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Dropped ball Kinematics motion

  1. Dec 12, 2013 #1
    1. The problem statement, all variables and given/known data

    You're atop a building of height, h, and a friend is poised to drop a ball from a window at h/2. Find an expression for the speed at which you should simultaneously throw a ball downward, so the two hit the ground at the same time.

    2. Relevant equations

    None

    3. The attempt at a solution

    1)H = vit + 0.5gt^2
    t1 = SQRT[(H - vit)/-4.9ms^-2]

    2)H/2 = vit + 0.5gt^2
    t2 = SQRT[(H/2-vit)/-4.9ms6-2]
    t2 is the time taken for ball to fall from H/2 to ground

    t2=t1
    so,
    set t1= SQRT[(H - vit)/-4.9ms^-2] to be equals to t2
    therefore,
    SQRT[(H/2-vit)/-4.9ms6-2] = SQRT[(H - vit)/-4.9ms^-2]

    vi = [H - 4.9ms^-2(h/2-vit)/(-4.9ms^-2)]/t

    Could someone give me a let up?
     
    Last edited: Dec 12, 2013
  2. jcsd
  3. Dec 12, 2013 #2

    TSny

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    Hello.

    Note that one of the balls is dropped. For this ball, try to find an expression for the time it takes the ball to reach the ground in terms of H and g.
     
  4. Dec 12, 2013 #3
    If the ball is dropped from H/2, then the assumption vi=0 can be made.
    Hence, H/2 = -4.9ms^-2t^2
    t = SQRT[(H/2)/-4.9ms^-2]

    As for ball dropped from H:

    H = vit + 0.5at^2
    H = t(vi + 0.5at) = vi + 0.5at
    t = [H -vi]/05a = [H - vi]/-4.9ms^-2
     
  5. Dec 13, 2013 #4

    TSny

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    OK, but it looks like you've run into a sign problem. You have an expression where you're taking the square root of a negative number.

    I suggest that you express your results in terms of the symbol g and not substitute a numerical value for g. (Since a value for H is not given, you won't be able to find a numerical answer to the question.)

    You can't just drop out the factor of t

    Is there any relationship between the t in this equation and the t you got for the ball that was dropped from H/2?
     
  6. Dec 13, 2013 #5

    H/2 = 0.5gt^2
    t = SQRT[H/g]


    What a terrible elementary blunder.

    It appears logical to substitute t = SQRT[H/g] into H = vit + 0.5gt^2. Then isolate vi from the other variables.
    The ball thrown from H and the ball dropped from H/2 must both touch the ground at t = SQRT[H/g]
    From this, it can be deduced that the distance H = vit + 0.5gt^2 must be achieved in t = SQRT[H/g]
     
    Last edited: Dec 13, 2013
  7. Dec 13, 2013 #6

    TSny

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    Sounds good!
     
  8. Dec 13, 2013 #7
    Solved
     
    Last edited: Dec 13, 2013
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