Asked by Anonymous
Due to a strong wind from the west of 65 km/h, a plane ends up traveling 330 km/h [W 15 S] relative to the ground. What was the magnitude of the velocity of the plane relative to the air in km/h?
Answers
Answered by
Damon
330 cos 15 = Vwest - 65
330 sin 15 = Vsouth
so
Vwest = 383.7 km/h
Vsouth = 85.4 km/hr
Vplane = sqrt (383.7^2+85.4*2)
tan angle south of west = 85.4/383.7
330 sin 15 = Vsouth
so
Vwest = 383.7 km/h
Vsouth = 85.4 km/hr
Vplane = sqrt (383.7^2+85.4*2)
tan angle south of west = 85.4/383.7
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.