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Projectile motion of a cannon ball

  1. Jan 18, 2009 #1
    1. A cannon mounted on a inclined plane launches a ball with initial speed v0 directly up the slope. The plane is inclined at angle phi above the horizontal, and the initial velocity is at angle theta above the plane. Using axes with x parallel to the plane and y normal to it, write out the x and y components of Newton’s second law, and show that the ball lands on the slope a distance R = (2V(o)^2 sin(theta) coa(theta + phi))/(g cos^2(phi)) from the cannon.

    I can almost get the direction equations as required from the first part, but all these sin Phi, cos Phi, sin theta, and cos theta terms are confusing me.

    Also, once you have these equations I'm kinda lost as what to do to get the distance R.

  2. jcsd
  3. Jan 18, 2009 #2
    Post the working that you've done so far. You need to show attempt of some kind first.
  4. Jan 18, 2009 #3
    I know the equations are:

    x = Vot(cos (theta)) + 1/2 gt^2(sin(phi))

    y = Vot(sin (theta)) + 1/2 gt^2(cos(phi))

    Trouble is I don't know how to get them. I started off by integrating ax and ay to get vx and vy, and then again ro get dx and dy. But its the angles I'm confused about.

    Shouldn't the y component always be sine, and the x component be cos? Plus I'm not sure if the x component is exactly up the incline, and y component perpenddicular to the surface, because thats the way the axis are pointing. Doen't that mean the x compnent should only have theta in the equation, because thats the angle between the inclined plane and the projectile?

    I think I now get why phi is in the terms with g and theta is in the terms with Vo, since the y component is angle phi with gravity, amf x component is theta to the incline. Still just a little confused about where the sin and cos shoulgd go in each equation, ie in the first or second terms.

    I'm really not sure how to start the seond part still.
  5. Jan 19, 2009 #4
  6. Jan 19, 2009 #5


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    If you show your expressions for [itex]a_x[/itex] and [itex]a_y[/itex] and what you got when you integrated them, I'll be able to see where you are going wrong an help you.
  7. Jan 19, 2009 #6
    Normally, I would say that ay=-g and that ax=0, but If you start with zero accelaration, you can't get the same expression for X.
  8. Jan 19, 2009 #7


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    But gravity points downward. Is the negative y-direction really downward? Don't you have to take into account the angle of the plane?:wink:
  9. Jan 19, 2009 #8
    ax=gcoes(phi), and ay=gsin(phi), then Vx=gtcos(phi), and vy=gtsin(phi). Finally, x=vot(cos(angle)=(1/2)gt^2(cos(phI), and y=vot(sin(angle)+1/2gt^2sin(phi).

    I belive "angle" is theta since thats the angle ox of the projectile with the rotated plane.
  10. Jan 19, 2009 #9


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    Good :smile: Of course, this assumes that g is negative right? (g=-9.8m/s2)

    Why aren't there any constants of integration?

    This does not follow from your previous expressions. :confused:
  11. Jan 19, 2009 #10
    Sorry, I'm so used to them not being important that I forget a lot of the time.

    Yeah, I've got cos and sin the wrong way round fpr th vot terms. Not sure why its the other way round though?
    Last edited: Jan 19, 2009
  12. Jan 19, 2009 #11


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    Include them and use your initial conditions to determine them :smile:
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