Recent content by JohnTravolski

Thank you.

Okay, I think that I finally understand. I reworked the problem and this time my answer has the correct units (which all cancel out). Here's my new work: https://i.imgsafe.org/f019ae72df.jpg I'm pretty sure this is the correct answer. I don't know if it can be simplified or not.

Screw it, let me just show you the work I've done so far. I found Phi but I have no idea if it's correct or not: https://s11.postimg.org/aap6ye6yb/IMG_1202.jpg

The x and the y components of velocity at the time the box hits the ground.

I thought I did calculate it but I have no idea if what I did was the correct method or not. I'm basically guessing my whole way through this problem. If what I did is correct, I'm still unsure of how to proceed.

None of it is lost. So that means that the velocity is related to the distances it travels.

The vertical distance it travels is D*Cos(theta). The horizontal distance it travels is D*Sin(theta). I'm still unsure of how to get velocity out of that since I'm not provided with any time interval.

Well, before I get that far, do I have to first somehow calculate the initial velocity at the moment the box leaves the incline? If so, how do I go about that?

I'm really lost on this one. I'm really not sure how to get started on this. I started it out as a force problem and solved for ax = g*cos(theta). I then integrated that (treating it like a constant) from zero to D to find my initial velocity at the time the block reaches the edge of the...

OH, now I understand! I didn't realize it was that simple. To tell you the truth, I actually haven't even had Physics class at all this year yet and I'm just trying to get a head-start on my homework, but as I said, none of the assigned reading has covered this. Hopefully my instructor will...

Okay, I almost understand, but there's still one detail I'm missing: Yes, I did find the derivative of the position equation that was given to me, but I still don't understand how that derivative I obtained correlates to an "x-component." In the derivative I obtained: Velocity = -2ct + b What...

Alright, I appreciate the rewording there, your version is less confusing. I suppose I'll have to look up how to use that formula, though, since I've never used it so far.

So I assume that this is the one that I have to use, then? One of the reasons I'm confused is that these equations have appeared nowhere in my assigned reading and I've had absolutely no experience with them yet.

Without the sheet, I would derive the equation given to me to get: Velocity = -2ct + b Meaning that, at T=0, the x component of velocity would = b since the velocity is just a line with a negative slope with the x-axis representing time and the y-axis representing the x-axis that the particle...

I don't understand what "the x-component of velocity" in the question is referring to given that none of the formulas given on the sheet involve right triangles. I don't know how to derive the formula for the x component given the equations on the sheet since I'm required to start there.