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How to solve projectile motion problems

  1. Sep 29, 2007 #1

    Chi Meson

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    You have two types of motion, vertical and horizontal. Both happen at the same time, but they are separate. In the following equations, the subscript x refers to the horizontal components and y for the vertical components.
    The horizontal motion happens at constant velocity, and has one equation takes care of that:
    [tex]d _{x} = v_{x} t[/tex]

    The vertical motion happens while undergoing uniform acceleration (due to gravity). In all of the following use
    [tex]a = -9.807 \frac {m}{s\ ^{2} }\ [/tex]

    Depending on what you are given, you will generally use one of the following four equations to find an unknown:
    [tex]v_{fy} = v_{oy}+at[/tex]

    [tex]v^{2}_{fy} = v^{2}_{oy} +2ad_{y}[/tex]

    [tex]d_{y}=v_{oy}t + \frac {1}{2}\ at^{2}[/tex]

    [tex]d_{y}=\frac {v_{oy}+v_{fy}}{2}\ t[/tex]

    Set up two columns, one for the horizontal information, and one for the vertical. Using angles given (usually the initial angle of velocity) find the horizontal and vertical components of the initial velocity

    [tex]v_{x}=v_{o}cos \theta[/tex]
    [tex]v_{oy}=v_{o}sin \theta[/tex]
    (here, theta is the angle that the initial velocity makes with the horizontal surface)

    Examine the problem for as many of the values that go with these equations. remember, there are three variables for horizontal motion (constant velocity), but five variables for vertical motion (uniform acceleration).

    Next, determine which unknown variable will answer (or help to answer) the question as stated in the problem.

    In your two columns, the one factor that is the same for both is the time, t. When you have solved for t in one column, then you can drag that value to the other column. You will be told (including inferences and assumptions) either two of the three values for the horizontal information, or three of the five variables fore the vertical information.

    Often, you need to find the t in one column, drag it to the other column and use that t to solve for the wanted unknown variable.
    Last edited by a moderator: Sep 30, 2007
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  3. Sep 30, 2007 #2

    Doc Al

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  4. Sep 30, 2007 #3


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    It might be useful to at least remind folks (but not necessarily derive) where these constant-acceleration equations come from...
    ...to emphasize the fundamental simplicity and unity of the underlying physics:
    [tex] \vec s =\vec s_0 + \vec v_0 t + \frac{1}{2} \vec a_0 t^2[/tex]
    (written out in component form, if needed).
    If necessary (say, for an algebra-based course), you may wish to include
    the velocity equation
    [tex] \vec v =\vec v_0 + \vec a_0 t[/tex]

    The complications one encounters in problem solving arise from geometry, constraints, in the specification of certain sets of initial conditions, and, of course, interpreting the physical situation and reformulating it as a mathematics problem.
  5. Sep 30, 2007 #4


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