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Biomechanics Quadratic Equation problem

  1. May 15, 2012 #1
    1. The problem statement, all variables and given/known data
    1.
    Neglecting the height of release, a ball is thrown vertically upwards at 20 m/s, find:
    d) time (s) that the projectile is at height 10.5 m.

    V1 = 20m/s
    a = -9.81m/s^2
    d = 10.5m

    2. Relevant equations

    d = v1*t + 1/2 a(t^2)



    3. The attempt at a solution

    subsitutiting i can get the formula to:

    10.5 = 20t + 1/2(-9.81)t^2
    10.5 -20t +4.905t^2 = 0
    4.905t^2 - 20t = -10.5
    t^2 - 20/4.905 t = -10.5/4.905

    However I am rusty on my quadratic equation work and do not know how to solve from here and cannot seem to figure it out? any help would be awesome Thanks in advance.
     
  2. jcsd
  3. May 15, 2012 #2

    tiny-tim

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    Homework Helper

    welcome to pf!

    hi dsm63! welcome to pf! :smile:

    (try using the X2 button just above the Reply box :wink:)
    either use the formula [-b ± √(b2 - 4ac)]/2a,

    or complete the square
     
  4. May 15, 2012 #3
    would i be right to assume that
    for

    [-b ± √(b2 - 4ac)]/2a

    a = 4.905
    b = -20
    c = 10.5

    when using this i get

    t = 20 ±√19.7747

    t = 24.44 or 15.55

    this answer does not make sense? i think i have some something wrong..

    using a calculator online i recieved

    t = .619 or 3.45s
     
  5. May 16, 2012 #4

    tiny-tim

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    (just got up :zzz:)
    that's correct :smile:

    but you've done -b ± √[(b2 - 4ac)/2a] ! :rolleyes:
     
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