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Energy conservation and projectile motion

  1. May 19, 2009 #1
    1. The problem statement, all variables and given/known data
    A spring is pulled back and launches a ball. The ball flies into the air at a perfectly horizontal angle. How far will the ball go before it hits the ground?

    Variables for the FINAL equation are:
    ∆x - distance the spring is pulled back
    g - acceleration of gravity
    m - mass of the ball
    k - spring constant
    ∆R - range the ball will fly
    h - height above the ground at the point that the ball passes the end of the launcher


    2. Relevant equations
    1/2mv^2 - kinetic energy
    1/2kx^2 - elastic potential energy

    v=√(kx^2/m) - velocity derived from the kinetic energy equation

    3. The attempt at a solution
    I broke it up into 2 separate parts, one dealing with conservation of energy and the spring, and the other dealing with kinematics and the projectile motion of the ball in the air. I managed to find the velocity of the ball when it leaves the launcher, but beyond that I really had no idea where to go, as I need to find some way to eliminate both the velocity and time variables from the final equation, which would be much easier to use if I was able to.
     
    Last edited: May 19, 2009
  2. jcsd
  3. May 19, 2009 #2

    rock.freak667

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    Just consider vertical motion to get the time parameter and then just put that into the equation for horizontal motion
     
  4. May 19, 2009 #3
    I would, but I need to somehow come up with a final equation using only those variables.
     
  5. May 19, 2009 #4

    rock.freak667

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    If R is the horizontal distance traveled, then [itex]R=vcos\theta t[/itex]

    Consider vertical motion now, use the fact that when the ball hits the ground, the vertical displacement is zero (this will get you the time parameter).
     
  6. May 19, 2009 #5
    Why would the vertical displacement be zero? The ball is assumed to be starting above the ground, in which case the vertical displacement would be a non-zero number (again, the ball is being shot at a perfectly flat angle, only completing one half of a parabola). Your math is also kind of going right over my head, is there a simpler explanation for what you mean? Also, what would I do with the time? I could solve this easily enough if it was an allowed variable, but it can't be used in the final equation for the problem.
     
  7. May 19, 2009 #6

    rock.freak667

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    Let's start over now that I've read the horizontal angle thing.

    You correctly got the equation for the velocity,v.

    We know that ∆R=vt, but you don't have t.

    We also have the equation s=s0+Vt-(1/2)gt2 for vertical displacement.

    The initial vertical displacement is h. When the ball hits the ground, won't it have a vertical displacement of zero? (When the ball is touching the ground, you would not say it has a vertical displacement).

    So you can now find t in terms of the parameters that you have. When you get that, put that into the formula for ∆R, and you will get the equation you need.
     
  8. May 19, 2009 #7
    I see what you mean by the displacement being 0. What does s represent in the equation you gave? The sine of the launch angle?
     
  9. May 19, 2009 #8

    rock.freak667

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    No no, s is just the vertical displacement and s0 is the initial vertical displacement
     
  10. May 19, 2009 #9
    So, just to keep the variables similar, the equation could be written as:
    h=hi+vt-(1/2)gt2
    where all variables correspond to those in my first post?
    Also, if h is the vertical displacement, can I use:
    ∆d=vi∆t+1/2a(∆t)2

    where vi∆t=0 because the initial vertical velocity is 0, and the rest can be solved as ∆t=√(2∆h/a)
     
  11. May 19, 2009 #10

    rock.freak667

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    Here is where you need to be careful, if you take the origin to be where the ball is launched, the initial vertical displacement is zero and you want to put s=-h.

    OR otherwise, you want s=0 and s0=h.

    and the v in this equation is the initial vertical velocity which is zero. Can you get t now?
     
  12. May 19, 2009 #11
    So t=√(2h/g) and the final equation is ∆R=√(kx2/m)*√(2h/g)
     
  13. May 19, 2009 #12

    rock.freak667

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    Yes and you have all of the variables you needed in this equation.
     
  14. May 19, 2009 #13
    Thanks for the help. I'll definitely come back to these boards in the future if I need anything else.

    EDIT: AAAARRRRRRGGGGGGG! I just realized that I had figured out the time equation earlier without realizing it was the right equation. I took the equation h=vi∆t+(1/2)g(t)2 and figured out that vi∆t=0 and that ∆d=(1/2)g(t)2 could become ∆t=√(2h/g). While I'm not sure that that method will work all the time it is aggravating to know that I would have still gotten the right answer even if my explanation was incorrect.
     
    Last edited: May 19, 2009
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