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Velocity, distance and time

  1. May 19, 2005 #1
    Could someone please try and help me figure out this problem? I keep getting very large numbers and I do not think it is correct! Thanks

    A stone is thrown with an initial velocity of 20 m/s straight upward from the edge of a cliff 100m above a canyon floor. The stone just misses the eliff's edge on its way down.

    1.) determine the time required for the stone to reach its maximum height.
    2.) determine the maximum height of the stone above the edge of the cliff.
    3.) how much time elapses as the stone falls from its maximum height to the level from which it was thrown?
    4.) what is the velocity of the stone upon returning to the level from which it was thrown?
    5.) Determine the velocity of the stone 6 seconds after it is thrown.
    6.) determine the position of the stone 6 seconds after it is thrown.

    Sorry its long, but pelase help!
     
  2. jcsd
  3. May 19, 2005 #2

    Doc Al

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    Staff: Mentor

    Show us your work so we can check it out.
     
  4. May 19, 2005 #3
    okay!!!!

    1.) Vi= 20m/s (given)
    Vf= 0m/s (because @ max height, Vf is 0)
    a = 9.81m/s^2

    a= change in v/t
    9.81m/s2=20m/s / t
    t = 2.04 seconds (I think i did that right.)

    2.) v=20m/s t=2.04s d=vt d=(20m/s)(2.04s) d= 40.8 m

    ((i have more work, i jsut want to make sure thats correct before i proceed to type it all))
     
  5. May 19, 2005 #4
    3.) how much time elapses as the stone falls from its maximum height to the level from wich it was thrown?
    a=9.81m/s2 d= 40.8m d=(1/2)at^2 40m=(1/2)9.81m/s2(t^2) t= 2.85s
     
  6. May 19, 2005 #5

    Doc Al

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    Right!

    Incorrect. The equation d = vt assumes the speed is constant, but that's not true here. The motion is uniformly accelerated. What are some useful kinematic equations for uniformly accelerated motion?
     
  7. May 19, 2005 #6

    Doc Al

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    Incorrect, because you are using the incorrect distance. (Also realize that the time it takes for the stone to rise to maximum height must equal the time it takes to fall back down to that same starting point.)
     
  8. May 19, 2005 #7
    Um, the equations i ahve are:
    a = v/t
    Vf= Vi+at
    d=Vit +(1/2)at^2
    Vf^2= Vi^2 + 2ad

    I think I would use the d= vit + (1/2)at^2 equation, right?

    Therefore, d= (20m/s)(2.04s) + (1/2)(9.81m/s^2)(2.04s)^2
    d= 40.8 + 20.4
    d= 61.2m (maybe?????)
     
  9. May 19, 2005 #8

    Doc Al

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    All good.

    Right!

    Careful about directions (which are represented by signs). If you take up as positive, then the initial velocity is +20 m/s (since it's upward) but the acceleration is -9.81 m/s^2 (since it acts down). Do it over with the correct sign for a.
     
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