91Ó°ÊÓ

Velocity is the rate of change in position of an item in motion as seen from a specific frame of reference and measured by a specific time standard.

The velocity of the plane relative to the earth will be considered resultant.

This problem is the relative velocity problem; we can use the relative velocity formula.

Hence the velocity can be calculated by the following equation.

$V_{pe}=V_{pa}+V_{ae}$

The component table will be as below:

Here we need to find the x component and the y component.

$$\begin{gathered} Xcomponent \\ {V}_{pax}=V_i\cos \mathrm{θ} \\ {V}_{pax}=\left(-70\right)\cos \mathrm{θ} \\ Ycomponent \\ {V}_{pay}=V_i\sin \mathrm{θ} \\ {V}_{pay}=\left(-70\right)\sin \mathrm{θ} \end{gathered}$$

Hence, solving the resultant equation to get the value of theta:

$\begin{array}{rcl}\left(-70\right)\cos \mathrm{θ}+10& =& 0\\ \left(70\right)\cos \mathrm{θ}& =& 10\\ \cos \mathrm{θ}& =& \frac{10}{70}\\ \mathrm{θ}& =& 81.8^{°}.\end{array}$

The angle of the direction will be$81.8^{°}$ .

Hence the resultant vector will be:

$\begin{array}{rcl}\stackrel{â‡\P Ä}{R}& =& \sqrt{{\left(\underset{}{∑}X\right)}^2+{\left(\underset{}{∑}Y\right)}^2}\\ \stackrel{â‡\P Ä}{R}& =& \sqrt{{\left(0\right)}^2+{\left(-70\sin 81.8\right)}^2}\\ \stackrel{â‡\P Ä}{R}\left(V_{pe}\right)& =& 63.3\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\end{array}$

The velocity of the plane relative to the earth is $63.3\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.