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Math Help - Interesting gravity problem.

  1. #1
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    Interesting gravity problem.

    Suppose the acceleration of gravity of planet X is one-half that on the earth. Determine the initial velocity of the ball on the earth (in terms of v_0) so that the maximum height attained is the same as on planet X.

    What I tought was, first to find what the maximum height in planet X is.
    So, having g as my constant for gravity I have:

    On Earth : a(t)=-g ; and on Planet X: a(t)=-g/2
    Then when setting the velocity on Planet X equal to zero, and plugging that t in the position function : s(t)=(-1/4)gt^2 + v_0t; supposing that the initial position is zero.

    I get: [(v_0)^2]/g as the maximum height on planet X.

    Once I get that answer I try to set it equal to the position function of Earth, but it doesn't seem to work out well. And I'm not even sure if that's what I should do.
    Any help would be greatly appreciated.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Arturo_026 View Post
    Suppose the acceleration of gravity of planet X is one-half that on the earth. Determine the initial velocity of the ball on the earth (in terms of v_0) so that the maximum height attained is the same as on planet X.

    What I tought was, first to find what the maximum height in planet X is.
    So, having g as my constant for gravity I have:

    On Earth : a(t)=-g ; and on Planet X: a(t)=-g/2
    Then when setting the velocity on Planet X equal to zero, and plugging that t in the position function : s(t)=(-1/4)gt^2 + v_0t; supposing that the initial position is zero.

    I get: [(v_0)^2]/g as the maximum height on planet X.

    Once I get that answer I try to set it equal to the position function of Earth, but it doesn't seem to work out well. And I'm not even sure if that's what I should do.
    Any help would be greatly appreciated.
    If v_{E0} is the initial velocity on earth, then the position of the ball on earth is given by s(t)=v_{E0}t-\tfrac{1}{2}gt^2.

    The maximum height is reached when t=\frac{v_{E0}}{g} (verify).

    Thus, the maximum height is \frac{v_{E0}^2}{2g} (verify).

    If the maximum heights on earth and planet X are to be the same, then \frac{v_{E0}^2}{2g}=\frac{v_0^2}{g}. Now solve for v_{E0}.

    Can you take it from here?
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  3. #3
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    That is actually what I considered, but it didn't seem to make sense at first. Now I see. Thanks.

    P.S. Your animations are hilarious.
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