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Cannon(Projectile Motion)

  1. Jun 7, 2009 #1
    1. The problem statement, all variables and given/known data
    A cannon with muzzle speed of 1000 m/s is used to start an avalanche on a mountain slope. The target 2000m from the cannon horizontally and 800m above the cannon. At what angle should the cannon be fired?



    2. Relevant equations
    A buttload of the projectile motion ones.


    3. The attempt at a solution
    Ok what I did was I tried to solve for t and get equations for the angle, but my lack of trig gets in the way. Ok for no better thing to use let :smile:=theta. I get 1000cos(:smile:)t=2000 so cos(:smile:)t=2. Then for the y component I get 800=sin(:smile:)t-4.9t^2. I solved here for t with the quadratic equation. Substituting the resulting gives me 19.6=500sin(2:smile:)+64sqrt[15625sin^2(:smile:)-245]. I don't know how to solve for :smile:. Help please! Thank you!
     
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  3. Jun 7, 2009 #2

    dx

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    Should be 800 = 1000sin(:smile:)t - 4.9t2.
     
  4. Jun 7, 2009 #3
    Yeah I just forgot to type that there, however I do not know how to solve the trig equation there.
     
  5. Jun 7, 2009 #4

    Pengwuino

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    Using your kinematic equations, you know...

    [tex]\begin{array}{l}
    y = \sin (\theta )v_0 t + \frac{{gt^2 }}{2} \\
    x = \cos (\theta )v_0 t \\
    \end{array}[/tex]

    If you isolate the trigonometric parts on the right side, you can divide the equations and find [tex]\frac{{y - \frac{{gt^2 }}{2}}}{x} = \frac{{\sin (\theta )v_0 }}{{\cos (\theta )v_0 }}[/tex]

    Now the problem is that t is still in there. I would suggest substituting [tex]t = \frac{x}{{\cos (\theta )v_0 }}[/tex]. Do some manipulation and you can determine your angle! If you need to use trigonometric identities, i highly suggest checking out http://en.wikipedia.org/wiki/Trigonometric_identities and really getting down the use of the identities.

    Note, there are 2 solutions to this problem. This is simply because there is a very large angle that you can shoot the cannon upwards such that it'll hit its target after its long but narrow arc is completed and it's falling downward.
     
    Last edited: Jun 8, 2009
  6. Jun 8, 2009 #5

    dx

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    Suppressing the constants, you should get an equation like

    [tex] \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos^2 \theta} = 1 [/tex]

    Wrtie sin(θ)/cos(θ) as tan(θ), and 1/cos2(θ) as 1 + tan2(θ).

    Putting constants back, you will get a quadratic in tan like tan2(θ) + B tan(θ) + C = 0.
     
    Last edited: Jun 8, 2009
  7. Jun 8, 2009 #6
    Solving for :smile: we get tan(:smile:)=0,-1. That is an angle of 0 degrees or 135 or 225? What?

    Oops sorry forgot about the constants..
     
    Last edited: Jun 8, 2009
  8. Jun 8, 2009 #7

    dx

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    What equation did you solve? This is the one you should have gotten:

    819.6 = 2000 tan(θ) - 19.6 tan2(θ)
     
  9. Jun 8, 2009 #8
    Um can you show me where you got those constants from?
     
  10. Jun 8, 2009 #9

    dx

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    You had the equation

    800 = 1000sin(θ)T - 4.9T2

    Now just substitue T = 2/cos(θ) in this.
     
  11. Jun 8, 2009 #10
    Ok I learned that 1/cos^2(:smile:)= Tan^2(:smile:)+1. That helped a lot so solving for :smile:, I finally get :smile:=22.36 or 89.43. Thanks a bunch both of you! BTW when did you guys learn these trig identities? I did not see that 1/cos^2(:smile:)=tan^2(:smile:)+1 on the wikipedia page.
     
  12. Jun 8, 2009 #11

    dx

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    I didn't see it at first either. I was breaking my head trying to solve it for cos(θ)! Im surprised 1/cos^2(θ)=tan^2(θ)+1 is not listed on wiki, it's a pretty standard identity, usually written in the form sec^2(θ) = 1 + tan^2(θ).
     
  13. Jun 8, 2009 #12

    Pengwuino

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