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Homework Help: Angle at which canon can be fired

  1. Jul 7, 2011 #1
    1. The problem statement, all variables and given/known data
    canon is firing at a mountain and it want the angle above the horizontal which it should be fired

    muzzle speed= 1000m/s
    x=2000m
    y=800m

    2. Relevant equations

    vxi=vicos(theta) vyi=visin(theta)

    3. The attempt at a solution

    Not sure how to start this.
     
  2. jcsd
  3. Jul 7, 2011 #2

    tiny-tim

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    welcome to pf!

    hi gwizz! welcome to pf! :smile:

    (have a theta: θ and try using the X2 icon just above the Reply box :wink:)

    write out the https://www.physicsforums.com/library.php?do=view_item&itemid=204" equations for the x and y directions (separately, with a = 0 and -g) …

    that gives you two equations with two unknowns (v and t), so eliminate t to find v …

    what do you get? :smile:

    (btw, a canon is a priest … you mean cannon! :wink:)
     
    Last edited by a moderator: Apr 26, 2017
  4. Jul 7, 2011 #3
    sorry, cannon. also, pardon for the equations, idk how to use the wrap stuff with the tool.

    so for X i used the kinematic equation: Xf=Xo + Vot + (1/2)at^2 and got t=2 seconds then plugged that in the Y kinematic equation Vy=Voy + ayt and got a vertical initial velocity of Voy= 19.6 m/s. Is this correct so far?
    If so, now what? thanks.
     
  5. Jul 7, 2011 #4

    SammyS

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    Acceleration in x direction is zero, so the x component of velocity is constant.

    Acceleration in y direction is -g = -9.81 m/s2, or whatever approximation is acceptable for you class.

    y = y0 + (v0)yt - (1/2)g(t2)
     
  6. Jul 8, 2011 #5
    SammyS, if i use the Y equivalent to the equation i had, does that mean what i did was incorrect?
     
  7. Jul 8, 2011 #6

    SammyS

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    Probably it's wrong.

    How did you get it from the information you have?
     
  8. Jul 8, 2011 #7
    i just thought those equations were the one to use based off tiny-tim's help. What do i do exactly, with what equations to get started then?
     
  9. Jul 8, 2011 #8

    SammyS

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    Use those equations along with what you know about the acceleration due to gravity.

    For the Y-components:
    (VF)Y = (V0)Y + aY · t

    YF = Y0 + (V0)Y · t + (1/2)aY · t2

    What is aY ?​

    For the X-components:
    (VF)X = (V0)X + aX · t

    XF = X0 + (V0)X · t + (1/2)aX · t2

    But gravity acts only vertically, so what is aX ?
    That will simplify the equations for the X-components.​
    What do the equations become when you replace aX & aY with the appropriate values for acceleration?
     
  10. Jul 8, 2011 #9
    ay is -9.8m/s^2 due to gravity and ax is 0 because it is always zero in the x direction. is my velocity correct for Vyin my previous post?
     
  11. Jul 8, 2011 #10

    SammyS

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    I don't think you can find (V0)Y unless you know the launch angle. (That is, unless you're a savant.)

    You do know V0, because it's given (muzzle speed). Use trig to relate the launch angle, θ, and V0 to (V0)X & (V0)Y.

    Use the kinematic equation for XF to find t in terms of θ.

    Plug that result into the kinematic equation for YF. Solve for θ. It's kind of tricky.
     
  12. Jul 8, 2011 #11
    [itex] Vx = V cos\alpha[/itex]


    [itex]Voy = V sen\alpha[/itex]

    [itex]x = Vx .t = V cos \alpha.t [/itex]


    [itex] y = Voyt - gt² /2 = V sen\alpha t - gt²/2[/itex]

    Isoling t

    [itex]t = \frac {x} {V cos\alpha}[/itex]

    [itex] y = x tan\alpha - g x²/2V²cos² \alpha [/itex]


    Replacing we find
    [itex]\alpha = 22°[/itex]
    [itex]\alpha = 89° [/itex]
     
  13. Jul 9, 2011 #12
    ya it is tricky, thanks for the help. and Jaumzaum, i got an angle of like 21 but the answer is 7.8 or something like that.
     
  14. Jul 10, 2011 #13
    First of AL, if g is down, tan > 800/1000, angle > 21º, otherwise muzzle will pass beneath
     
  15. Jul 11, 2011 #14
    well i'm just restating the answer i got from the answer sheet
     
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