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Particle Movement w/o Knowing a or v

  1. Sep 13, 2006 #1
    Object is dropped from a cliff of unknown height. I'm on another planet without the knowledge of the planet's gravitational acceleration. The object takes 4.15s to reach the bottom. Again, from the top of the cliff I toss the object upwards to a height of 2m then it falls back to the cliff bottom. This takes 6.3s.
    I solved for the height of the cliff using the standard x=x+vt+1/2at^2 and yeilded -8.6115s^2(a). How can I move on with upward then downward equation with only knowing a=cliff height/8.6115s^2 and not knowing the initial velocity of the object being thrown upward. Any suggestions on what to start solving for first?
     
  2. jcsd
  3. Sep 13, 2006 #2
    When dealing with an object tossed upwards that is under a constant acceleration downwards, it is often easier to think of it as two problems:

    In the first, the ball is thrown upwards, and stops 2m up. In the second, it is dropped freely from 2m above the clifftop. In the second you do know the initial velocity...
     
  4. Sep 13, 2006 #3
    You have 3 unknowns, x, the height of the cliff, a, the acceleration, and u, the initial velocity in part 2. So you need 3 equations to find them. The first one, which you have, is [itex]x=\frac{1}{2}at^2_{1}[/itex], where t1=4.15s, the second one uses that same equation you quoted, but with different values for x, v and t, (use the second situation) and the third one is for the initial velocity u in terms of acceleration and distance only. You don't want to include the time, so do you know the equation you'll need? Think of a part of situation 2 where you know the distance and the final velocity. (Think about what the velocity of a body thrown upwards is at its maximum height.) Then you need to rearrange and substitute until you have the unknown you want. I hope this isn't too cryptic, I have a feeling it is. Just say if you don't follow me.
     
  5. Sep 13, 2006 #4
    So am I right to solve for a in the first equation? Doest the second look like 2m=(8.6115s^2)a+vt+[(x/(8.6115s^2)t^2]/2 or am I way off?
     
  6. Sep 14, 2006 #5
    Quick question. If I have 5m*s^2,then I square that do you get 25m^2*s^4 ? Reason I ask is the following. I setup an equation for the upward motion as follows. 0=V^2 + 2a(2meters) and wound up with V=2m(-a)^(1/2)
     
    Last edited: Sep 14, 2006
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