Recent content by Cipherflak

Yup. Those 4 basic equations of motions is what you use when you start out with physics, but you see clearly where they naturally come from first when you learn about derivates and integration. :)

OUCH! i simply forgot to write down the half! Pretty amazing that no-one noticed! Sorry about that, dude! :)(the half actually appears from taking the anti-derivative of of t, resulting in 0.5(t)^2.)

With 0 acceleration in the horionztal I take it. That's actually what I said in my equations.

well you can't solve this problem with only one equation.

It's like hearing my own voice. Everything came easy until i started on the university, where I faced the shocking and hard truth of simply not understanding it. The realization that, hey, you might not be able to. Well, after quitting half way into the first semester i tried again, working...

start with setting up all the forces working just as you release the spring. At first, the spring force F(x)=-kx, (x in the direction of the plane) will be greater than the projection of gravity, but as x gets smaller, gravity will win (or even the spring might contract and push back)...

Tha navier-stokes equation sounds a bit overkill for this problem, but in lack of a better suggestion atm, I'll fold. Btw, V(r=0) is greater than Vo, right, and V(r=R)=0. By continuity, the flux Vo*R at the initial transect should be exactly equal to the flux through a transect far down stream...

s = vt + 1/2a(t^2). here you have a solution for s. if you plug in s=0 (the ground where the projectile starts), you probably have two different t that satisfy this. One is obviously t=0, but the other?. As for the maximum height; well, what you have here is distance as a function of time...

Funny that they call it a rocket. Rockets are usually self propulsive by hurling mass backwards. This is a normal projectile, right? :)

if you break the problem into a horizontal and a vertical part, the sum of forces should be 0 in each of these. in the horisontal, the force on cable 1 must be equal to the HORIZONTAL PART of the force of cable 2. In the vertical the VERTICAL PART of cable 2 must balance gravity. Agree?

The tension is just the force exerted by the cables. Have you made a drawing and applied the three forces working?

don't know about no algorithm, but solving it with a computer seems very straightforward.

what do you mean with ax? acceleration times position? or is it street lingo for acceleration? kids these days..

we just told you that you should have a plus in there, not a multiplication! you do something wrong when you insert the substitution.

yup, you switched a + for a *. also, it seems you added a minus to the 6. Drop the minus. (I hope it wasn't because you misread my last post, the -6m thing was just an answer to your hypotethical situation where you stand 6m above ground and just clear a "fence 0 meters high".)