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*sniff* Thanks

  1. Feb 28, 2004 #1
    No one answered my burning questions about imaginary tides...these tides that have me in a bind, so to speak.

    So, *sniff, sniff* I will just blunder through the essay questions as best I can.

    Thanking everyone who has helped me all these weeks...I have a test Monday and feel confident because of all the wonderful help from the forums and because the professor suddenly relented and is going to write formulae on the board for us to use.

    If anyone has a last-minute answer for this, I'd appreciate it:
    Q. A projectile is fired straight upward at 141m/s. How fast is it moving at the very top of its trajectory? Suppose that instead it were fired upward at 45 degrees? What would be its speed at the top of the trajectory then?
    A. I have it going 0 m/s at the very top of any trajectory. We ignore air resistance in our class.

    Thx again y vaya con Dios, amigos y amigas. Bueno!
     
  2. jcsd
  3. Feb 28, 2004 #2

    chroot

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    The projectile has zero vertical velocity at the top of its trajectory, because gravity acts on it vertically to slow it down. The projectile's horizontal motion never changes throughout the entire trajectory, though. If the projectile starts with some horizontal velocity, it will have that same horizontal velocity at all times, including at the top of its trajectory.

    What is the horizontal component of velocity for a projectile fired at 141 m/s at an angle of 45 degrees?

    - Warren
     
  4. Feb 29, 2004 #3
    So, the horizontal velocity at the top of the trajectory is 141m/s?
     
  5. Feb 29, 2004 #4

    chroot

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    No. If the projectile is fired at 45 degrees, some of its velocity is vertical, and some of it -- not all of it -- is horizontal.

    - Warren
     
  6. Feb 29, 2004 #5

    turin

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    What imaginary tides? Are you talking about the ones in the lakes?
     
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