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Finding Angles in Projectile Motion

  1. May 20, 2013 #1
    1. The problem statement, all variables and given/known data

    A fire hose held near the ground shoots water at a speed of 6.8 m/s. At what angle(s) should the nozzle point in order that the water land 2.0 m away? Why are there two different angles?

    ax=0
    xo=0
    x=2.0 m

    ay= 9.80 m/s/s
    y0=0
    y= I'm guessing 0, because it lands on the ground at approximately the same level..



    2. Relevant equations

    I'm guessing:

    x = Vx0t

    Vx0 = 6.8 cosθ

    Vyo= 6.8 sinθ

    y= y0 + Vyot - 1/2ayt2



    3. The attempt at a solution

    Oh my god. I have no idea why I'm having so much trouble with this problem. I've been trying to solve this for about two hours. I've re-written the equations in my book multiple times and I don't know how to solve this. I've been trying to solve for the time but there are so many variables.. I'm not even really familiar with cosine and sine. The problem implies that there are two angles in the answer. So I'm guessing it's because there's some sort of quadratic equation involved. I don't know. :/

    I even tried to find a way to use substitution and find the time, but it was no use. Please, help! :'(
     
  2. jcsd
  3. May 20, 2013 #2

    mfb

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    You can use the last equation to find the time (as function of θ, everything else is known), and find x based on that. As y and y0 are 0, you do not have to solve a (real) quadratic equation.
     
  4. May 20, 2013 #3
    I don't understand.. how would I do that?? I already know that x is 2.0m.. Right? The final position is 2.0 meters away.. If I don't know Vyo, then how can I find t?? I'm so confused. Maybe it's the math :/
     
  5. May 20, 2013 #4

    mfb

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    Replace Vx0 and t by their formulas - you have the one for Vx0, you just need one for t.
    Exactly, and you can use this to solve for the angle afterwards.

    Anyway, try to use the last equation first. Plug in everything you know about the parameters there, simplify, try to solve for t. If you get stuck somewhere, post your steps here please.
    It is just mathematics, indeed.
     
  6. May 20, 2013 #5
    When I plug everything in, should I have:

    0=6.8sinθ - 1/2(-9.80)(2.0/6.8cosθ)2

    ? Because t= x/Vx0
     
  7. May 20, 2013 #6
    Uhm, so I did this:

    0 = 6.8sinθ + 4.90[4.0/(6.8cosθ)2]

    =6.8sinθ+ 19.6/(6.8cosθ)2

    =(6.8sinθ)(6.8cosθ)2+ 19.6/(6.8cosθ)2

    But I don't know what to do now.. Do I just remove the 6.8cosθ2?
     
  8. May 21, 2013 #7

    mfb

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    Don't replace t in the last equation, that makes it more complicated.

    0 = Vyot - 1/2ayt2
    Can you find solutions for t in this equation? Something like t=...?
    One solution is given by t=0 - it has to be, as you start at the ground. But there is another solution.
     
  9. May 21, 2013 #8
    So like:

    4.9t^2 + 6.8sin(theta)t = 0

    t(4.9t + 6.8sintheta)= 0

    t=0 or t= -6.8sin(theta)/4.9

    ?? Thank you so much for helping me, by the way. If what I did is right, then do I put that into 2.0= Vx0t??
     
  10. May 21, 2013 #9

    mfb

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    Right. This will give you an equation where just θ is unknown.
     
  11. May 21, 2013 #10
    Okay but now the problem is that I don't know what to do with the cos and sin.. Is there a number for each? Is sin theta just multiplying the angle by something? How would I even multiply cos and sin together ? :/
     
  12. May 21, 2013 #11

    mfb

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    It is not.

    It could be enough to give that equation a computer (WolframAlpha can solve this, for example).
    If you have to show more steps, you can use ##\sin^2(\theta)+\cos^2(\theta)=1## to get rid of either sin or cos, and solve the equation for sin(θ) (or cos(θ)) afterwards.
     
  13. May 21, 2013 #12
    Well, when subbing your value for t into the equation for the displacement in the x direction, isn't there a trig identity that'd be much simpler to use?
     
  14. May 21, 2013 #13

    verty

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    There is yes, one of the rules you need to know. It is actually in reverse.
     
  15. May 21, 2013 #14
    I'm not actually the one asking the question. I was just trying to make the asker aware that there's a relatively simple solution available.
     
  16. May 21, 2013 #15
    Okay so from:

    9.8 = (6.8cosθ)(-6.8sinθ)

    How would I put in

    cos^2θ = 1 - sin^2θ ?

    If I have cosθ, how would I make that cos^2θ to put it in? I'm confused.. I'm in a college algebra class and it's basically one of the pre-requisites in math for graduating. It's the lowest college level math class and we are doing mostly functions. I don't know pre-calculus or anything. The class just started a week ago.. so yeah. :/ My physics class has no pre-requisite.. except intermediate algebra.
     
  17. May 21, 2013 #16

    cepheid

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    Keep in mind that cos2θ is just a shorthand. It means (cosθ)2. Therefore$$\cos\theta = \sqrt{\cos^2\theta}$$

    Well then you're in luck, because sine and cosine are functions! Trigonometric functions. It sounds like you need to brush up on trigonometry. I highly recommend doing some reading about sine and cosine online before proceeding further with this type of problem.
     
  18. May 21, 2013 #17
    Well, there are quite a few identities out there that can come in handy when solving variables which are part of trigonometric functions. Have a look around and I'm sure you'll find one that'll help you out. It's one of the more common ones.
     
  19. May 23, 2013 #18
    I think I might just give up on this problem. It's just making me anxious. I can't really "brush up" on trigonometry... I've never really learned it.
     
  20. May 23, 2013 #19
    Have a think about the identity sin(2x) = 2sinxcosx.
     
  21. May 23, 2013 #20

    verty

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    I would learn Trig, there are plenty resources available.

    PS. ##sin^2 + cos^2 = 1## was never the right formula to use. Why deal with those horrid squares and square roots?
     
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