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Physics Experiment Question; Projectile Motion

  1. Feb 15, 2005 #1
    Hello, I have a question that says' Derive an equation for calculating the theoretical range.

    I did like half of it, I got

    t = sqrt[ (2y) / a ]

    from the kinematic equation ( X = X0 + V0 t + 1/2 a t^2 ). How would I find the theoretical range after this????

    And how to prove that the theory is consistent with the experiment?

    please help.
     
  2. jcsd
  3. Feb 15, 2005 #2

    Doc Al

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    Staff: Mentor

    Range of what? A projectile with a certain initial speed? Note that the range--the horizontal distance traveled--will depend on the angle that the projectile is launched.

    You'll need to combine equations for both horizontal and vertical motion.
     
  4. Feb 15, 2005 #3
    If you combine the two equations for projectile motion

    [tex]y = y_0 + v_it + .5at^2[/tex]
    [tex]x = x_0 + v_it +.5at^2[/tex]

    into one equation, the highest range would be the vertex of the parabola, which is half of the total time I believe. I would suggest solving for t, then divinding that by 2 to get the time when the highest point is reached, which is what I'm assuming the "range" means.
     
  5. Feb 15, 2005 #4

    Doc Al

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    Staff: Mentor

    Two comments, Jameson:
    (1) Only the vertical component of the motion is accelerated.
    (2) "Range" generally means horizontal distance traveled. (If you shot the projectile straight up, it would fall straight down. Range = 0.)

    But, yes, the highest point attained is the vertex of a parabola.
     
  6. Feb 15, 2005 #5


    You'd need to break your theoretical initial velocity and angle of projection down into the initial X and Y components.
    (R being initial velocity magnitude, O being initial angle)

    Vox R x cosO
    Voy R x sinO

    Then, you'll also have to break down your equations for motion into X and Y, I'm going to assume by range you mean distance which is X-motion...
    (Xo being initial X position -usually 0-; Vox being the x-component's initial velocity)

    X X(t) = Xo + Voxt + 0.5at^2

    (Remember that the acceleration for the x-component is ZERO!)
    Now, since you have a theoretical Vox, you can plug it into the above eq.

    *edit* You'd also have to do the same for the Y component, and derive that equation. Then set that equal to zero, because that will tell you the two time the projectile will cross the X-axis. Pick the positive time and substitute it into your position-time function for X.

    I believe that the above is right as for range, now as for the experimental proving bit, that I can't help with.
     
    Last edited: Feb 15, 2005
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