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Calculating time of falling object

  1. Sep 10, 2007 #1
    1. The problem statement, all variables and given/known data

    A person standing at the edge of a cliff kicks a stone over the edge with a speed of 20m/s. The cliff is 51m above the water. Gravity acceleration is 9.8m/s

    how long does it take the stone to fall and hit the water, answer in seconds

    2. Relevant equations

    D= Vi t + 1/2A tsquared

    3. The attempt at a solution

    51m = D

    Vi = 20m/s

    A = 9.8m/s

    t= ???

    when i solve for that

    i get 2.551 = t + tsquared

    how do i solve for t at that point?
     
  2. jcsd
  3. Sep 10, 2007 #2

    learningphysics

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    does he kick the stone horizontally?
     
  4. Sep 10, 2007 #3
    it doesnt say it just says kicks the stone over the cliff at speed of 20m/s
     
  5. Sep 10, 2007 #4
    if he kicks it straight down does my equation i used work?

    if he kicks it horizontally what equation is then needed?
     
  6. Sep 10, 2007 #5

    learningphysics

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    I think we are to assume it is kicked horizontally... horizontal velocity has no effect on vertical displacement.

    So if it is kicked horizontally... vi = 0 (vertical direction).

    Are there other parts to the question?

    If it is kicked straight down, then your equation works, but you made an algebra mistake.
     
  7. Sep 10, 2007 #6
    what algebra mmistake did i make... and if it is kicked horizontally why would they give the 20m/s info
     
  8. Sep 10, 2007 #7
    since it is 51m down and gravity is -9.8 does it matter if i do them as negatives or positives in this case?
     
  9. Sep 10, 2007 #8

    learningphysics

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    the coefficient of tsquared here:

    seems wrong to me.

    I don't know... I'm wondering if there are more parts to the question that use the 20m/s. Or maybe it is a trick question.
     
  10. Sep 10, 2007 #9

    learningphysics

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    Be consistent... if you take up as positive and down as negative, then stick with that... that would mean displacement is -51m. acceleration is -9.8.

    You can also choose down as positive, and up as negative if you want... but whatever you choose, be consistent...
     
  11. Sep 10, 2007 #10
    ok solving with 20m/s as the initial velocity

    it looks like this

    51=20t + 1/2 9.8 tsquared
    51=20t + 4.9 tsquared
    51/4.9 = 20t/4.9 +4.9tsquared/4.9

    10.08= 4.08t + tsquared
    2.551 = t + tsquared (then i am stuck and not sure how to solve for this)

    or if i solve with initial velcoity at zero i get 3.226 seconds
     
  12. Sep 10, 2007 #11
    does the horizontal velocity have anything to do with the time it would take to fall to the water?
     
  13. Sep 10, 2007 #12

    learningphysics

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    The last line is wrong... It's a quadratic equation, so you can use the quadratic formula.

    But I really doubt the 20m/s is straight down...
     
    Last edited: Sep 10, 2007
  14. Sep 10, 2007 #13

    learningphysics

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    No it doesn't. Are there other parts to the question?
     
  15. Sep 10, 2007 #14

    learningphysics

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    do they give an angle by any chance?
     
  16. Sep 10, 2007 #15

    learningphysics

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    yes, this is right if the 20m/s horizontal.
     
  17. Sep 10, 2007 #16
    they do not give an angle and the only other part of the question is at what speed does the rock hit the water?
     
  18. Sep 10, 2007 #17

    learningphysics

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    Oh... ok, well this part requires you to know the 20m/s. how would you approach this part?
     
  19. Sep 10, 2007 #18
    ok it was obviously 20m/s horizontal since my answer of 3.226 was correct

    now to solve for the speed at impact i would use the equation

    Vf squared= Visquared +2AD?

    Vi = 0 or 20 m/s
    A = 9.8 (negative of positive)
    D = 51m
    solve for Vf???
     
    Last edited: Sep 10, 2007
  20. Sep 10, 2007 #19

    learningphysics

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    yup, that's the equation, but don't forget the squares it's: Vf^2 = Vi^2 +2AD

    that will give you the vertical velocity at impact... you need to use the horizontal velocity also, and get the magnitude of velocity.
     
  21. Sep 10, 2007 #20
    i need the magnitude of velocity to solve for Vf?

    do i use 0 as the intitial velocity or 20?
     
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