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Simple (was thought to be) Gravity Question

  1. Nov 5, 2007 #1
    [SOLVED] Simple (was thought to be) Gravity Question

    1. The problem statement, all variables and given/known data
    A ball is rolled up a constant slope. After 3.6 s it reaches its maximum displacement of 2.6m, and then begins to roll back down. What was the initial velocity of the ball when it started up the slope?

    2. Relevant equations

    3. The attempt at a solution
    i know that d=vt of course, but, 2.6=v(3.6), and V would equal 2.6 divided by 3.6 which is .722, and the answer in the key is 1.4m/s?? In all likelihood, its probably something extremely obvious?
  2. jcsd
  3. Nov 5, 2007 #2
    You can't use d=vt because there is an acceleration due to gravity
    use d=vt-.5at^2
  4. Nov 5, 2007 #3
    you mean d=Vit+.5at^2, emphasis on plus sign?
  5. Nov 5, 2007 #4
    well yes, in this case the acceleration is negative but the formula you gave is the more general form.
  6. Nov 5, 2007 #5
    edit: thanks by the way, the question screwed me up w/ the word constant, plus im taking my very first physics course (phys. 11)----one more question is i might have quite a few questions today.....will mods get mad for me making quite a few threads? (maybe 10?)
  7. Nov 5, 2007 #6
    If they're the same types of problems I'd just stick them in the same thread.
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