A model airplane is flying north with a velocity of of 15m/s. A strong wind blowing east at 12m/s What is the airplanes resultant speed (magnitude of vector)

What is the airplanes heading (direction of velocity vector

1.) 19.21 m/s

2.) 57 degree

Explanation: Given that a model airplane is flying north with a velocity of of 15m/s. A strong wind blowing east at 12m/s

The airplanes resultant speed can be calculated by using pythagorean theorem

R = sqrt ( 15^2 + 12^2 )

R = sqrt( 225 + 144 )

R = sqrt( 369 )

R = 19.21 m/s

The magnitude of the resultant vector is 19.21 m/s

The direction of velocity vector will be:

Tan Ø = 15 /12

Tan Ø = 1.25

Ø = tan^-1(1.25)

Ø = 57.04

Ø = 57 degree.

Therefore, the airplanes is heading 57 degree in the horizontal direction of velocity vector

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