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Calculating Initial Velocity

  1. Aug 24, 2014 #1
    1. The problem statement, all variables and given/known data

    For archery practice, a knight's squire drops sandbags from a 12.0 meter tower. At exactly the same time the sandbag is dropped, the knight shoots an arrow up at the sandbag from the base of the tower. If the arrow strikes the sandbag at exactly 1.1 seconds, calculate a) how far from the ground the arrow strikes the sandbag, and b) the arrow's initial velocity

    2. Relevant equations

    d = ½(vi + vf)t
    vi = (d/t) – ((a x t)/2)

    3. The attempt at a solution

    a) d = ½(-9.8)1.1
    d = -5.39

    12 - 5.39 = 6.61
    The arrow strikes the sandbag at 6.6 m from the ground

    b) vi = (6.61/1.1) – (((-9.8) x 1.1)/2)
    vi = 6 – (-5.39)
    vi = 0.41 m/s

    Does this look right?
     
  2. jcsd
  3. Aug 24, 2014 #2

    SteamKing

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    You might want to check this calculation:

    Remember, the sandbags are free-falling after being dropped, so their velocity is accelerating as they fall.

    Don't forget to indicate the correct units for your answers.
     
  4. Aug 24, 2014 #3
    d = vi X t + 1/2a X t2

    that would be a better formula because (-9.8) is the rate of acceleration, not velocity. Thank you for that!

    d = 1.1 + 1/2(-9.8) X 1.12
    d = (-4.829)
    d = (-4.8) m

    12 – 4.829 = 7.171m

    a)7.2 m

    b)vi = (d/t) – ((a x t)/2)
    vi = (7.17/1.1) – (((-9.8) x 1.1)/2)
    vi = 11.9 m/s

    Does that look right now?
     
  5. Aug 24, 2014 #4

    SteamKing

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    You're veering out of control now. You fixed one problem and have introduced another into your calculations:

    This is the correct formula. So far, so good.

    What is vi for the sandbags when they are dropped from the tower?
     
  6. Aug 24, 2014 #5
    I thought that vi would be zero for the sandbags.

    should it be (-9.8) m/s?
     
  7. Aug 24, 2014 #6

    Nathanael

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    It would be zero.

    This seems to imply that the initial velocity is 1 m/s (zero times 1.1 is zero, not 1.1)
     
  8. Aug 24, 2014 #7
    face palm!!

    d = vi X t + 1/2a X t2
    d = ½(-9.8) X 1.12
    d = (-5.929) m

    12 – 5.929 = 6.071 m

    a) 6.1 m
    b) vi = (d/t) – ((a x t)/2)
    vi = (6.071/1.1) – (((-9.8) x 1.1)/2)
    vi = 5.519 – (-5.39)
    vi = 10.9 m/s

    Thank you very much!
     
  9. Aug 25, 2014 #8
    just to be certain, can you confirm that what I've done looks right?

    thanks again!
     
  10. Aug 25, 2014 #9

    Nathanael

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    Yes you're correct, and you're welcome.
     
  11. Aug 25, 2014 #10
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