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Trajectory of a squirt of water to hit a fly

  1. Sep 25, 2009 #1
    Paradoxical Algebra: Trajectory of a squirt of water to hit a fly

    Can ANYONE explain to me why 3 perfectly algebraically sound procedures all give me a different answer, where only one gives me the correct answer? Also, can you disprove the 2 wrong procedures (IE: As I said before, tell me why they don't work)?

    1. The problem statement, all variables and given/known data
    http://media.wiley.com/product_data/excerpt/19/04717580/0471758019.pdf
    Problem #45

    Upon spotting an insect on a twig overhanging water, an archer fish squirts water drops at the insect to knock it into the water. Although the fish sees the insect along a straight-line path at angle phi and distance d, a drop must be launch at a different angle theta if its parabolic path is to intersect the insect.

    If phi = 36 degrees, d = .900 m (direct path from frog to fly; hence, x = .9cos36 and y = .9sin36), and the launch speed is 3.56 m/s, what angle theta is required for the drop to be at the top of the parabolic path when it reaches the insect?


    2. Relevant equations
    y = tan[tex]\theta[/tex]x - ((g)(x2))/(2vi2)(cos[tex]\theta[/tex])2
    Quadratic Formula
    [tex] v^2 = v_0^2 + 2 a \Delta x[/tex]

    3. The attempt at a solution

    y = vertical distance
    y = .9sin36

    x = horizontal distance
    x = .9cos36

    g = 9.8
    VInitital: 3.56

    .9sin36 = tan[tex]\theta[/tex](.9cos36) - ((9.8)(.9cos36)^2)/(2(3.56)^2)(cos^2[tex]\theta[/tex])

    .529 = .726tan[tex]\theta[/tex] -((5.20)/(25.3))(1+tan^2[tex]\theta[/tex])
    .529 = .726tan[tex]\theta[/tex] - ((.206)(1+tan^2[tex]\theta[/tex]))
    .529 = .726tan[tex]\theta[/tex] - (.206 + .206tan^2[tex]\theta[/tex])
    .529 = .726tan[tex]\theta[/tex] - .206 - .206tan^2[tex]\theta[/tex])
    .206tan^2[tex]\theta[/tex] - .726tan[tex]\theta[/tex] + .735 = 0

    Quadratic formula: No Real Answer

    Doing the same thing above, but changing up the way I tackle the equation slightly... which is algebraically sound, but provides a different answer (Wrong answer albeit):

    Here I even tried number crunching it differently:

    y = tan[tex]\theta[/tex]x - ((g)(x2))/(2vi2)(cos[tex]\theta[/tex])2

    y/x = tan[tex]\theta[/tex] - ((g)(x))/(2vi2)(cos[tex]\theta[/tex])2

    (.9sin36/.9cos36) = tan[tex]\theta[/tex] - g(.9cos36)/(2(3.56)2)*(1(cos[tex]\theta[/tex]))

    .4755 = tan[tex]\theta[/tex] - .2815(1 + tan2[tex]\theta[/tex]

    -.2815tan2[tex]\theta[/tex] + tan[tex]\theta[/tex] -.2815 - .4755

    -.2815tan2[tex]\theta[/tex] + tan[tex]\theta[/tex] -.757

    x = 3.24, .308

    Why do I get an answer (the wrong answer) doing it THIS way which is essentially the EXACTLY same thing I just did?

    How to get the "right answer", but why do I get a different answer than the previous 2 solutions?

    [tex] v^2 = v_0^2 + 2 a \Delta x[/tex]

    0 = 3.56sin[tex]\theta[/tex]^2 + 2(-9.8)(.9sin36)
    0 = 3.56sin[tex]\theta[/tex]^2 - 10.37
    10.37 = 3.56sin[tex]\theta[/tex]^2
    3.22 = 3.56sin[tex]\theta[/tex]
    3.22/3.56 = sin[tex]\theta[/tex]
    = 64.8 degrees (The REAL answer)

    Once again, back to square one of how these 3 formulas give me seperate answers, but are all sound to me in theory.
     
    Last edited: Sep 25, 2009
  2. jcsd
  3. Sep 25, 2009 #2

    djeitnstine

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    I have to see the definition of d but as far as I can see, if it is to be at the top of the parabolic path it should be [tex]x=\frac{d}{2}[/tex]
     
  4. Sep 25, 2009 #3
    "d" = .9 is the diagonal path from the fish to the fly.

    Therefore, I can consider it to be the hypotenuse.

    Therefore, shouldn't x = .9cos36 and y = .9sin36 in my trajectory equation?

    I seriously can not think of ANY reason why my method produces a non-real answer...
     
  5. Sep 25, 2009 #4

    djeitnstine

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    Only other questions I can ask are: what is the definition of [tex]\phi[/tex]

    and what are the units of v.
     
  6. Sep 25, 2009 #5


    [tex]\phi[/tex] = 36 = the angle inbetween the D and the surface of the water. Once again, hence:

    x = .9cos36
    y = .9sin36

    Velocity is standard m/s... apparently I skipped writing that down in the problem.

