A small airplane takes of at an airspeed of 180km/h, at an angle of inclination of 14 degrees toward the east. A 15 km/h wind is blowing from the southwest. Determine the resultant ground velocity.
i think i will b able to solve it
It's something to do with the unit vectors, but after I get them, I'm at a loss because I don't know what's next to do with them. The one with the initial air speed's unit vectors are: x=180scos14, and y=180sin14
But after that, I'm lost
so u wanna know the resultant ground velocity
i think wind velocity will b \[15i-15j\]
and the result of ur question will b \[180 \cos 14+15 \cos 45\]
m i right
do u wanna c details
Details please. Sorry for the late reply
ok
the ground component of the plane speed of 180 km/h is 180cos14
and this ground component is along east
next the component along east of wind speed (15) is 15 cos 45
then i just add those
And it's 45 degrees because of the fact that it's coming STRAIGHT from the SW right?
ya i think so
how would you get the angle?
i guessed
Alright. I'll try to get it in some way :P. Thanks for the help
thanks for medal
anyway if u wanna know just the resultant velocity of ur plane then the result will b different
How would you get the resultant plane velocity?
i wanna introduce unit vector here
i and j being x and y respectively
ya i,j,k will b along X,Y and Z axis
unit vector
And then, how do you get an absolute answer from those?
ok
i was wrong
actually the plane velocity will b 180 cos 14 i+180 sin 14 k
Alright, then you just add those and get the answer?
ya
n we can get absolute number too
now wind velocity will b 15 cos 45 i+15 sin 45j
so the resultant velocity of ur plane will b 180 cos14i+15cos45i+15sin45j+180sin14k
absolute is \[r=\sqrt{x ^{2}+y ^{2}+z ^{2}}\]
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