Maximum velocity of buoyant object

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To determine the maximum velocity of a 2.75" diameter, 16-gram plastic ball submerged at 6 feet, one must consider the balance between buoyant force and viscous drag. Terminal velocity is achieved when these forces equalize, which may not occur within the 6-foot depth. The dynamic equation ma = f_b + f_drag should be used, with f_b representing buoyancy and f_drag the drag force, assuming Stokes flow for simplicity. Tethering 30 balls together with fishing line at 1/2" spacing will complicate the drag calculations, as each sphere's interaction must be accounted for. Understanding these principles will provide insights into the ball's velocity behavior upon release.
ralphamale
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I have a plastic ball 2.75" in diameter weighing 16 grams submerged at 6 feet depth. I am trying to determine the maximum velocity achieved when I release the ball? Also, if I tether (30) balls together with fishing line at 1/2" spacing how will that affect velocity? I really appreciate any help and input on this. Thank you!
 
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Terminal velocity is reached when the buoyant force equals the viscous drag force, which may not occur over the 6 feet. I would start with the full dynamic equation: ma = f_b + f_drag, where f_b is buoyancy and f_d the drag term. Assume Stokes flow for the drag term, and solve for the velocity as a function of depth. You will also have to assume that Stokes flow hold for each sphere in the chain in order to keep this problem easy.
 

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