Time it takes for submerged object to rise to surface

In summary, to calculate the time it takes for an object to reach the surface of the water, you would need to use Archimedes' principle to find the buoyant force, and the weight of the object to find the downward force due to gravity. Then, using the net force, you can calculate the acceleration of the object and use linear motion formulas to solve for the time. However, this calculation does not take into account the resistance of the medium, which would significantly affect the results. Additionally, if there is tension in a cable attached to the object, it would have to be
  • #36
The second integration is even easier: [tex]\frac {1} {\alpha} \int_0^t \tanh \alpha a t dt =\frac {1} {\alpha} \int_0^t \frac {\sinh \alpha a t} { \cosh \alpha a t }dt = \frac {1} {\alpha} \int_0^t \frac { (\frac {\cosh \alpha a t} {\alpha a})'} { \cosh \alpha a t }dt = [/tex][tex] = \frac {1} {\alpha^2 a} \left[\ln \cosh \alpha a t\right]_0^t = \frac {1} {\alpha^2 a} \ln \cosh \alpha a t = \int_{x_0}^{x} dx = x - x_0[/tex]
 
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  • #37
Cool, thanks for sharing the way you worked them out! I think I prefer the u substitution and tanh^-1 identity over using logarithms and Euler's formula, but mostly just because I already have all the hyperbolic identities and their inverses memorized from differentiation. That's still an interesting way to see it done though, and I'm going to try to remember how to do it. I saw that last identity yesterday (Euler's formula) in my book when looking for hyperbolic identities, but didn't think it could be applied until I saw how you first expanded it and integrated into logarithms.

After the first integration was explained yesterday, the second integration became obvious, but still, thanks for walking me through the steps you took for both!
 

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