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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


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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.

  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.

  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

    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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