Answer to Question #240378 in Physics for Morgan Hopkin

Question

An airplane is traveling at a ground velocity of 190 km/h [N 35° E]. The velocity of the wind relative to the ground is 41 km/h [S]. What is the velocity of the airplane relative to the air?

Answer 1

From the right triangle in the figure:

$$\theta = 90\degree +35\degree = 125\degree$$

According to the classical velocity-addition formula (see https://en.wikipedia.org/wiki/Velocity-addition_formula), the velocity of the airplane relative to the air is given as follows:

$$\mathbf{v} = \mathbf{v_{ne}-v_s}$$

Using the cosine law, find the magnitude of $\mathbf{v}$:

$$v = \sqrt{v_{ne}^2 + v_s^2-2v_{ne}v_s\cos\theta}\\ v = \sqrt{190^2 + 41^2-2\cdot 190\cdot41 \cos125\degree} \approx 216km/h$$

The direction with N axis can be found using sine law:

$$\dfrac{v}{\sin\theta} = \dfrac{v_{ne}}{\sin\alpha}\\ \alpha = \arcsin\left(\dfrac{v_{ne}\sin\theta}{v} \right) = \arcsin\left(\dfrac{190\cdot \sin125\degree}{216} \right)\approx 46\degree$$

Answer. 216 km/h [E 46° N].