter
     
  11. Dec 25, 2007 #10

    arildno

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    Well, first off:
    Do not mix m/s^2 as you do in g, and feet as you do in your destinations!
    Furthermore, what was your choice of initial angle?
     
  12. Dec 26, 2007 #11
    My initial angle is from atan(b/a), is this correct? Also, am I reading the denominator of your equation correctly? "(2*cos^2)*((a*tan)-b)"
     
  13. Dec 26, 2007 #12

    arildno

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    Eeh, no.
    What do you mean by:
    "My initial angle is from atan(b/a), is this correct?"
     
  14. Dec 26, 2007 #13
    The arc tangent of b/a
     
  15. Dec 26, 2007 #14

    arildno

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    But that is precisely the angle you CANNOT use!
    Do you understand why?
     
  16. Dec 26, 2007 #15
    Because the denominator turns to zero, I assume? So then, how do I find my initial angle? My only thought would be to use another formula to find the apex of the flight...
     
  17. Dec 26, 2007 #16

    arildno

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    I already told you.

    You cannot "find" your intial angle, it is an independent parameter.

    There are MANY combinations of initial angle&initial speed that enables you to reach the point (a,b), not just a single one! Infact, there are infinitely many such combinations, each of them a solution of the equation I gave you.

    If you CHOOSE an initial angle, and insert it in the equation I gave, you can determine what your initial speed must be in that particular case.
     
  18. Dec 26, 2007 #17
    Ahh, now I am begining to understand this reasoning a bit more. So if a choose an angle that "looks" appropriate, and the tangent of that angle is greater than b/a, then it should have a valid solution? Thanks for the quick replies!
     
  19. Dec 26, 2007 #18

    arildno

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    You can pick any angle lying strictly between arctan(b/a) and 90 degrees, yes.

    Note that the closer you get either of these limits, the required initial speed goes to infinity.

    This should tell you there exists an "ideal" angle [itex]\theta^{*}[/itex] somewhere in between, ideal just meaning this choice will require the LEAST initial speed.

    For your benefit, in terms of your coordinate parameters "a" and "b", we have that:
    [tex]\tan(\theta*)=\frac{b+\sqrt{b^{2}+a^{2}}}{a}[/tex]
    and the required initial speed associated with this angle choice is:
    [tex]v_{0,min}=\sqrt{g(b+\sqrt{b^{2}+a^{2}})}[/tex]
     
  20. Dec 26, 2007 #19
    Thanks for adding those extra formulas to find the most efficient path, just to clarify:

    Continuing with my previous example, if a = 20, b = 10, then: My ideal angle would be 58.3 degrees? Which would require a minimum velocity of 17.8m/s?

    Thanks so much for your help!
     
  21. Dec 26, 2007 #20

    arildno

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    Possibly. Remember consistent choices of units for length.
     
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