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Relative velocities

  1. Feb 24, 2006 #1
    Well, I was hanging out with a friend the other day in a shopping mall's foodcourt and this guy I know from school comes up to me, asks me if I have a pen and a paper, and writes out a physics question for me to solve (talk about being a weirdo :biggrin:) and it went something like this:

    A man is juggling while riding on a horizontal escalator; he is moving at 1m/s relative to the escalator. The escalator is moving 0.5m/s relative to the ground. The maximum vertical point a ball reaches is 2m and another man, walking by the escalator, notices that the horizontal range of the ball is 1m. What is the velocity of the other man? (Relative to the ground)

    I guessed I'd first write down what's given relative to the ground:
    v ball x = 1.5m/s
    y = 2m
    xP = 1m
    Then, finding out how much time it takes for that one ball to go through half a cycle (i.e. in order to find ACTUAL horizontal range):
    0 = vy^2 - 19.6*2
    v ball y = 6.26m/s
    0 = 6.26 - 4.9 t
    t = 1.27775s
    And to find the actual horizontal range,
    x = 1.5 * 1.27775 = 1.9166m
    And then, because everything in the x direction is going at constant velocity (and we're assuming so is the other man), so velocity is directly proportional to distance, and because the constant k, 1/t, is constant (because the time intervals are the same), we can write:
    k xP / kx = vP / v ball x = xP / x
    v P = 0.7826m/s

    Where v P is perceived velocity by the walking man and x P is the perceived range.
    Next, because he sees it as 0.7826m/s, we can write:
    V + 0.7826 = 1.5
    V = 0.717m/s
    And what I got is that he's walking at 0.717m/s.
    Can anyone tell me if I'm right or wrong? Thanks.
  2. jcsd
  3. Feb 24, 2006 #2
    I got the same number for one of the possible velocities, the other is 2.28 m/s. I did it a little differently though.

    I found the ammount of time it would take the ball to fall from its maximum height of 2 meters, then multiplied by two to get its total flight time. You can do this because the ball's velocity curve must be symmetric about the maximum point due to conservation of energy. I then found the velocity relative to the man/ball system which would see the ball as having a horizontal distance traveled as 1m for the flight time then added and subtracted that number to the man's velocity relative to the ground.

    Could you explain what you're doing in the first few lines of your solution? I don't know what vbally is, is that the velocity of the ball when it reaches/leaves the man's hand?
  4. Feb 24, 2006 #3
    vball y is the vertical velocity of the ball as it leaves the man's hand.
    vball x is the actual horizontal velocity of the ball.
    First I used the formula v^2 = v0^2 + 2ad to find the inital vertical velocity...
    Then, I found total flight time: v = v0 + at, but I substituted v0 from what I got in the previous equation, 0 for v, -9.8 for a and (t/2) for t, thus getting:
    0 = v0 - 4.9t
    And solving for t...
    Then finding the ACTUAL horizontal range of the ball by using x = v ball x * t (Since t is the total flight time)
    And then I used the proportions to get the relative velocity and... yeah.
    Maybe you can point out where I made a mistake? :)
  5. Feb 24, 2006 #4

    Doc Al

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    Staff: Mentor

    Your answer and method look good to me.
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