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Simple kinematics problem I can't figure out!

  1. Oct 7, 2015 #1
    1. The problem statement, all variables and given/known data
    A ball is thrown down with an initial velocity of 2.0 m/s from a height of 1.75 m, determine how long until it hits the ground.

    initial velocity = 2 m/s
    distance = 1.75 m
    acceleration = 9.8 m/s2
    time = ?

    2. Relevant equations
    d=vit+1/2a(t)2

    3. The attempt at a solution
    1.75=2t + 1/2(9.8)t2
    1.75=2t + 4.9t2
    Square root of both sides???
    1.32 = 1.4t + 2.21t
    1.32 = 3.61t
    All divided by 3.61???
    t = 0.36
    .... I know the solution to be 0.43 seconds, but can't get there!

    Any help is greatly appreciated!
     
  2. jcsd
  3. Oct 7, 2015 #2

    RUber

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    That is where you went wrong...
    Solve this as a quadratic equation.
     
  4. Oct 7, 2015 #3
    In your solution,you took square root of both sides which is incorrect.Instead,solve the quadratic equation formed in t.You will get the time taken.
     
  5. Oct 7, 2015 #4
    Thanks RUber and Ellispson, really appreciate it, can you give any more detail, I'm not sure how to do that?
     
  6. Oct 7, 2015 #5

    RUber

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    If you have an equation of the form : At^2 + Bt + C = 0, then
    ##t = \frac{ -B \pm \sqrt{B^2 - 4AC}}{2A}##
    Only one of the solutions will be a positive time...choose that one.
     
  7. Oct 7, 2015 #6
    but what does C represent in the quadratic equation? Sorry, it has been a long time since I have used the quadratic equation.
     
  8. Oct 7, 2015 #7
    I mean, what does it represent in terms of the problem I posted?
     
  9. Oct 7, 2015 #8

    RUber

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    You have 1.75 = 2t + 4.9t^2 . Can you put that into the form of At^2 + Bt + C = 0? C is the number that is not multiplied by t or t^2...I get -1.75.
     
  10. Oct 7, 2015 #9
    The constant term.Which means the term not having the variable t in it.
     
  11. Oct 7, 2015 #10
    Oh of course, I see. I'll try that now
     
  12. Oct 7, 2015 #11
    That worked for me! Thanks very much to both of you. Happy physics-ing!
     
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