An airplane pilot wishes to fly due west. A wind of 80.0 km/h (about 50.0mi/h) is blowing toward the south. (a) If the airspeed of the plane (its speed in still air) is 320.0 km/h (about 200.0 mi/h), in which direction should the pilot head? (b) What is the speed of the plane over the ground? Draw a vector diagram.

The given data can be listed below,

The speed or velocity of the object with respect to a stationary or traveling object or person is known as its relative velocity.

The vector diagram for velocity vectors is shown as below,

From diagram, the direction of plane is calculated as,

$\sin \theta =\frac{v_{AG}}{v_{PA}}$

Here, v_AG is the speed of air and v_PA is the speed of the plane.

Substitute all the values in the above,

$\begin{array}{rcl}\sin \theta & =& \frac{80km/h}{320km/h}\\ \theta & =& {\sin }^{-1}\left(\frac{1}{4}\right)\\ & =& {14}^o\end{array}$

Thus, the plane will head towards northwest direction at 14^o.

BY, Pythagoras theorem, the velocity of the plane relative to ground is given by,

$v_{PG}=\sqrt{{\left(v_{PA}\right)}^2-{\left(v_{AG}\right)}^2}$

Here, v_AG is the speed of air and v_PA is the speed of the plane.

Substitute all the values in the above,

role="math" localid="1659681921352" $\begin{array}{rcl}{v}_{PG}& =& \sqrt{{\left(320km/h\right)}^2-{\left(80km/h\right)}^2}\\ & =& \sqrt{\left(102400\right)-\left(6400\right)}km/h\\ & =& \sqrt{96000}km/h\\ & =& 310km/h\end{array}$

Thus, the velocity of plane relative to ground is 310km/h.