Wednesday, 25 March 2015

A jet takes off at an air velocity of 450 km/h with a direction of [E 12° N]. A 15-km/h wind is blowing from the east. Determine the resultant...

Hello!


The resultant ground velocity of a jet is a vector sum of the jet's air velocity and wind ground velocity. To find its magnitude and direction, we consider the projections on the N-S and W-E axes.


Please look at the attached diagram. The jet's air velocity is in green, the wind velocity is in red and the resultant velocity is in blue. The projections of the air velocity are


`450*cos(12^@) approx440.17 (km)/h`  and  `450*sin(12^@)...

Hello!


The resultant ground velocity of a jet is a vector sum of the jet's air velocity and wind ground velocity. To find its magnitude and direction, we consider the projections on the N-S and W-E axes.


Please look at the attached diagram. The jet's air velocity is in green, the wind velocity is in red and the resultant velocity is in blue. The projections of the air velocity are


`450*cos(12^@) approx440.17 (km)/h`  and  `450*sin(12^@) approx93.56 (km)/h.`


The projections of the wind velocity is more obvious, `-15 (km)/h` and `0.`


Hence the resultant velocity has the projections about `425.17 (km)/h` and `93.56 (km)/h.` The magnitude is equal to


`sqrt(425.17^2 +93.56^2) approx435.34 (km)/h.`


The direction is `tan^(-1)(93.56/425.17) approx 12.41^@.`


The answer: the resultant ground velocity is about `435.34 (km)/h` in the direction of about E `12.41^@` N.


If the wind velocity is actually `15 m/s,` than the idea is absolutely the same, but with the different value `(15 m/s =54 (km)/h).`


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