An airplane flies between two points on the ground that are 500 km apart. the destination is directly north of the origination of the flight. the plane flies with an air speed of 120 m/s. if a constant wind blows at 10.0 m/s due west during the flight, what direction must the plane fly relative to the air to arrive at the destination?

 Plane must fly 4.78⁰ due north to east at a speed of 120 m/s to reach destination.

Explanation:

  Let the east point towards positive X-axis and north point towards positive Y-axis.

  Speed of plane = 120 m/s

  Wind speed  = 10.0 m/s due west  = -10 m/s

  So plane must have an horizontal speed of 10 m/s to cancel this wind speed

  Horizontal speed of plane = u cos θ = 120*cos θ = 10

  Angle θ =85.22⁰ from positive horizontal axis.

  Angle from North axis towards East required = 90 - 85.22 = 4.78⁰

 So plane must fly 4.78⁰ due north to east at a speed of 120 m/s to reach destination.

the components of a vector are the parts of the vectors that are to each other.

b perpendicular

snell's law

refractve index = sin incidence / sin refraction