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Projectile motion and theta

  1. Aug 25, 2009 #1
    1. The problem statement, all variables and given/known data

    A projectile is launched from O with speed u at angle theta to the horizontal. In its subsequent trajectory it just clears two obstacles of height h at horizontal distances d and 2d from O, respectively.

    http://img403.imageshack.us/img403/4369/projectile.jpg [Broken]

    Show that tan (theta) = 3h/2d

    2. Relevant equations



    3. The attempt at a solution

    I do not see anyway to get tan (theta) here. From class I have the following equations:

    Solution of the equation of motion:
    r(t)=(utcos(theta))i + (utsin(theta) - ½gt²)j

    trajectory is parabola with equation:


    y=xtan(theta)-x²(g/2u²)(1+tan²(theta))

    but I cannot see how I begin to find just tan(theta)

    I just need to know where to start. My guess is I'm missing something that's taken to be 'obvious' but what my professor thinks is 'obvious' is usually obscure for me :)
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 25, 2009 #2

    kuruman

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    You need to relate theta to d and h. You also have an equation for the parabola y(x). Start by saying that y is equal to h at x = d and x = 2d. That's the condition for "just clearing" the obstacles.
     
  4. Aug 25, 2009 #3
    So I get a pair of simultaneous equations for h:


    (1) h=(d)tan(theta)-(d)²(g/2u²)(1+tan²(theta))
    (2) h=(2d)tan(theta)-4(d)²(g/2u²)(1+tan²(theta))

    Multiply (1) by -4:

    (1a) -4h=(-4d)tan(theta)+4(d)²(g/2u²)(1+tan²(theta))

    Add (1a) and (2):

    (3) -3h=(-2d)(tan(theta)) as required.

    LDO :p
     
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