91Ó°ÊÓ

The net and y- component of velocity, considering the east direction as x-component,

$$\begin{gathered} V\left(or{V}_{est}\right)=70.0\mathrm{km}/\mathrm{h} \\ {\mathrm{V}}_{\mathrm{y}}\left(\mathrm{or}{\mathrm{V}}_{\mathrm{south}}\right)=20.0\mathrm{km}/\mathrm{h} \end{gathered}$$

The velocity is a vector quantity. The vector has magnitude as well as direction. We can add the vectors using the law of vector addition. The magnitude of the resultant vector is found by adding the square of the components of the resultant and finding the square root.

We are given two perpendicular velocities; hence, we can find the resultant velocity of aircraft by using the resultant velocity equation.

Formula:

The resultant velocity of a body in motion,$V=\sqrt{V_x^2+V_y^2}$ (i)

Here,v is resultant velocity,$V_x$ is x component of the velocity,$V_y$ is y component of the velocity.

Using equation (i), we get the resultant velocity.

$$\begin{gathered} V=\sqrt{V_x^2+V_y^2} \\ {V}^2=V_x^2+V_y^2 \\ {V}_x^2=V^2-V_y^2 \\ V=\sqrt{V^2-V_y^2} \end{gathered}$$

Substitute the value of resultant velocity and y component of the velocity in the above equation.

$$\begin{gathered} V_x=\sqrt{{\left(70\mathrm{km}/\mathrm{h}\right)}^2-{\left(20.0\mathrm{km}/\mathrm{h}\right)}^2} \\ =67.08\mathrm{km}/\mathrm{h} \end{gathered}$$

The speed of aircraft relative to the ground is 67.08 km/h .