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#325 - EK Physics: Math Error?

Discussion in 'MCAT Study Question Q&A' started by ilovemcat, Jan 18, 2011.

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  1. ilovemcat

    ilovemcat Banned

    665
    3
    Apr 16, 2010
    I'm have trouble trying to figure out what I'm doing wrong here:

    The question states:

    A 2 kg ball is thrown upwards with a speed of 40 m/s. At what height will the ball be, when its kinetic energy is equal to its potential energy?

    I know that during half the trip (half the height), Potential Energy = Kinetic Energy. In order to solve for height, I use one of the kinematic equations.

    But here's where the problem lies, I get two different answers using different equations but can't seem to understand why. They should be equal regardless.

    h = hi + vit + .5at^2
    h = 0m + (40m/s)(4s) + (5 m/s)(4^2)
    h = 240 meters

    or using this equation:

    vf^2 = vi^2 + 2ah and solving for 2
    2gh = vi^2
    2(10m/s)h =1600 (m/s)^2
    h = 1600 (m/s)^2 / 20
    h =80 meters

    Any idea what I'm doing wrong to get different answers?
     
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  3. ilovemcat

    ilovemcat Banned

    665
    3
    Apr 16, 2010
    By the way, the correct answer is 40 meters. (half of 80 meters)
     
  4. ilovemcat

    ilovemcat Banned

    665
    3
    Apr 16, 2010
    Ah, nevermind! I just realized I forgot to subtract in the first kinematic equation (160 - 80). I need to rest my brain. I've been working too hard today lol.
     
  5. shffl

    shffl

    130
    0
    Sep 20, 2010
    Remember acceleration is a negative number. If you used -10m/s^2, then your answer would be h = 80m as well
     
  6. TwoPaddles

    TwoPaddles

    231
    0
    Sep 6, 2010
    I think the 2nd one saves much more time. At maximum height, final velocity is zero.
    vf^2 = vi^2 + 2ax
    vf=0
    x = vi^2 / 2a
    x = 1600 / 20 = 80m so this is the maximum height (all kinetic energy converted back to potential energy)

    Halfway height is where KE = PE; 40m.
     

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