Recent content by RenaltJ

alright thanks, definitely did not understand this one at all... I appreciate your help. I was able to get the answer I believe.

I'm not at all confused by the chain rule, I'm confused by how to employ it from the starting conditions. I've successfully solved the equation, which looks like 1/k*ln[(cosh(1/k*arctanh(v/√gk))] evaluated from v0 to v1 I cannot imagine this is what the prof is looking for though (even though...

How do you derive dv/dx and dx/dt from equations with neither of these terms? ie. dv/dt = g-mv dy/dt grabbing the already solved equation (y would be the falling problem) yields dy/dt = \sqrt{g/k}tanh(\sqrt{g/k}t dv/dy = 0? I see no way to convert unless I'm missing key steps here if you sub...

Here's what I get: t + C = 1/2*\sqrt{k/g}[ln\frac{v-√g/k}{v+√g/k}] When I work out the rest nothing simplifies. I've also tried it using your first suggestion stuckAgain. Using some info in this book (taylor classical) I got v = \sqrt{g/k} tanh(\sqrt{g}t/\sqrt{k}) but this is the general equation?

Homework Statement A particle of mass "m" whose motion start with downard velocity V0 in a constant gravitational field. The drag force is quadratic and proportional to kmv2. What is the distance s through which the particle falls in accelerating from v0 to v1. Give your expression for s in...