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Time taken for body under free fall

  1. Sep 2, 2012 #1
    1. The problem statement, all variables and given/known data
    A ball is dropped from the top of a cliff. If time taken for half of the descent is 3s, then what is the time taken for the rest of rest of the descent? Take g=10m/s2


    2. Relevant equations
    T=√(2h/g)
    v=u-gt
    s=ut-gt2/2

    3. The attempt at a solution
    T=√(2h/g)
    h=T2g/2=45m

    ∴height of cliff=90m

    v=u-gt=0-(10*3)=-30ms-1

    s=ut-gt2/2
    45=-30t-5t2
    t=-3

    :frown: Looks wrong to me. Time can't be negative, can it? Also, shouldn't the answer be less than 3?
     
  2. jcsd
  3. Sep 2, 2012 #2

    Doc Al

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    Staff: Mentor

    You're making a sign error. Use the sign convention consistently. If down is negative, what should be the sign of s?
     
  4. Sep 2, 2012 #3
    If the origin is on the ground, the distance (which is above the origin) will be positive, right?
     
  5. Sep 2, 2012 #4

    Doc Al

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    Staff: Mentor

    No. What you're calling 's' is really Δs, which is negative as the object is falling.

    The full statement of position as a function of time would be (using y instead of s, for clarity):
    y = y0 + ut - 1/2gt2

    In your case, y = 0 (the final position, measured from the ground) and y0 = 45.

    Let me know if that's clear.
     
  6. Sep 3, 2012 #5
    Okay. So y=yo+ut-gt2/2
    0=45-30t-5t2
    t=1.242s
     
  7. Sep 3, 2012 #6

    Doc Al

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    Staff: Mentor

    Looks good.
     
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