Recent content by Daniel Ivanov

Of course you're right regarding mg, missed it for some reason. Regarding the Reynolds number I also agree, thanks for pointing it out. I will take a few moments to recalculate it

From what I know in fluid dynamics, v^2 is the case when the velocity of the object is high. Given that I'm dropping a sphere\jumping myself from 10 meters, the velocity is going to be not very large (~ 14m/s). Newton's second law in my case looks like: mv' + bv = -rho*g*V. According to...

You're right, but the thing is that after jumping into the pool there are a few seconds when we go down and then up, so after literally touching a bottom of a pool I became curious.

It's not important I think. Physically it barely changes something, for now I assumed that the "body" is just a sphere but in general of course I'm interested in a human's body to get a more "realistic result".

Yes, I have read about it (terminal velocity). Then I think the second stage does not help me at all. Yes, I totally ignored it, thanks for your responce ! In general I thought to take it but wasn't sure how to do it properly. Now I think that I need to answer a slightly different question...

First of all, obviously, we need to get the velocity before entering the water from the energy conservation 1/2mv^2 = mgh -> v = sqrt(2*g*h). After entering the water, we still have a gravity force and also drag force. Since the velocity is low, I take a drag term as "-bv", where b is a constant...