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Calculating vertical velocity

  1. Feb 5, 2012 #1
    How do I calculate vertical velocity? I feel very silly for asking this question, as I should know this and is very basic. I don't seem to get it from any of my books and it is the quick,, thus, I can't contact my teacher or friends as I don't have their email or skype. If I can't still work it out, then I will post the question.
  2. jcsd
  3. Feb 5, 2012 #2
    Vertical velocity of what?
    What information do you have?
  4. Feb 5, 2012 #3
    Ok. I'll thenpost my question:

    A bowling ball of mass 7.50kg travelling at 10.0 m s-1 rolls off a horizontal table 1.00 m high.

    *skipping a) as it's done and not posting others as I'm not up to them*

    b) What is the vertical velocity of the ball as it strikes the floor?

    Thanks in advance :smile:
  5. Feb 5, 2012 #4

    What equations do you know how to use and of them, which do you think you should use here?
  6. Feb 5, 2012 #5
    Well I'm required to use these three Equations of Motion:

    v = u + at
    s = ut + [itex]\frac{1}{2}[/itex]a[itex]t^{2}[/itex]
    [itex]v^{2}[/itex] = [itex]u^{2}[/itex] + 2as

    This is what I believe I am to use. It's meant to be primarily focused on projectile motion. I believe, correct me if I am wrong, that the first one is for vertical velocity. I do feel that I am wrong as I don't have acceleration nor can I calculate it. The answers in book don't have solutions to follow and the worked examples aren't of much use.

    I seriously feel silly for even asking for this, it is simple, yet for some reason ... I'm not getting it !!! It's probably a right in my face answer that will give me a silly "Ah-ha" moment. Yet thankyou for your help.
  7. Feb 5, 2012 #6


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    Those equations of motion pertain to motion in general, and hold as long as the acceleration is constant. There's no particular equation for vertical velocity as such (but of course, all those equations also apply).

    For this problem, realise that:

    1) horizontal and vertical motion are independent of each other.

    2) vertical velocity starts as zero, and is subjected to a constant acceleration of g downward.

    3) the only other quantity you're given is the height through which the ball descends, so use the appropriate equation.
  8. Feb 5, 2012 #7
    okay, I'll explain these equations to you!

    The standard notion for this kind of stuff is
    v = final velocity
    u = initial velocity
    s = displacment
    a = acceleration
    t = time elapsed

    These are all vector equations, so these apply for each component of velocity or displacment
    The first equation
    [itex]v=u + at[/itex]
    This tells you that the final velocity is equal to the initial velocity plus a term due to acceleration, which should be expected.

    The second equation
    [itex]s = ut + \frac{1}{2} a t ^2 [/itex]
    This equation tells you that the displacment is equal to the initial velocity times time, which again should be expected, plus a term which involves acceleration (once you know yourself some calculus you'll know where the half and the second power come from)

    The third equation
    [itex]v^2=u^2 + 2 a s[/itex]
    Isn't really anything new, it's really just the first two equations
    It tells you that the square of the initial velocity is equal to the square of the final velocity plus a term due to acceleration.

    Now, with what you are told in the question, which of these do you think you should use?
    We have an TWO initial velocities here, one is stated explicitly, the other is between the lines, we also have a displacment.
    And we're trying to find a final velocity.

    Which of these equations do you think will help us with this?
  9. Feb 5, 2012 #8
    Yes, I do get that the other one is 9.8 ms; but the answers say otherwise and I want to know how to get to it. I initially thought the answer was 9.8, it's what made sense.

    WAIT, hang on! Would 9.8 go in the acceleration?! So with what we have, it would be the third equation. Yet I have tried this option before, and like before, the answer came out wrong (I did square-root my answer to ensure it was correct).

    See people, I do seem silly and your probably face-palming yourselves. Sorry, but I don't know why. I'm sure my teacher taught me correctly.
  10. Feb 5, 2012 #9


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    Approximately, g = 9.8 [itex]ms^{-2}[/itex] (note the units). Yes, that's the acceleration.

    Use the third equation, but show your working. Then we can see what you're doing wrong.
  11. Feb 5, 2012 #10
    [itex]v^{2} = 10^{2} + 2x9.8x1[/itex]

    [itex]v^{2} = 119.6[/itex]


    = 10.94
  12. Feb 5, 2012 #11


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    The initial *vertical* velocity is zero. So why are you putting u = 10?
  13. Feb 5, 2012 #12





    You have no idea how long I have dwelled on this !!!!!!! Yes, it is sad. * facepalm x100 *

    Many blessing for you my friend!

    What can I say, I am soooooo relieved.

    The answer is correct

    HOW CAN I, IN MY RIGHT MIND, EVEN NOT REALISE THAT. I can now finally move on with my work and stay up to speed ... hopefully, this doesn't happen again.




    P.S it's ok, I'm not crazy no desperate or mad ... I'm just trying to express my gratitude in the limited methods of communication we have access to (ie. text on monitor ... no face-to-face or voice). I have learned well from this crazy dilemma.

    By the way - How can I resolve his thread ... yes, another silly question. I can't find the button on the page nor my control panel.
    Last edited: Feb 5, 2012
  14. Feb 5, 2012 #13


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    After all that, a mere "you're welcome" seems somehow inadequate. But you're welcome. :smile:
  15. Feb 5, 2012 #14


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    You could try:
    (You're welcome.)3141
  16. Feb 5, 2012 #15


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    Or I could offer him a nice slice of 3.141.
  17. Feb 7, 2012 #16
    Gosh you all are funny ;)

    Both shall suffice and I also say them too :D

    I got my point across and gratitude ... this is a different circumstance as it felt so SIMPLE yet I could not work it out for someone insane reason -_-
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