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Arrow Loses Mechanical Energy

  1. Nov 13, 2016 #1
    1. The problem statement, all variables and given/known data
    A 125 - G arrow is shot vertically upwards with a Vi of 28/ms. Assuming a 20% loss in Mechanical Energy while ascending, what maximum height above the position it was shot does it reach?

    2. Relevant equations
    Vi= 28m/s
    Vf= O m/s
    D= ?
    a= -9.8 m/s

    Vf^2= Vi^2 + 2aD
    Eg = mgh
    Ek = 1/2mv^2
    EM = Eg + Ek (not sure}
    3. The attempt at a solution
    I did not really understand this question but I tried to make a solution using what I know.
    I tried using one of the 5 equations but that did not help so i tried using work and energy equations.

    So i did Ek first
    Ek= 1/2 mv ^2
    Ek = 1/2 (0.125 kg) (28 m/s)^2
    Ek = 98 J

    Next I tried finding the Eg
    Eg = Ek x 0.70
    (30% loss of mechanical energy, this is what i am thinking)
    Eg = 68.6 J

    Eg = mgh
    Eg= 0.125 (9.8) h
    68.6 = 1.225 h
    68.6/1.225= h
    h= 55.51 m

    Did I do this question right or wrong? I am very confused. Some help would be appreciated!
  2. jcsd
  3. Nov 13, 2016 #2


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    Homework Helper
    Gold Member

    Where does the 30% come from? Your method is correct.
  4. Nov 13, 2016 #3


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    Staff: Mentor

    Check that calculation. Did you divide by two?
  5. Nov 13, 2016 #4


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    Science Advisor
    Homework Helper

    Yes it's the right approach just check the working as others have suggested.
  6. Nov 13, 2016 #5
    i meant 30%
    Last edited: Nov 13, 2016
  7. Nov 13, 2016 #6
    (In the post i meant 30% loss, not 20 % thats why i did multiplying by 0.7.)
    I divided by two so the number is 49 J , multiplying it by 0.7 would be 34.3 J
    34.3 J / 1.225 = h
    H= 28 M exactly
    Is this method now correct?
  8. Nov 13, 2016 #7


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    Staff: Mentor

    Yes. Although "exactly" depends upon your choice of value for g :smile:
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