Recent content by fisiks

So, how would I work from there?

Decreasing from with height. So, there's no way to just calculate as if density is just modifying D, which would be a function? And if so, how should I recreate the equation?

Forgot that \rho is proportionate to displacement. So, I'm guessing my next step would be solving another differential equation, \frac{dD}{dt}= ...?

Integrate t?

http://en.wikipedia.org/wiki/List_of_integrals_of_rational_functions (2nd one) It looks right to me (factoring out -1/c^2, to make it match the form).

I tried looking around. I couldn't find the problem, but I must admit, I'm kinda confused.

So, was the second integration also incorrect?

So: t=\frac{tan^{-1}(cv_i)}{cg} tcg=tan^{-1}(cv_i) v_i=\frac{tan(tcg)}{c} Doesn't really feel right. Tried integrating from v_i to v: t=\frac{tan^{-1}(cv_i)}{cg}-\frac{tan^{-1}(cv)}{cg} tcg=tan^{-1}\frac{cv_i-cv}{1+c^2v_iv} \frac{cv_i-cv}{1+c^2v_iv}=tan(tcg)...

Okay. So would the first scenario be something like \int_{0}^{t}dt'=\frac{1}{g}{}\int_{v_i}^{0}\frac{dv'}{-1-c^2v'^2} where c=\sqrt{\frac{k}{mg}}?

Okay, modifying my equation so that it's negative. Forgot about that. dt=\frac{dv}{-g- \frac {kv^2}{m} } About the integration: am I supposed to get an indefinite integral, or integrate from 0 to the variable?

Sorry, new to this. Probably should've mentioned that this is actually my older brother's homework, which I thought I'd try my hand at. Would this be right for the velocity part? ma=m\frac{dv}{dt}=mg+kv^2 (where k=\frac{1}{2}\rho C_D A) ∴\frac{dv}{dt}=g+\frac{kv^2}{m} dt=\frac{dv}{g+ \frac...

Apologies, this is most likely basic to most of you, and there are probably much better ways to go at this than I have. Homework Statement An 8mm wide cone of a hypothetical substance which does not melt, with a mass of 2.7 grams, is shot straight up at a 90 degree angle. Only accounting...

Tried to calculate what's required to reach that height first, but I don't know how to work with when a variable depends on itself. V=V_i-((\rho*C_d*A*V^2)/2M - G)T D=\int_0 (V_i-((\rho*C_d*A*V^2)/2M - G)T)dT

Yes, fries would be good. Anyways, I've tried deriving it myself, but *cough* I'm embarrassed to admit it, but I got stuck deriving the kinematics part >.> 2Vi*t+\sqrt{3}*Vi*t-\sqrt{3}*a*t^2-d=0

... factoring in fluid dynamics, not just Newtonian physics... At a given height, the object will possesses 75% of the kinetic energy of when it initially began moving. Mass is irrelevant. Given variables: -Drag coefficient -Fluid density -Cross-sectioned area -Displacement -Initial and final...