Given data:
The speed of aircraft is u=630 mi/h.
The speed of wind is v=16 mi/h.
The angle of aircraft with north is θ=118⁰.
The wind is out of the north, it means the wind is blowing from the south.
The diagram of the given situation is shown below,
Considering north and east as positive direction, then west and south will be negative.
The velocity of airplane in terms of unit vector can be written as,
[tex]u=630\cos 118\degree(j)+630\sin 118\degree(-i)[/tex]
The velocity of wind in terms of unit vector will be,
[tex]v=16(-j)[/tex]
The actual velocity of airplane will be,
[tex]\begin{gathered} u+v=630\cos 118\degree(j)+630\sin 118\degree(-i)+16(-j) \\ u+v=-295.76(j)+556.25(-i)+16(-j) \\ u+v=295.76(-j)+556.25(-i)+16\mleft(-j\mright) \\ u+v=556.25(-i)+16(-j)+295.76\mleft(-j\mright) \\ u+v=556.25\mleft(-i\mright)+311.76\mleft(-j\mright) \end{gathered}[/tex]
The magnitude of velocity of actual velocity of airplane will be,
[tex]\begin{gathered} \lvert u+v\rvert=\sqrt[]{(556.25)^2+(311.76)^2} \\ \lvert u+v\rvert=637\text{ mi/h} \end{gathered}[/tex]
The direction of actual velocity will be,
[tex]\begin{gathered} \tan \alpha=\frac{311.76}{556.25} \\ \tan \alpha=0.56 \\ \alpha=29\degree \end{gathered}[/tex]
The actual velocity of plane on the diagram is shown below,
The direction from the north will be,
[tex]\begin{gathered} \alpha=90\degree+29\degree \\ \alpha=119\degree \end{gathered}[/tex]
Thus, the final velocity of airplane is 637 mile per hour and the direction from the north is 119⁰.
