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Acceleration And Distance

  1. Sep 9, 2008 #1
    1. The problem statement, all variables and given/known data

    If a ball is dropped from rest and in the last second travels three quarters of the distance, from what height is the ball dropped.

    I actually haven't taken physics for about a year, and I'm not taking it in university as a course, but one of my friends showed me this question, and, majoring in math, I was intrigued. Where should one begin on a question like this?
    Also, I'm sorry that I don't have any formulas or anything, again, I haven't been to any university classes and am therefore unsure if the formulas remain the same.
    Last edited: Sep 9, 2008
  2. jcsd
  3. Sep 9, 2008 #2


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    Homework Helper

    Welcome to PF.

    Here are some basic kinematic equations that might be able to get you going.
  4. Sep 9, 2008 #3
    Well, lets see...
    I know that acceleration is constant, being 9.81.
    I would then need to find the distance, and in doing so, I would assume I need the time that the ball is falling. I might even need the final velocity, knowing that the initial velocity is zero.
    But therein lays the problem I am faced with: I only have two variables to work with, and those formulas seem to require at least knowing the time, as well as the acceleration. For instance, the distance formula requires me to know 1/2at^2, and I only know the acceleration, and therefore have two unknown variables. I'm not sure where to start with these formulas with the limited information I have been given...

    And thank you for the welcome. :D
  5. Sep 9, 2008 #4


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    But you are provided with the additional information that 1 sec is 3/4 of the distance.

    X2 = X1 +a*t1 + 1/2 a*12

    t1 is the initial time, that along with acceleration gives initial speed for the time to drop the remaining 3/4.
    X1 is the initial distance

    X2 is 3 times X1
    And you also know X1 = 1/2 a * t12

    You end with a quadratic. Not too much for a math major then I should hope.
  6. Sep 10, 2008 #5


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    It occurred to me that the solution is much easier. You can in this case make use of the observation that since distance is proportional to the square of the time and the distance is 4:1 ... and the Time is T+1 that T must be 2 as the square of 2 is 4. This makes the height calculation remarkably easier. It is simply given by

    X = 1/2 a * 22

    A much more satisfying road to solution.
  7. Sep 11, 2008 #6
    Sorry, but how did you determine T to be 2? I understand the T+1 part, but I'm unsure of how you reached got 2 as the time...

    And I did in fact get 19.62 as an answer with the "long route". Thanks for your help on that.
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