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Two Dimensional Motion

  1. Oct 1, 2005 #1
    Is there an equation that gives the horizontal distance travelled of an object projected at an angle (theta) with an initial velocity (V0) and given that the object lands (y) above/below where it was launched?

  2. jcsd
  3. Oct 2, 2005 #2


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    Of course. Just use the same equations you would if the landing point were on the same level:

    x= v0cos(θ)t, y= v0sin(θ)t- (g/2)t2.

    Let y0 be the height of the landing point relative to the starting point (0 is same level, negative if below, positive if above). Solve
    y= v0sin(θ)t- (g/2)t2= y0 for t. (You will get two solutions the time for the landing is the larger of the two).
    Since that is a quadratic equation, use the quadratic formula:
    [tex]t= \frac{v_0sin(\theta)+\sqrt{v_0^2sin^2(\theta)-2gy_0}}{g}[/tex]
    Now put that value of t in the equation for x (use the positive sign for the root to get the larger of the two solutions):
    [tex]x= v_0cos(\theta)\frac{v_0sin(\theta)+\sqrt{v_0^2sin^2(\theta)-2gy_0}}{g}[/tex]

    A bit more complicated than the case y0= 0 but still a valid formula. Notice that when y0= 0 that reduces to
    [tex]x= \frac{2v_0sin(\theta)cos(\theta)}{g}[/tex]
    the usual formula.

    I would consider it easier to solve the equations than to memorize that formula.
    Last edited: Oct 2, 2005
  4. Oct 2, 2005 #3


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    The last equation (where y0 = 0) should be
    [tex]x= \frac{2v_0^2sin(\theta)cos(\theta)}{g}[/tex]
    (notice the v02)
    Which simplifies to:

    [tex]x= \frac{v_0^2sin(2 \theta)}{g}[/tex]
    Last edited: Oct 2, 2005
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