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Firing cannons projectile motion

  1. Apr 26, 2014 #1
    1. The problem statement, all variables and given/known data
    A cannonball is fired with initial speed v0 at an angle 30° above the horizontal from a height of 38.0 m above the ground. The projectile strikes the ground with a speed of 1.3v0. Find v0. (Ignore any effects due to air resistance.)
    ScreenShot2014-04-26at71948PM.png

    2. Relevant equations
    Time= 0=v0T-½gT2 T<0
    t= (vOy+√(vOy2-2gy))/g
    distance/range x=x0+V0xt
    x/y velocity- v0x=Vcos(θ); v0y=vsin(θ)


    3. The attempt at a solution

    This is a practice problem (with different values than original), and the provided answer is 32.9m/s

    i set the origin at the cannon
    Since the velocity magnitude = √(Vcos(θ)2+Vsin(θ)2)

    I tried variations along the line of V0cos(30)2+V0sin(30)2= 1.3 V02
    in order to split the velocity into xy components.

    Conceptually i understand it takes him -38m to reach 1.3 times his original velocity, but in trying to plug the provided data into the equations i end up with too many variables.

    Any and all help very appreciated, thank you.
     
  2. jcsd
  3. Apr 26, 2014 #2

    SteamKing

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    I'm not sure what you are trying to do here. The angle at which the projectile strikes the ground may not be equal to the initial firing angle, i.e. 30 degrees.

    The best way to approach this problem is to find the maximum altitude which the projectile reaches after it is fired. At this point, the vertical component of the velocity is zero. After reaching the maximum altitude, the projectile falls under the force of gravity. Knowing the magnitude of the final velocity, you should be able to determine the duration of the fall and the resulting vertical velocity component. Remember, since there is no air resistance, the horizontal component of the velocity remains unchanged during the entire flight of the projectile.
     
  4. Apr 26, 2014 #3
    That is brilliant, thank you. However, Without knowing distance (x), time, or hard value for velocity (since it's expressed as a relation to the initial velocity) i'm not sure how to go about this.
     
  5. Apr 26, 2014 #4

    SammyS

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    Try using conservation of energy.
     
  6. Apr 26, 2014 #5
    Hmmm I don't think we've learned that yet..
     
  7. Apr 26, 2014 #6

    SammyS

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    In that case,

    What can you conclude about the x component of the velocity?


    What kinematic equations do you know regarding the vertical component of the velocity ?
     
  8. Apr 26, 2014 #7
    X component will be the same through it's entirety.

    y=y0+v0y(t)+½at2
    or
    Vy=V0y+gt
    Vy=1.3V0y-9.81t
    But Theoretically if we were starting at the maximum, V0y would = 0,

    Vy/-9.81=t?

    I'm getting stuck down either avenue, I do think i'm missing something intuitive here..
     
  9. Apr 26, 2014 #8

    SammyS

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    vy ≠ 1.3 v0y . I'm assuming that you mean vy to be the y component of the final velocity.

    How is speed obtained from components of velocity ?


    Also, there's another kinematic equation. One that doesn't involve time, t.

    Edited:
    [STRIKE](vy)2 = (v0y)2 - g(Δy)[/STRIKE]

    (vy)2 = (v0y)2 - 2g(Δy)
     
    Last edited: Apr 27, 2014
  10. Apr 27, 2014 #9
    Oh yes, i made a false assumption about the y component.
    Do you mean speed as in magnitude of velocity?

    Also, should (vy)2 = (v0y)2 - g(Δy)
    have a -2g(Δy)?

    I apologize for being slow, but i'm not sure how to use this either since I don't know the change in height from the maximum
     
  11. Apr 27, 2014 #10

    SammyS

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    Right. There should be a 2 in there.


    Then, ...

    I'll ask again,
    How is speed obtained from components of velocity ?​
     
  12. Apr 27, 2014 #11
    Since speed is the magnitude of velocity,
    by components of velocity I imagine you mean Vx and Vy,
    or √(Vcos(θ)2+Vsin(θ)2)?
     
  13. Apr 27, 2014 #12

    SammyS

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    Yes. Of course, sin2(θ) + cos2(θ) = 1

    Anyway, ##\ v_0=\sqrt{(v_{0x})^2+(v_{0y})^2}\ ,\ ## so that ##\ (v_0)^2=(v_{0x})^2+(v_{0y})^2\ .\ ##

    And in general ##\ v^2=(v_{x})^2+(v_{y})^2\ .\ ## Right?


    Now, what do you know regarding the x component of velocity for this projectile?
     
  14. Apr 27, 2014 #13
    sorry, that's what i meant when i put √(V0cos(θ)2+V0sin(θ)2), that it's = V0.

    I know that it doesn't change, but i'm not sure how to find its' magnitude in this situation.
     
  15. Apr 27, 2014 #14

    SammyS

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    So then ##v_{0x}=v_x\ ## .

    Regarding your kinematic equation with the ##(v_{y})^2\, ##, what is it missing for it to have v2 and v02 in it ?
     
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