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

AI Thread Summary
The discussion revolves around a physics problem involving the relative velocities of a falling package as observed by an immobile person and a pilot flying horizontally. The pilot perceives the package as falling vertically, suggesting that his horizontal speed matches the package's horizontal motion. The participant initially considers vector addition to solve for the pilot's speed relative to the ground but later realizes it can be approached as a right triangle problem using the Pythagorean theorem. The correct relationship is established as V3 = sqrt(V1^2 - V2^2, where V3 is the pilot's speed, V1 is the hypotenuse, and V2 is the vertical speed of the package. The conclusion confirms that the participant's understanding improved through the discussion, leading to the correct application of physics principles.
exi
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Homework Statement



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?

Homework Equations



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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?
 
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exi said:
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.
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?
 
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?
 
Now you've got it.
 
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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