Recent content by StNowhere

  1. S

    2-Dimensional motion problem once again

    That depends on the shape of the cliff as defined in the original problem. I imagined it as a straight drop-off at the end, in which case the base of the cliff is directly below the point where the car leaves the cliff, in which case, no, you don't need to add on, since the question asks for the...
  2. S

    2 questions one wave one delta function

    2nd question: Delta function: \int^{\infty}_{-\infty} f(x)\delta(x-a) dx = f(a) Using that, it looks to me like your value is 6 I'll look at the first question a little more before I hazard a guess on it.
  3. S

    2-Dimensional motion problem once again

    The time the car spends in the air is also the only time the car has to travel horizontally. You have the initial velocity and the angle, so you should be able to calculate the initial horizontal velocity. Once you have that, velocity times time equals ...?
  4. S

    2-Dimensional motion problem once again

    As usual, it depends on which direction you're looking. Since the (vertical) velocity is directed down, the acceleration is down, and the displacement is down, I chose "down" to be the positive direction. So use the positive value for g.
  5. S

    Calculating Acceleration at t = 2.20 s for a Particle in 2D Motion

    I think you're on the right track, but it looks like you're trying to use time as a vector here. remember, it divides both components of your \Delta v vector.
  6. S

    Physics rock off a cliff question

    To see how this is going, how much time did you get for the rock in the air? That should speak volumes.
  7. S

    Physics rock off a cliff question

    I don't think we came up with the same horizontal distance. Make sure you know that there are two separate parts to the time, the time it takes for the rock to impact, and the time it takes for the sound waves of the impact to reach the kicker.
  8. S

    2-Dimensional motion problem once again

    If the car was on the cliff for 5 s, you have the acceleration, 4 m/s^2. Thus, v_0 = gt = 4 \ m/s^2 \ * \ 5 \ s = 20 \ m/s The car now falls under gravity from a height of 30 m. 30 = 20 \sin{24} \ * \ t + \frac{1}{2} gt^2. That should give you the hangtime. Horizontal motion is...
  9. S

    2-Dimensional motion problem once again

    If you have the time it took to go down the cliff, then you should be able to get it's velocity once it leaves the cliff. Now it's gravity's turn. Note, of course, the angle with which the car leaves the cliff, which means it has initial velocity in both x and y.
  10. S

    Physics rock off a cliff question

    I'll take a stab at this: Assuming air resistance is neglected or neglectable (I assume this is true unless there's something more we need to know), the time until the rock hits the water is based on the height and the acceleration due to gravity. \frac{1}{2}gt^2 = 40 should give you the...
  11. S

    Calculating Force of Friction on a Bicyclist

    2nd part: I may be wrong here, but because force is a vector quantity, and since the normal force and friction do not act in the same direction, you cannot simply add to find the total force.
  12. S

    Solve Griffiths' QM Problem 1.16: Time Derivative of Inner Product

    You know, now that I think about it, that does make a lot of sense. I had a feeling one of my assumptions was wrong. Ahh. Now I can move on with a clear conscience. Thank you all very much.
  13. S

    Solve Griffiths' QM Problem 1.16: Time Derivative of Inner Product

    I was pretty sure I had already tried that, but let me try to clarify my position just to make sure. (Beware my use of LaTeX, I'm not very good with it.) Limits of integration suppressed: they are -\infty to \infty Problem starts out: Show that d/dt \int \psi_1^*\psi_2 \ dx = 0 given that...
  14. S

    Solve Griffiths' QM Problem 1.16: Time Derivative of Inner Product

    I guess this isn't really homework; I had homework problems from this book assigned in my QM classes in college, but this one was never assigned, probably because it's too trivial and I'm just trying too hard. Recently, though, I've decided to go through all of my old physics textbooks and do...
Back
Top