A small body of density 0.800 g/cm^3 is dropped from a height of 10.0 m into a l

Question

A small body of density 0.800 g/cm^3 is dropped from a height of 10.0 m into a lake of density 1.00 g/cm^3. Neglect all frictional and other dissipative effects, except the buoyancy force. What is the acceleration of the body while it is under water? What is the maximum depth to which the body will sink prior to returning to float on the surface? Why, in real life, does the small body not descend all the way to the maximum depth?

Explanation / Answer

V=sqrt(2gh)=14m/sec a=g(d1-d2)=9.8*(1-0.8)=0.2g Fbuy*x= dgvx=0.5*v*d2*v^2==>9.8*x=0.5*0.8*196==>x=8m it can't go all the way down as it is less denser than water so the buyoncey will be more at one point so it wil floats back

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