Vectors and Angles in Helicopter Flight

  • Thread starter Thread starter gric122
  • Start date Start date
  • Tags Tags
    Angles Vectors
Click For Summary

Homework Help Overview

The problem involves a helicopter's flight path described in terms of distances and headings, requiring the calculation of a displacement vector to return to the launch pad. The subject area includes vectors and angles in a two-dimensional plane.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss drawing the problem and using trigonometric functions to resolve the distances into components. There are attempts to calculate horizontal and vertical distances, with some participants questioning the accuracy of their results and the interpretation of angles.

Discussion Status

The discussion is ongoing, with participants sharing their calculations and questioning each other's methods. Some guidance has been offered regarding the use of trigonometric functions, but there is no explicit consensus on the correct approach or results yet.

Contextual Notes

Participants are navigating potential confusion regarding angle measurements and the direction of their calculations. There is mention of using degrees and the need to clarify the setup of the problem.

gric122

Homework Statement



Bob has a helicopter and from the launch pad he flies the following path. First he travels from the launch pad a distance of 17 kilometers at heading of 63 degrees East of South. Then he flies 45 kilometers heading 73 degrees West of North. After this he flies 34 kilometers heading 79 degrees West of North. Now he is ready to return to the launch pad.

What is the displacement vector that he needs to take to return directly to the launch pad from his present location (for the heading give the number degrees north of east - your answer may be greater than 90 degrees)?

Homework Equations


none

The Attempt at a Solution


So I've attempted to draw this problem out and then solve it with trig, but I'm stuck on where the angles point to. From what I've drawn, it looks like a slanted parallelogram, and I don't think I can trig it
 
Physics news on Phys.org
I would just work out the displacement in the north/south and east/west direction for all the given distances, and then add those all up.
The final leg of the jouney will be in the opposite direction of the result.
 
I calculated the horizontal distance and got -61.26 km, but for some reason my vertical is 8.66 which seems really low, am I doing somethign wrong? For x I did 17cos(333) + 45cos(163) + 34cos(169), and for y I just replaced cosine with sine
 
gric122 said:
17cos(333) + 45cos(163) + 34cos(169), and for y I just replaced cosine with sine
When I do that I get 11.9.
 
haruspex said:
When I do that I get 11.9.
How did you get that? Are u talking about the x or y value? I keep typing it in the calculator and x doesn't come out to be 11.9. I'm using degrees
 
gric122 said:
How did you get that? Are u talking about the x or y value? I keep typing it in the calculator and x doesn't come out to be 11.9. I'm using degrees
I'm referring to the y value, obtained as you posted:
gric122 said:
17cos(333) + 45cos(163) + 34cos(169), and for y I just replaced cosine with sine
I.e. just replace all of those cos functions with sine functions. And, yes, angles in degrees.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 19 ·
Replies
19
Views
8K
Replies
6
Views
2K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K