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    illGeno's Avatar
    illGeno Posts: 1, Reputation: 1
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    #1

    Oct 15, 2007, 08:47 AM
    Hard physics
    A diver jumps vertically with a velocity of 4.0m/s from a platform and enters the water 1. seconds later. If she falls with an acceleration due to gravity of 9.8 m/s^2, what is her final velocity?
    DoctorK's Avatar
    DoctorK Posts: 15, Reputation: 1
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    #2

    Oct 15, 2007, 11:26 AM
    Vf= Final Velocity ?
    Vo= Original Velocity 4 m/s
    a= Acceleration 9,8 m/s^2
    t= Time 1 s

    Vf=Vo+a(t) 13,8 m/s
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #3

    Oct 15, 2007, 11:31 AM
    Quote Originally Posted by DoctorK
    Vf= Final Velocity ?
    Vo= Original Velocity 4 m/s
    a= Acceleration 9,8 m/s^2
    t= Time 1 s

    Vf=Vo+a(t) 13,8 m/s
    I believe the question requires v0 to be set as -4m/s, not +4m/s.
    DoctorK's Avatar
    DoctorK Posts: 15, Reputation: 1
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    #4

    Oct 15, 2007, 11:53 AM
    Quote Originally Posted by Capuchin
    I believe the question requires v0 to be set as -4m/s, not +4m/s.

    As the question states "A diver jumps vertically" I assumed the diver jumped DOWN in which case the original value is correct, and V0= +4.

    If the diver jumps UPWARDS, then your appreciation is correct, and V0= -4.
    asterisk_man's Avatar
    asterisk_man Posts: 476, Reputation: 32
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    #5

    Oct 15, 2007, 12:21 PM
    DoctorK, please demonstrate how to jump down. I'd love to see it. :)
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #6

    Oct 15, 2007, 01:40 PM
    Quote Originally Posted by asterisk_man
    DoctorK, please demonstrate how to jump down. I'd love to see it. :)
    Had to spread the rep, but this is exactly the point I was trying to make (albeit subtly). Plugging numbers into an equation is fine and dandy, but not at the cost of common sense. :)

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