Recent content by chris61986

Great post! So if I were to solve it in reverse, it would be PE = PE + KE. All variables are known at that point, except velocity. Makes sense! It's also extremely simple... It also means that I have to re-evaluate this problem tomorrow.

Took a break, but I've solved for d: So if I know the velocity in the Y, it turns into a trig problem. So I can find the x using tanθ=6.32/a So the velocity in the x-axis is 3.65m/s. Using that, I can easily solve for D. 3.65*4 = 14.6m Now I just need to solve for θ2: All I need...

Okay... While I don't know the initial velocity, I know that it's going to hit the top of the building with a certain velocity. The velocity at 20m off the ground will be the same for both sides of the curve. So I can use KEi=PE so v=sqrt(2gh) V (at 20m) = 6.32m/s

I'm going to be a tad upset if I just figured this out: Using kinematics from the top of the parabola: for the Y Vf=Vi + at Vf= 0 + 10*4 Vf= 40m/s Plug into equation from my first post: mgH= 1/2*mv^2 + mgh H= (1/2*v^2 + gh)/g H= 100m Edit: No, this isn't right either! It's not 4 seconds...

Well, I took my Physics final today, and I passed it but I didn't do quite as well as I'd hoped. Anyway, there was one problem I spent tons of time trying to figure out. I've tried giving it a shot again at home, but it's still eluding me. Some guidance would be great! Sorry if this is...

Do you think you could be a little less cryptic?

Homework Statement https://dl.dropbox.com/u/92857534/physicsproblem.png Note: Solving for part D and EHomework Equations Unsure. language in the problem seems ambiguous. Perhaps (KE + PE)b = (KE + PE)a The Attempt at a Solution So I think it's done this way. From A to B and B to C, the...

Hello, new to these forums! I'm working on my UPI homework, and I just want to verify if I'm doing something correctly. The problem: The position vector r of a particle moving in the x - y plane given by r = (2t^3)i + (6-7t^4)j, where r is in meters and t is in seconds. Calculate r v a...