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Projectile Motion Problem and Muzzle Velocity Help Needed

  1. Mar 2, 2009 #1
    1. The problem statement, all variables and given/known data
    A cannon is discharged towards a castle tower (near the time when the ancient world was
    meeting the modern world). The cannon ball flies 5 m above the top of the 10 metre tall tower and lands on level ground (same level as cannon) behind the tower. The cannon was positioned 300 m from the tower and was angled at 50[tex]\circ[/tex] to the horizontal. What was the muzzle velocity of the cannon ball? How far behind the tower did it land? [up + and right +]

    2. Relevant equations
    [tex]\Delta[/tex]d[tex]_{x}[/tex] = v[tex]_{1x}[/tex][tex]\Delta[/tex]t
    [tex]\Delta[/tex]d[tex]_{y}[/tex] = v[tex]_{1y}[/tex][tex]\Delta[/tex]t - [tex]\frac{1}{2}[/tex]g[tex]\Delta[/tex]t[tex]^{2}[/tex]
    v[tex]_{1y}[/tex] = v[tex]_{sin}[/tex][tex]\Theta[/tex]
    v[tex]_{1x}[/tex] = v[tex]_{cos}[/tex][tex]\Theta[/tex]

    3. The attempt at a solution
    In my attempts to find the muzzle velocity I figured I could use the equation
    [tex]\Delta[/tex]d[tex]_{x}[/tex] = v[tex]_{1x}[/tex][tex]\Delta[/tex]t and isolate the variable of time. Then I figured I could plug it into the equation
    [tex]\Delta[/tex]d[tex]_{y}[/tex] = v[tex]_{1y}[/tex][tex]\Delta[/tex]t - [tex]\frac{1}{2}[/tex]g[tex]\Delta[/tex]t[tex]^{2}[/tex]
    So I rearranged the whole thing fully and correctly to solve for [tex]\vec{v}[/tex]. After plugging in my numbers and such however, I ended up with ~35.5m/s where I should have ~56m/s. I have no idea what I'm doing wrong, please help, someone?

    EDIT: Don't know why my subscripts always turn out as superscripts...
     
  2. jcsd
  3. Mar 3, 2009 #2

    rl.bhat

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    your second equation reduces to
    y = x*tan(theta) - g*x^2/2v^2*cos(theta)^2
    Substitute the values and solve for v.
     
    Last edited: Mar 3, 2009
  4. Mar 3, 2009 #3

    lanedance

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    Hi AudioGeek

    sounds like a reasonable approach, i tried it your way & got 55.8m/s.... can you show your working solving for Dt?
     
  5. Mar 3, 2009 #4
    Thanks, I think I will try it this way! I may have made everything too complicated for myself by isolating v before plugging in the numbers. In order to get remove the half from g, and the square from cos, is that because you were able to cancel them out from each other?

    Great, then I know it does work eventually! I'm able to grasp the concepts, philosophy and ideologies of physics quite easily, and I love it, but I am very far behind in my math skills.
     
  6. Mar 3, 2009 #5

    lanedance

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    the half is still in there & i think it should probably read

    y = x*tan(theta) - (1/2).g.x^2/(v.cos(theta))^2

    generally i've always found the longer you can leave things in equations before putting numbers, the more you get used to it and the easier it gets
     
  7. Mar 3, 2009 #6

    rl.bhat

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    y = x*tan(theta) - (1/2).g.x^2/(v.cos(theta))^2

    Yes. You are right. Typing error.
     
  8. Nov 7, 2010 #7
    where did you guys get tan form? nd also why is the last part squared?
    can somone please solve this step by step for me? like with the rearranging of the formulas to get that last formula and then how to use that to solve for teh muzzle velocity?
    PLEEASe
     
  9. Nov 8, 2010 #8

    lanedance

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    hi schoolisew

    this post is over a year old, you really start a new post

    Show us what you can do or have tried & you'll get help on where to go, generally you won't just get a worked solution though
     
  10. Nov 9, 2010 #9
    Start with:

    [tex]V_{ix} = V \cdot \cos \theta[/tex]

    [tex]V_{iy} = V \cdot \sin \theta[/tex]

    [tex]X = V_{ix} \cdot t[/tex]

    [tex]Y = V_{iy} \cdot t - \textstyle \frac{1}{2} \displaystyle \cdot at^2[/tex]


    Then combine...


    Note that [tex]\frac{\sin \theta}{\cos \theta} = \tan \theta[/tex]
     
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