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Projectile Motion

  1. Jan 26, 2008 #1
    [SOLVED] Projectile Motion

    1. The problem statement, all variables and given/known data
    You are shooting a cannon from a height of 11.23m off the ground. The target is 20.36m away. The angle of projection is 30 degrees. What does the initial speed need to be in order to hit the target?


    2. Relevant equations
    When I look at this problem I just cannot solve it. I personally don't think I have enough information but it has really been bugging me.


    3. The attempt at a solution
    I know I need to figure out one of the velocities, either for the x or the y axes in order to solve for the initial velocity. The only information I have is:
    [tex]\Delta[/tex]y=11.23
    a=-10

    [tex]\Delta[/tex]x=20.36
    a=0

    I tried using the range formula before remembering that it can only be used when the initial and final velocities are the same.
     
  2. jcsd
  3. Jan 26, 2008 #2
    You have everything you need.

    [tex]x=(v_0\cos\theta)t[/tex]
    [tex]y=(v_0\sin\theta)t-\frac 1 2 gt^2[/tex]

    Solve for t in the x and plug it into y, then solve for v initial.
     
  4. Jan 26, 2008 #3
    Thanks very much, that makes perfect sense. I'll keep a lookout for problems like these in the future
     
  5. Jan 26, 2008 #4
    Welcome. You may also want to write those formulas down in your book for future reference so you don't have to go through the annoying algebra everytime.
     
  6. Jan 26, 2008 #5

    Yea

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    I just tried it and the answers does not make sense.
     
  7. Jan 26, 2008 #6
    What did you get?
     
  8. Jan 26, 2008 #7

    Yea

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    I got vo= 72
     
  9. Jan 26, 2008 #8
    Final equation should be ...

    [tex]v_0=\frac{x}{\cos\theta}\sqrt{\frac{g}{2(x\tan\theta-y)}}[/tex]
     
  10. Jan 26, 2008 #9
    I got the same problem... maybe our algebra is the problem
     
  11. Jan 26, 2008 #10
    [tex]y=v_0\sin\theta\left(\frac{x}{v_0\cos\theta}\right)-\frac{1}{2}g\left(\frac{x}{v_0\cos\theta}\right)^2[/tex]

    [tex]y=v_0\tan\theta-\frac{gx^2}{2(v_0\cos\theta)^2}[/tex]
     
    Last edited: Jan 26, 2008
  12. Jan 26, 2008 #11
    This is the equation with all the numbers plugged in, and its still not turning out right

    (should be cos 30 and tan 30, but i couldnt get it to work)
     
  13. Jan 26, 2008 #12

    Yea

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    what is up with this problem, why doesn't the answer make any sense?
     
  14. Jan 26, 2008 #13
    The algebra looks all right, so I don't understand how we could be getting such a wrong answer. I have a feeling it's a simple little mistake
     
  15. Jan 26, 2008 #14

    Yea

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    i have a feeling that this problem cannot be that complicated.
     
  16. Jan 26, 2008 #15
    What's the actual answer?
     
  17. Jan 26, 2008 #16
    we forgot the negative on the delta y....
     
  18. Jan 26, 2008 #17

    Yea

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    ahhhh thats the key
     
  19. Jan 26, 2008 #18

    Yea

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    ahh thats the key
     
  20. Jan 26, 2008 #19
    You're the man or woman! :-]
     
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