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Homework Help: Answer check: A ball on a spring launched into the air. find speed/height

  1. Nov 11, 2012 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    Since energy is conserved, MEinital = MEfinal

    3. The attempt at a solution
    1/2 mvi2 + mgyi + 1/2kxi2 = 1/2 mvf2 + mgyf + 1/2kxf2

    After finding what equals 0, im left with:
    1/2kxi2 = 1/2mvf2 + mgyf

    I found yf by: mgyf = 1/2kxi2
    yf = 3.63m

    Just checking to see if these answers are correct. THank you for the help!!!!
  2. jcsd
  3. Nov 11, 2012 #2


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    I get a different result. Please post more details of your calculation.
  4. Nov 12, 2012 #3
    First I found yf, which is the answer to part b.
    Since energy is being conserved, I came up with PEf = KEi. This seems weird because usually there is no potential energy at the final position but it was the only thing I could come up with.

    mgyf = 1/2kxi2
    y= (1/2kxi2) / g


    1/2kxi2 = 1/2mvf2 + mgyf

    sqrt((kxi2 - mgyf) / m) = v
    v = 5.97m/s
    Last edited: Nov 12, 2012
  5. Nov 12, 2012 #4


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    Compare those two equations. Shouldn't you get vf=0? ...
    .. but you didn't because you dropped a factor of 2.
    For what point of the trajectory do you think you should be finding the velocity in part (a)? What is the value of y at that point.
  6. Nov 12, 2012 #5
    oh ok. i understand. the fastest the ball will be going is right after its released from the spring, not at its highest position in the air.

    So its,
    1/2mvi2 + 1/2kxi2 = mgyf
    vi = sqrt( 2gyf - kxi2 )

    But now I'm stuck not knowing what it's final position is.
  7. Nov 12, 2012 #6
    Using yfinal = 3.63m. I got the velocity to be 7m/s
  8. Nov 12, 2012 #7
    I am wrong. The inal position is .15m because that's how far the spring is compressed an that's how far the ball moves before its released from the spring

    V= 8.3m/s for part a
    Y= 3.6m for part b

    Thanks for the help!!!
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