Question

A commercial passenger jet is flying with an airspeed of 185 miles per hour on a heading of 036°. If a 47-mile-per-hour wind is blowing from a true heading of 120°, determine the velocity and direction of the jet relative to the ground.

Answer 1

Answer:

195.5 mph, 49.85 degree

Explanation:

velocity of jet with respect to wind = 185 mph at 36 degree

V (j, w) = 185 (Cos 36 i + Sin 36 j) = 149.67 i + 108.74 j

velocity of wind with respect to ground = 47 mph at 120 degree

V(w,g) = 47 (Cos 120 i + Sin 120 j) = - 23.5 i + 40.70 j

Velocity of jet with respect to wind = velocity of jet with respect to ground - velocity of wind with respect to ground

V (j, w) = V (j, g) - V (w, g)

149.67 i + 108.74 j = V(j,g) + 23.6 i - 40.7 j

V(j,g) = 126.07 i + 149.44 j

Magnitude of velocity of jet with respect to wind

= $\sqrt{126.07^{2}+149.44^{2}}$ = 195.5 mph

Let it makes angle θ from + X axis

tan θ = 149.44 / 126.07 = 1.1185

θ = 49.85 degree

Answer 2

Answer:

Either 186.1 mph, 069° or 188.5 mph, 021°

Most likely 186.1 mph, 069°

Explanation:

I wanted to comment this but there's no way for me to comment :/

Both 186.1 mph, 021° and 195.6 mph, 069° are wrong.

Most likely 186.1 mph, 069° because 186.1 mph is what I got from my calculations.

Attached was my method. It's wrong tho so