What is the Required Heading of the Plane in a Crosswind?

[kalupahana]

Relative Motion prob, please help

Homework Statement

A pilot of an plane has to get to a point 320 km due North in 1 hour. Ground Control reports inform that there is a cross wind of 80 km/h towards 37^o South of West. What is the required heading pf the plane?

Homework Equations

V_PE = V_PW + V_WE

The Attempt at a Solution

Horizontal component

V_PE = V_PW + V_WE

0 = xcosᵝ - 80cos37
xcosᵝ = 80 cos37 --------------(1)

Vertical component

V_PE = V_PW + V_WE

320 = xsinᵝ - 80 sin37
xsinᵝ = 80(4 +sin37) -------------(2)

Angle (heading angle)

(2)/(1)

tanᵝ = (4+sin37)/cos37 = (4 + 0.601)/0.7986 = 5.761

ᵝ = tan^-1(80.15)


Please tell me is this right

 

[anathema]

?

Your approach seems correct, but there are a few errors in your calculations. First, in your horizontal component equation, you have a negative sign in front of the 80cos37. This should be positive since the crosswind is towards the South of West. Also, in your vertical component equation, you have a typo where you wrote 4 + sin37 instead of 80 + sin37. Finally, when you take the inverse tangent to find the heading angle, make sure you are using the correct units (radians or degrees) and that you are taking into account the quadrant in which the angle lies. Overall, your approach is correct but make sure to double check your calculations.