Recent content by djalle12

Ohhh that makes sense. Ok so the final velocity of the down hill section should(as far as solving this problem is concerned) equal the initial velocity of the level section. I just realized that it makes no sense to automatically jump back to a slower velocity. Thanks a lot!

without the cos theta^2 I get 107.24, which I believe is what you got, right? From what I see the cos^2 is necessary though.

Oh and I realized I've been using 26° for theta this whole time. With 25° I actually get 88.08 meters.

Well the 2 is a mistake and I took it out and the cos theta came from finding the initial velocity in the level ground direction from the final velocity of the downhill which was in the direction of theta below the x axis.

What I have is [(cosΘ^2)(Δx_1)(sinΘ-μcosΘ)]/μ and I've looked through my derivation and redone it and get the same thing each time, and this equation results in 91.23 meters.

Yes, g canceled out for me.

Nevermind I got 91.23 half of my 182 answer because I believe I initially forgot the 2 in v^2=vi^2+2ax

OK I believe I finally have it :) Looked through again and found a mistake. Now I got 108.23 meters.

My final equation was Δx_2=[2(cosΘ)^2(Δx_1)(sinΘ-μcosΘ)]/μ.

OK now my answer is 182.45 meters, how does that match up with you BvU?

OK now I got 182.45 meters. How does that match up with you BvU?

I just started deriving my equation more thoroughly and noticed a mistake. I'll post another answer in a bit.

My final answer was 101.50 meters. If anyone else did it, did you get the same?

Ok now from the equation (v^2)=(vi^2)+2a(Δx), I figured v=√(2g(sinΘ)-μ(cosΘ)x_1) So I can now equate that to being an intial velocity vector Θ° below the horizontal at the start of level ground. Then solve for for initial velocity in the x direction in relation to the level ground. After...

I believe I made some progress. According to my force diagram then the following should be true F=mg(sinΘ)-μmg(cosΘ) if you set that equal to ma because of F=ma, then a=g(sinΘ)-μg(cosΘ). Right?