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Conceptual question on velocity (double-checking my own work)

  1. Jun 12, 2007 #1

    exi

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

    1: An immobile person sees a package falling to the ground. It appears to be falling (a) at an angle, and (b) at speed V1.

    2: A pilot flying horizontally at constant speed sees the same package. It appears to be (a) falling vertically, and (b) at speed V2.

    The question is: What's the speed of the pilot relative to the ground?

    2. Relevant equations

    --

    3. The attempt at a solution

    If the pilot is flying horizontally and sees the package falling straight down, it would suggest to me that his own horizontal velocity matches that of the package. I attempted to sketch this out, and it strongly resembles a vector addition problem where the magnitude would be equal to sqrt(x^2 + y^2), which leads me to believe that the correct answer would be sqrt(v1^2 + v2^2) (C, below).

    I'm not 100% on this, though, since I seem to be having a hard time visualizing this beyond that.

    This is a multiple-choice question:

    A. v1 + v2
    B. sqrt(v1^2 - v2^2)
    C. sqrt(v1^2 + v2^2)
    D. v2 - v1
    E. v1 - v2

    Am I going about this the proper way / am I right on this one, or am I a bit... off?
     
  2. jcsd
  3. Jun 12, 2007 #2

    Doc Al

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    Staff: Mentor

    Exactly.

    But I suspect your diagram is a bit off. It should resemble a right triangle. Which side equals V1? Which side equals V2? Which side is the speed of the plane?
     
  4. Jun 12, 2007 #3

    exi

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    I take my answer back. After looking at my little drawing again, it's just a simple Pythagorean question.

    If this is drawn out as a series of velocities, V2 would be one side (vertical to the ground), whereas V1 would be the hypotenuse, leaving the third side (parallel to the ground) equal to V3, which is what I'm after.

    So, if a^2 + b^2 = c^2, then:

    V2^2 + V3^2 = V1^2

    V3^2 = V1^2 - V2^2

    V3 = sqrt(V1^2 - V2^2)

    Am I thinking about this correctly?
     
  5. Jun 12, 2007 #4

    Doc Al

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    Staff: Mentor

    Now you've got it.
     
  6. Jun 12, 2007 #5

    exi

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    Thanks for the mental jogging. Having not seen geometry, trig/algebra, or physics since early 2003, I'm a little slow on some of this these days. :redface:
     
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