14.3 Resistive forces - high speed case

[@federicomiceli1344]

At 5:12 it is suggested to use integration by parts to integrate 1/(1-u²), but it's probably a lot easier to solve it observing that 1/(1-u²)=1/2*[1/(1+u) + 1/(1-u)] Also, at 5:34 there is a t missing. The integration should give -g sqrt(β/mg) t, instead of just -g sqrt(β/mg). The t term is added again a couple of lines later (when we exponentiate).

[@tanmay2340]

This one was the most complicated so far.

[@PaulSinnett]

At what speed would you change from the low speed to the high speed model?

[@talibeilm21c]

When the drag forces due to the high speed and low speed model become same. If the velocity is greater than this critical speed it's high speed model below which a low speed model. Now you could equate both these drag force types and arrive at a formula for the critical speed.

[@cafe-tomate]

I think he didn't mention the fact that for ln (at the end) to be safe we need v to be between -√(mg/β) and √(mg/β) I mean this "constraint" is satisfied by the final answer, but I feel like it's worth noticing and pointing it out

[@cafe-tomate]

Also, from a more physical point of view, we can see that the velocity we are given at the end can't outreach a maximum. I would have expected a formula valid for all high speeds...

[@erdos73]

Wouldn't an object moving faster than √(mg/β) just decelerate to √(mg/β) due to F_res being greater than mg? and an object with v

[@mortezakhoshbin]

great explanation