Ok so I didn't get it when I plugged it in with your equations (but there was a lot of scribbling so it was probably calc error)
but I got it using these equations
Vi*sin(90-theta)=2Vf
and
Vf^2=Vi^2-GM/R
two equations and two unknowns just make sure your units are right
Thanks for all the...
oh dangit this is what it should be (I think)
(1/2)m*vi^2-((GMm)/R)=(1/2)*m*vf^2-(GMm)/2R
and for the other conservation law would it be orbital momentum? our teacher hasn't really gone over it but we have homework on it, so I get the idea I just have no idea how to use that kind of momentum...
So I'm having trouble on this problem as well:
I have trouble visualizing it so is this the full work energy equation:
(1/2)m*vi^2+((GMm)/R)=(1/2)*m*vf^2-(GMm)/R
I want to say this is right but I don't know, and if it is what is the other conservation equation? I think it would be momentum but...
You're on the right track and close!
You just have to find the force of friction
so from the sum of forces in the x direction and the torque equation, respectively, we have this:
mg sinθ - friction = m a friction * R = I a / R
Since the moment of inertia for a hoop is mr^2...
ok I have no idea how this makes sense, I think something's wrong but I somehow got it all right using these equations for each step
for Delphi51's Triangle I used similar triangles of water velocity vector and the triangle he actually wanted to swim. so it'll look something like this...