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Free Falling Object question

  1. Sep 19, 2007 #1
    1. The problem statement, all variables and given/known data[/
    A ball is thrown directly downward, with an initial speed of 7.15 m/s, from a height of 31.0 m. After what time interval does the ball strike the ground?



    2. Relevant equations
    All the kinematics equations dealing with velocity and height and accelration. need to solve for time.



    3. The attempt at a solution
    I tried this one couple of times. The first thing I did was use the equation Yf=Yi+Viyt+1/2(a)t^2.

    I chose the top to be postive.
    0= 31.0m+ 7.15 m/s(t)+1/2(-9.80)t^2
    I saw that this look like a qudratic equation so i work it out
    ax+bx+cx
    -7.15 m/s +/-[(-7.15-4(-4.90)(31.0)/2(-4.90)]^1/2
    it came out to be
    -7.15+7.53 = 0.38
    -7.15-7.53= -14.68
    i think im not doing something right, need some help with this one.
     
  2. jcsd
  3. Sep 19, 2007 #2
    I didn't personally check it, but it seems fine. The ball would hit the ground within a couple seconds or so (approx sqrt(6)) if you just dropped it, so if you throw it down it will be even quicker, like the half a second you got.
     
  4. Sep 19, 2007 #3

    learningphysics

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    Viy = -7.15m/s since it is thrown downward.
     
  5. Sep 19, 2007 #4
    oh ok i going compute this out then ty.
     
  6. Sep 19, 2007 #5
    hmm its still not comeing out right i check my work but i keep getting the wrong anwser. I dont know what else i can do.
     
  7. Sep 19, 2007 #6

    learningphysics

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    What time do you get?
     
  8. Sep 19, 2007 #7
    when i caluclated everything t came out to be 15.34 if I chose the positve and -1.04 with the negative.
     
  9. Sep 19, 2007 #8

    learningphysics

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    can you show your calculations?
     
  10. Sep 19, 2007 #9
    0= 31.0m + (-7.15 m/s)t + 1/2(-9.80 m/s^2)t^2
    X= -(-7.15)+/-[-7.15^2-4(-4.90)(31.0)/2(-4.90)]^1/2
    X=-(-7.15)+/-[-51.12-(-608)/-9.80]^1/2
    X=-(-7.15)+/-[557/-9.80]^2
    X=-(-7.15)+/-[-56.8]^1/2 hmm i made a mistake with my signs man thse quads are tough hehe. i guess the solutions i got before dont reall exsit because of the negative rooot. hmmmm
     
  11. Sep 19, 2007 #10

    learningphysics

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    Hmmm... I don't think you're applying the quadratic equation properly. the /2a part doesn't go inside the square root... and you should have (-7.15)^2 = 51.12 not -7.15^2=-51.12
     
  12. Sep 19, 2007 #11
    wow i was messing that up all kinds of ways i finally got the right anwser. thanks alot for your help you save me on endless path hehe. the right anwser was 1.89s using the positive side.
     
  13. Sep 19, 2007 #12

    learningphysics

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    cool! no prob!
     
  14. Sep 19, 2007 #13
    [​IMG]

    sometimes it helps to reduce error and make life easier to plug in the x = vi t + 1/2 at^2 formula into the quadratic equation first and simplify before plugging in the actual values... this approach makes life easier for some ppl
     
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