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Homework Help: Position in Time

  1. Feb 9, 2005 #1
    If a ball is thrown upward at 64 ft/sec at an initial height of 80 ft, how would you get the position function that finds the height as a function of t? Do you just integrate?

    Thanks :smile:
  2. jcsd
  3. Feb 9, 2005 #2


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    Of course.Or you may use the equation of motion for a constant gravity field.

  4. Feb 10, 2005 #3

    Andrew Mason

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    What are you proposing to integrate?

  5. Feb 10, 2005 #4


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    One would actually integrate TWICE. For this single "z" dimensional problem, one would use:

    [tex] v(t) = \int a(t) dt [/tex]

    [tex] z(t) = \int v(t) dt [/tex]

    where a(t) is vertical acceleration, v(t) vertical velocity, and z(t) vertical displacement. Given constant gravitational acceleration a(t)=g, this would yield:

    [tex] v(t) = \int g dt \ = gt + v_0 [/tex]

    [tex] z(t) = \int (gt + v_0) dt \ = (1/2)gt^2 + v_0t + z_0 [/tex]

    Initial conditions would determine constants v0 and z0.

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