You are in a hot-air balloon that, relative to the ground, has a velocity of 9.5 m/s in a direction due east. You see a hawk moving directly away from the balloon in a direction due north. The speed of the hawk relative to you is 2.9 m/s. What are the magnitude and direction of the hawk's velocity relative to the ground? Express the directional angle relative to due east.
2007-02-05 14:03:57 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Be nice to shamalamadingdong
This is a simple vector problem
x component = 9.5m/s E
y component = 2.9 m/s N
V=sqrt(9.5^2 + 2.9^2)=9.93m/s
angle = arcTan(2.9/9.5)=16.98 degrees
That is all folks.
2007-02-05 14:37:42 · answer #1 · answered by Edward 7 · 0⤊ 0⤋
You know, any time I answer a 'homework' question, the asker never seems to pick ANY one as a 'best answer'. So, I'll say that the answer is:
The magnitude is 3.2 x 4 mph and the direction of the velocity is: 43 degrees, due east.
2007-02-05 22:15:14 · answer #2 · answered by ANSWER MY QUESTION!! 6 · 0⤊ 0⤋