Random info on PS3

I had an email exchange with one of you regarding tomorrow's problems, so I thought I'd reproduce it here in case it helped someone else.

In my notes i have the position function for the inclined plane problems as x(t)=.5(a_x)(t^2)-(h/sin(theta)). My question is why is the last term there? It is the same as the hypotenuse of the plane, so it could be the x initial if I rotate the x-axis to be parallel to the plane, but wouldn't it make more sense to make the top of the plane the origin giving an x initial of 0? Or is the origin placed at the end of the ramp in order to more easily compute the trajectory after the block leaves the ramp/table? Sorry, I'm just a little confused about it
all.

If you call the length of the ramp along the tilted part L, then you would say sin(theta) = h/L, or L=h/sin(theta). So, the h/sin(theta) part is just the hypotenuse. We put it in this form because for the problem we did in class, we were given the height above ground level, not the ramp length.

You are also right that if you choose the origin to be the starting point, it is just x(t) = 0.5(a_x)(t^2) for the position along the ramp. I chose the end point of the ramp to make the trajectory easier, as you suggest, but it really doesn't make that much of a difference. You can put the origin at the starting position, and for the trajectory it isn't that much harder either way.

What you do need to keep in mind is that once the thing hits the bottom of the ramp, you should define a new x-y coordinate system to do the trajectory - it will be easier for that step with x & y vertical and horizontal as usual, the rotated coordinate system will just make things harder. At that point, you could actually just redefine the origin of the new system as well to be at the end of the ramp.

So basically, you are right. It is really easier to treat it as two separate problems, and give each their own coordinate system and origin - first do the ramp, then start over with the velocity you found and do the projectile.