How High Will a Lighter-Than-Water Ball Shoot Above the Surface?

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A small ball released from a depth of 0.600 m in water, with a density of 0.300 that of water, will rise due to buoyancy. The discussion emphasizes using energy principles to equate the work done to submerge the ball with the gravitational potential energy gained upon release. The negligible drag force simplifies the analysis, allowing a focus on buoyant forces. The concept of buoyancy is further illustrated by comparing human density variations based on body composition. Understanding these principles is crucial for determining how high the ball will shoot above the water's surface.
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Homework Statement


Suppose that you release a small ball from rest at a depth of 0.600 m below the surface in a pool of water. If the density of the ball is 0.300 that of water and if the drag force on the ball from the water is negligible, how high above the water surface will the ball shoot as it emerges from the water? (Neglect any transfer of energy to the splashing and waves produced by the emerging ball.)

Homework Equations


F(b) = m(f)g (f is for fluid) - in this case water


The Attempt at a Solution


No idea how to set it up!
 
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It would probably be helpful to think in terms of energy when doing this problem: The work required to push the ball to a depth of 0.6m should equal the gravitational potential energy that it acquires when you release it.
 
Humans are made mostly of http://www.ionizerscompared.com" and are thus very close to being neutrally buoyant. However, if you take a large breath and hold it you can become positively buoyant since your overall volume increases thus lowering you overall density. Fat cells are less dense than water so people who carry a lot of extra fat can float more easily.
 
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Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
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