A small airplane is flying due north at $150\,\rm km/h$ when it encounters a wind of $80\,\rm km/h$ from the east. A small airplane is flying due north at $150\,\rm km/h$ when it encounters a wind of $80\,\rm km/h$ from the east.  what is the resultant ground velocity of the airplane?
 A: To solve this problem, you have to do a vector addition. Let $\vec v=150\,\rm km/h\,\,\text{(north)}$ and let $\vec u=80\,\rm km/h\,\,\text{(east)}$. If we project those vectors into a Cartesian coordinate system, and assuming that the unit vectors $\hat i=(1,0)$ and $\hat v=(0,1)$, we get that the the vector $\color{red}{\vec v}+\color{green}{\vec u}=\color{blue}{\vec w}$ is: $$\color{blue}{\vec w}=\color{red}{\begin{pmatrix}0\\150 \end{pmatrix}}+\color{green}{\begin{pmatrix}-80\\0 \end{pmatrix}}=\color{blue}{\pmatrix{-80\\150}}.$$
So, by the distance formula: $$\|\color{blue}{\vec w\,}\|=\sqrt{(-80)^2+(150)^2}=\sqrt{\color{white}{\overline{\color{black}{6400+22500}}}}=\sqrt{\color{white}{\overline{\color{black}{28900}}}}=170.$$
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$\color{white}{\text{this post assumes that $\gamma=0$ and so ignores the effects of special relativity and uses only the traditional velocity addition formula.}}$
A: If the plane's velocity relatively to the air remains constant, the resultant velocity can be found by vector addition.
Suppose $(1,0)$ means one km/h to the east and $(0,1)$ one km/h to the north then simply 
$$\vec v=(0,150)+(80,0)=(80,150).$$
Its speed can then be calculated by $$v=|\vec v| =\sqrt{80^2+150^2}=170.$$
A: If you draw the velocities in, you get a right-angled triangle as follows.
Now solve for $x$ (use Pythagoras' theorem).

A: Since the direction of the movement of plane and wind is perpendicular, then you can use Pythagoras theorem to calculate the resultant velocity.
$$
v_R=\sqrt{v_p^2+v_w^2}=\sqrt{150^2+80^2}=170\text{ km/h}.
$$
In general, the resultant velocity can be evaluated using formula
$$
v_R=\sqrt{v_1^2+v_2^2+2v_1v_2\cos\theta},
$$
where $\theta$ is the smallest angle between $v_1$ and $v_2$.
