EK physics question help

[inaccensa]

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I'm confused by this question-A ballon travels 6km upwards while the wind blows its 10km north and 8km east. What is the final displacement from its initial position.

To be honest, my mind was wrapped around the idea that 6km north+10 km north and 8km east. The term upwards for a ballon should be north as per my stupid deductions. However when I saw the explanation for it, the diagram was a little off. I'm wondering if any one has any ideas about this question and can explain it correctly. Thanks

 

[ilovemcat]

I'm confused by this question-A ballon travels 6km upwards while the wind blows its 10km north and 8km east. What is the final displacement from its initial position.

To be honest, my mind was wrapped around the idea that 6km north+10 km north and 8km east. The term upwards for a ballon should be north as per my stupid deductions. However when I saw the explanation for it, the diagram was a little off. I'm wondering if any one has any ideas about this question and can explain it correctly. Thanks

Don't worry, that question got me too! 🙂 This a 3 dimensional problem so you have to keep track of displacements in each axis (x,y, and z).

Let's call UP the "Z" axis. Let's call North the "Y" axis and Let's call East the "X" axis. The displacement between X and Y is: 10km North and 8km east. Because displacement is a vector, we have to use the Pythagorean Theorem to find the net displacement between "x" and "y":

sqrt(164km^2) = 12km North East. This is the displacement between x and y. Now we need to take into consideration the z-axis, which you approach in a similar fashion as done previously:

sqrt(36km^2 + 164km^2) = sqrt(200km^2)
net displacement = ~14km NE & Upwards

 

[bbottle]

To visualize what ilovemcat did, you can draw two right triangles. The hypotenuse of the first triangle in the xy plane becomes one of the sides for the second triangle (z is the other side).

 

[inaccensa]

You explained it really well. thanks a ton. I think conjuring up a 3-D diagram for this problem is key step and also realizing that the resultant of the former two is perpendicular to the z axis,although I'm still struggling with the second part.

 

[ilovemcat]

You explained it really well. thanks a ton. I think conjuring up a 3-D diagram for this problem is key step and also realizing that the resultant of the former two is perpendicular to the z axis,although I'm still struggling with the second part.

If conceptualizing it seems difficult, you could always use XY coordinates again. For instance, we can take the magnitude of the net displacement for the North East direction (which we found to be 12km NE) and label that as the x-axis (Convention for positive x-axis direction is usually east, but here we specified it as NE). For y-axis, label the magnitude of the UP direction (Convention for positive y-axis direction is usually north, but here we specified it as UP). Then just do the same exact thing you did earlier. This way works too.