    So, again... there needs to be no unit conversions in my trajectory equation, and my coordinates should be right.

    g = 9.8
    x = .9cos36
    y = .9sin36
    VInitital: 3.56

    (1/cos^2[tex]\theta[/tex]) = (1 + tan^2[tex]\theta[/tex])
     
  7. Sep 25, 2009 #6
    Here I even tried number crunching it differently:

    y = tan[tex]\theta[/tex]x - ((g)(x2))/(2vi2)(cos[tex]\theta[/tex])2

    y/x = tan[tex]\theta[/tex] - ((g)(x))/(2vi2)(cos[tex]\theta[/tex])2

    (.9sin36/.9cos36) = tan[tex]\theta[/tex] - g(.9cos36)/(2(3.56)2)*(1(cos[tex]\theta[/tex]))

    .4755 = tan[tex]\theta[/tex] - .2815(1 + tan2[tex]\theta[/tex]

    -.2815tan2[tex]\theta[/tex] + tan[tex]\theta[/tex] -.2815 - .4755

    -.2815tan2[tex]\theta[/tex] + tan[tex]\theta[/tex] -.757

    x = 3.24, .308

    Why do I get an answer (the wrong answer) doing it THIS way which is essentially the EXACTLY same thing I just did?
     
  8. Sep 25, 2009 #7
    You are making this far more complicated than it is. There's no need to try to model a parabola shape for the trajectory of the drop. If the drop is fired at an angle upwards, the force of gravity will act on the vertical component, creating a parabolic trajectory. The only equation you need is [tex] v^2 = v_0^2 + 2 a \Delta x[/tex].
     
  9. Sep 25, 2009 #8
    How does that help though? I don't know the final velocity OR the angle at which the projectile is fired at initially which with basic kinematics or whatever, I can't use the components of VFinal or VInitial.

    I think you are mixing up the fact that I don't know the angle at which the projectile is fired at because phi is a different angle.

    Look here:
    http://media.wiley.com/product_data/excerpt/19/04717580/0471758019.pdf

    Question #45

    As you can see, that equation doesn't help me -- at all.

    This question just is bugging me because I am seriously finding NO USE for the trajectory equation because every single time my teacher either uses it himself, or I try to use it -- it ends up either making no sense HOW he uses it, or the answer doesn't work in my case.

    Either way, if I can somehow make basic kinematics work, it still does't prove how my sound algebra above doesn't work because I modelled the trajectory perfectly theoretically.

    The Vf thing, though, won't work:

    Vfsin[tex]\theta[/tex] = 3.56[tex]\theta[/tex] + 2a(.9sin36)

    I'm left with the theta variable and the final velocity.

    I can't solve for final velocity, though, because I have no idea what time is or what theta is for the X components to help.
     
    Last edited: Sep 25, 2009
  10. Sep 25, 2009 #9
    You do know Vf, what must it be if the drop hits at it's maximum height. You can ignore the horizontal component, it doesn't change. Think of the problem in only the y dimension.
     
  11. Sep 25, 2009 #10
    Oh... right. Okay, then it works, but I would still like to know WHY my algebra is wrong (assuming this gives me the correct answer).

    I can't find any argument to why my algebra is not giving me the correct answer. What exactly DID I solve for? Also, why exactly are BOTH my answers different when I used the same numbers and non-illegal algebraic maneuvers.
     
    Last edited: Sep 25, 2009
  12. Sep 25, 2009 #11
    [tex] v^2 = v_0^2 + 2 a \Delta x[/tex]

    0 = 3.56sin[tex]\theta[/tex]^2 + 2(-9.8)(.9sin36)
    0 = 3.56sin[tex]\theta[/tex]^2 - 10.37
    10.37 = 3.56sin[tex]\theta[/tex]^2
    3.22 = 3.56sin[tex]\theta[/tex]
    3.22/3.56 = sin[tex]\theta[/tex]
    = 64.8 degrees (The REAL answer)

    Once again, back to square one of how these 3 formulas give me seperate answers, but are all sound to me in theory.
     
    Last edited: Sep 25, 2009
  13. Sep 25, 2009 #12
    You have to square Vo.
     
  14. Sep 25, 2009 #13
    Jebus, the equation I used way up is the formula for the trajectory of a projectile... I'm going to be posting another question somewhat similar that my prof did (even though I believe that his algebra is paradoxical at best).

    Yeah, okay... I did some minorly bad algebra there, but this question is frustratingly dumb as the trajectory formula I think is wonky at best.
     
  15. Sep 25, 2009 #14

    djeitnstine

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    Rogan, your equation of parabolic motion is fine, my guess is that equations are saying the velocity given is not fast enough to hit a target at that height.
     
  16. Sep 25, 2009 #15
    Yes, but as you can see (look back to my original post as I edited in the new equations)... It can reach that height because using the same equation, but changing up how I do the algebra, I can produce an answer (but the wrong answer).

    Also, as you can see in the third part with the basic kinematics can find the answer the text book wants.

    As far as I'm concerned, all 3 of these equations should solve for the same thing.
     
  17. Sep 25, 2009 #16
    If you plug all the values into your original equation it doesn't even come out correct, so something is wrong.
     
  18. Sep 25, 2009 #17
    That much I know, but if you look at the other question I just posted, my professor did something very similar and got a correct answer.

    The text book does no job explaining why my theoretical formulas for projection don't work.
     
  19. Sep 25, 2009 #18
    Furthermore, look at the following:

    Wikipedia:
    http://en.wikipedia.org/wiki/Trajectory_of_a_projectile

    Angle [tex]\theta[/tex] required to hit coordinate (x,y)

    That gives me the same answer that I did in attempt #2 at the equation which isn't the right answer... but that CLEARLY is a formula I should be able to use.

    Again, I also did this for another similar equation and the same thing happened. I think trajectory is a false formula.
     
    Last edited: Sep 25, 2009
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