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Need help deriving a formula

  1. Jul 11, 2005 #1
    Hey could anyone help me out there to know how you get to the formula of free fall displacement. D= 1/2 x g x t^2? I really have no clue of how to get it using other formulas. I started using the D=Vi x T + 1/2 x A x T^2 but then I got stuck I didnt know which other formula to bring in to cancel the variables I NEED HELP PLEASE :D
     
  2. jcsd
  3. Jul 11, 2005 #2

    Doc Al

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    Staff: Mentor

    If you know the formula for uniformly accelerated motion ([itex]D = v_i t + 1/2 a t^2[/itex]), then you can get the formula you need by setting the initial speed to zero and the acceleration to g. That gives you [itex]D = 1/2 g t^2[/itex].
     
  4. Jul 11, 2005 #3

    dextercioby

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    It's also important to make the wise choice of the sense of the coordinate axis along the displacement. That is, assuming the [itex] \vec{g} [/itex] field is pointing downwards, choose the vertical axis with the same sense.

    Daniel.
     
  5. Jul 11, 2005 #4
    There is no formula here . As Doc Al put it, the initial velocity has to be set to zero for the formula to be applicable , so the formula seems to be forced.It is better to get hold of the three basic kinematic equations rather than learning formulae for each and every situation.

    BJ
     
  6. Jul 11, 2005 #5
    Try to draw the velocity-time graph. The area under the curve will give you the displacement.
     
  7. Jul 11, 2005 #6
    Thanks

    Hey thanks to all you guys

    I never did know you could set the initial velocity to zero! that solved the entire problem thank you. Now im wondering why I never thought of that well anyways thanks guys
     
  8. Jul 11, 2005 #7
    yes initial velocity can be set to zero if the object stops of an instant, or if it starts from rest.
     
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