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    machinegunmax's Avatar
    machinegunmax Posts: 4, Reputation: 1
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    #1

    Jan 6, 2008, 03:52 PM
    Physics Buoyancy need elp
    currently I am doing coursework where I am dropping a buoyant ball from different heights above a water tank into the water tank. I am then measuring the depth. I lack results so I was wondering if I plotted a graph of height vs depth - would it be a straight line? I think it is because I can say that 1/2 mv^2=fd where force is the buoyant force of the ball which will be constant. But I'm just worried that there's something I've missed out here - so yeah would the graph be a straight line??
    machinegunmax's Avatar
    machinegunmax Posts: 4, Reputation: 1
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    #2

    Jan 6, 2008, 04:00 PM
    ^^^ missed something out I meant to put mgh instead of 1/2 mv^2...
    id assume h/d = f/mg... and f/mg is constant
    terryg752's Avatar
    terryg752 Posts: 197, Reputation: 4
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    #3

    Jan 6, 2008, 05:25 PM
    Depth of What?

    Depth of tank cannot change, so please clarify.
    machinegunmax's Avatar
    machinegunmax Posts: 4, Reputation: 1
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    #4

    Jan 7, 2008, 08:21 AM
    How deep the ball goes... not the depth of the tank. I measure the height above the tank I drop the ball... then I drop the ball film it - freeze frame it to see how deep below the waterlevel the ball gets.
    tjsail's Avatar
    tjsail Posts: 12, Reputation: 4
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    #5

    Jan 7, 2008, 06:52 PM
    One other thing to consider. The ball has another force on it that consumes its energy. Drag. If you had a ball with neutral buoyancy, it would slow down from drag in the water. And finally come to a stop and just hang there. That drag will be proportional to velocity in the water. So a plot of depth of penetration in the water versus the height it was dropped from, shouldn't be a straight line.

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