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Velocity vectors

  1. Jul 6, 2009 #1
    1. The problem statement, all variables and given/known data
    In a local bar, a customer slides an empty beer mug down the counter for a refill. The bartender is momentarily distracted and does not see the mug, which slides off the counter and strikes the floor 0.90 m from the base of the counter. The height of the counter is 0.840 m.

    2. Relevant equations
    r=vot + 1/2(a)t2
    Vf=Vo + at

    3. The attempt at a solution

    Tried finding time for the y component but ended up with square roots of negative numbers and I've been stuck on this for awhile. I have a lot of other problems like this so if I know how to do this one it would really help me out on the rest.
  2. jcsd
  3. Jul 6, 2009 #2


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    The time is simple.

    H = 1/2*g*t2

    No negative numbers involved at all.

    If you know the time and how far away, then v = x/t.
  4. Jul 6, 2009 #3
    OK thanks I was actually doing those two equations but I kept using -9.8. Never occurred to me to make it positive!
  5. Jul 6, 2009 #4


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    Well you could write it formally. If 0 is the ground, and positive is up ...

    0 = H + 0*t - 1/2*g*t2

    Rearranging ...

    H = 1/2*g*t2

    If you set the top of the bar as 0 then you still have

    - H = 0 + 0*t - 1/2*g*t2

    yielding the same result.
  6. Jul 6, 2009 #5
    Thanks for the explanation but I forgot to post the second part of the problem. The second part asks:

    (b) What was the direction of the mug's velocity just before it hit the floor?

    I have to give the answer in ° (below the horizontal).

    At first I thought it meant that the floor was the base of the angle and the angle was from where the mug hit the floor to the top of the table where it left but that angle was wrong. Other questions also ask me to give degrees against the horizontal but I'm not sure what that's talking about either.
  7. Jul 6, 2009 #6


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    You should have the horizontal velocity. Add then the vertical velocity. They are at right angles so Pythagoras gives you the magnitude and the ratio (draw a diagram and confirm) gives the angle from the tan-1.
  8. Jul 6, 2009 #7
    Thanks, just one more question.

    A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 8.00 m/s at an angle of 20.0° below the horizontal. It strikes the ground 4.00 s later.

    (b) Find the height from which the ball was thrown.

    Found the y velocity which is 2.736 and did -h = 2.736t - 4.9t2 to get 67.456 for h. Found out it was wrong and then I realized thats the height for the curve of the ball plus the building. I tried using the formula (82sin220o)/2g to get just the height of the curve and subtracting it from 67.456 but that's wrong too.
  9. Jul 6, 2009 #8
    Get your signs correct in the equation! Which way is up? Which way is down?
  10. Jul 6, 2009 #9
    Ok so it should be h = 2.736t + 4.9t2 and that will get me the right answer? I only have one more try so I want to be sure....
  11. Jul 6, 2009 #10


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    If you are using g as -, then 20 degrees below the horizontal is also -.
  12. Jul 6, 2009 #11


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    Substituting 4 for t should yield the right answer.
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