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Velocity as a function of time.

  1. Sep 24, 2007 #1
    Hi all. This is my first post here and I have a feeling this is going to become a regular site for visiting. Anyway, I have this problem to do:

    1. The problem statement, all variables and given/known data

    [​IMG]
    Right now I'm just looking for help on part A.

    2. Relevant equations

    see above :)

    3. The attempt at a solution

    I know that I need to get a function which, when the time is plugged in, will return the velocity at any given moment. As of now I have the answer v(t) = (VoCos(theta) + VoSin(theta))î - ((1/2)gt^2)ĵ, but this doesn't seem right, at least the î section. I do believe the ĵ section to be correct, though. If anybody could point me in the right direction without giving me the answer it'd be greatly appreciated. Working with vectors really confuses me. Thanks!!
     
  2. jcsd
  3. Sep 24, 2007 #2

    Doc Al

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    Staff: Mentor

    Try this: First write expressions for horizontal and vertical speed separately, then combine them into a vector expression with i & j.
     
  4. Sep 24, 2007 #3

    Astronuc

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    Try this instead

    v(t) = (VoCos(theta))î + (VoSin(theta) - (1/2)gt^2)ĵ

    vx(t) is constant, if we neglect air resistance.

    vx(t) changes under the influence of gravity, which act always in the -ĵ direction.

    A good back reference is http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html
     
  5. Sep 24, 2007 #4
    It now makes perfect sense that (VoCos(theta))î is the X component, but could you possibly elaborate one why (1/2)gt^2 is subtracted from VoSin(theta)? Is this to account for the velocity for the time spent above the edge of the cliff? I'm trying to gather my thoughts to expand on this post...

    Thanks!
     
    Last edited: Sep 24, 2007
  6. Sep 24, 2007 #5

    Doc Al

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    Are you familiar with the kinematic equations for accelerated motion? In this case, V0sin(theta) represents the initial velocity.
     
  7. Sep 24, 2007 #6
    Wouldn't it represent the initial velocity in the y direction? Yes, I'm familiar with them. I have a feeling I'm overlooking something blatantly obvious =/.
     
  8. Sep 24, 2007 #7

    Doc Al

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    Absolutely. And we are talking about the y-direction (also known as the j direction).
     
  9. Sep 24, 2007 #8
    Thanks. I've got a bunch of thoughts going through my head right now...I'll update once I've sorted them all out :).
     
  10. Sep 24, 2007 #9
    This is where I'm standing at the moment...I'm assuming there's no wind resistence in this problem. I know this standard equation:

    v(t) = v(0) + at

    I think in the above equation I made things more complicated for myself by using (1/2)gt^2, which gives the same answer when numbers are substituted in as a*t, so I'll use the above equation and model it after the equation listed above.

    I know vxo=vocos(θ) and since this is projectile motion the x (or i) component of the velocity is constant, therefore no acceleratio, so my first part of v(t) is just vocos(θ)î.

    For the second part all I need to do is plug and chug information in and vyo=vosin(θ), but since this y vector has gravity acting on it I must include the + at on the end, therefore the second part of v(t) is (vosin(θ) + at)ĵ.

    So, after combining it all I get v(t)=vocos(θ)î + (vosin(θ) + at)ĵ.

    Does the above seem correct? My problem with physics is my common sense is always contradicting itself and I overthink even simple things. Thanks a lot for the help!
     
    Last edited: Sep 24, 2007
  11. Sep 24, 2007 #10

    Doc Al

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    Perfect. Just put the acceleration in and you are good to go.
     
  12. Sep 24, 2007 #11
    Do you mean the a in the j vector isn't sufficient to represent the acceleration or I just need to plug in -9.8?
     
  13. Sep 24, 2007 #12

    Doc Al

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    Yes and no. Don't leave it as just "a", but don't plug in numbers either. Use what they told you: a = -g.
     
  14. Sep 24, 2007 #13
    Ah okay, thanks a lot for the help! I love it when things click.
     
  15. Sep 24, 2007 #14
    Back again...I'm really struggling with this problem. I've been playing around with how to get part b and I started with the equation yf = yi + viy*t - (1/2)gt^2 becuase I know yi = h0 and yf = 0, but I can't get it in terms of t=, the math just won't work. Could anyone point me in the right direction? Thanks!
     
  16. Sep 24, 2007 #15
    Mybe it is just my computer, but I do not see a problem posted in thread #1.
    It says "See above" ...but I do not see anything above.

    Casey
     
  17. Sep 24, 2007 #16
    Last edited: Sep 24, 2007
  18. Sep 24, 2007 #17
    i guess it's te sign of direction that's the problem. at first, accelration was down, and therefore negative, while, velocity was upwards, therefore velocity. but this changes as soon as from when velocity becomes momentarily zero. you ahve to break it in two parts.

    hope this helps.
     
  19. Sep 24, 2007 #18

    learningphysics

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    Use the quadratic formula to solve for t.
     
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