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Aggiebound13
You are in a hot-air balloon that, relative to the ground, has a velocity of 8.0 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 2.0 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.
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|dw:1317129261354:dw| as the awk seems to be moving due north with respect to the ballon it is stationary in the east direction with respect to the ballon and hence have a velocity 8m/s due east so magnitude of hawks velocity is the vector sum of its velocities in east and north dir = v= sqrt(8^2+2^2) v=sqrt(68) now the directional angle of velocity with east is tan^-1(1/4)