Two airplanes leave an airport at the same time. The velocity of the first airplane is
670 m/h at a heading of 23.6°. The velocity
of the second is 610 m/h at a heading of 183º.
How far apart are they after 2 h?
Answer in units of m.

2446.4 m between the two airplanes

Explanation:

The distance is function of the horizontal velocity,

To find the horizontal velocity, we need this relations:

sinθ = co/h = y velocity

cosθ = ca/h = x velocity

x2 = 670 * cos(23.6º) = 614 m/h * 2h = 1228 m

x1 = 610 * cos(183º) = -609.2 m/h * 2h = -1218.4 m

D2 - D1 = 1228 m - (-1218.4 m) = 2446.4 m between the two airplanes

f is prop to m1xm2 double m1 double force

it doesn't. it was imagined up in the middle ages.