They tell you that the wind is blowing at 42 km/h so we need to convert that to m/s: (42km/h)(1000m/1km)(1h/3600s) = 11.67 m/s. Now find the horizontal and vertical components of the wind speed; use SOHCAHTOA: vx = 11.67cos(20) = 10.97 m/s. vy = 11.67sin(20) = 3.99 m/s.
XXX XXXX XXXXXXXX did the XXXXX travel XXX to just the XXXX: d=XX. XXXXX, break it into XXXXXXXXXX: dx=vxt XXX dy=vyt. (XX.97m/s)(900s) = 9873m and (X.99m/s)(900s) = 3591m.
XXXXX having moved XXXXX in the horizontal direction and XXXXX in the XXXXXXXX direction from XXX starting XXXXX, how do you XXXX XX move to be 55000m due North of XXX XXXXXXXX XXXXX? Well we need XX XXXX an addditional 51409m north (55000m-3591m) and XXXXX XXXX. XX XXXXXXX know the XXXX, so we just need to XXX the equation v=x/t to XXXXX XXX the component speeds, XXXX combine XXXX into one XXXXXX XXXXX.
(XXXXX)/(XXXX) = 10.91 m/s and (51409m)/(XXXX) = XX.XX m/s
Finally, XXXXXXX them the the XXXXXXXX c=SQRT((A^2)+(B^X)), where 10.XX and 57.12 XXX X and X respectively. This is just XXX Pythagorean Theorem XXXXXXXXXX, btw.
SQRT((10.XX^2)+(XX.XX^2)) = XX.XX m/s
XXXXX If you wanted direction of XXX plane too, XXX XXXXXXXXX just like XXXXXXX in XXX problem. tanθ = 57.XX/XX.91. The XXXXX is moving ~XX.XX degrees west of north.
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