Answer:
The resultant velocity of the helicopter is [tex]\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)[/tex].
Explanation:
Physically speaking, the resulting velocity of the helicopter ([tex]\vec v_{H}[/tex]), measured in meters per second, is equal to the absolute velocity of the wind ([tex]\vec v_{W}[/tex]), measured in meters per second, plus the velocity of the helicopter relative to wind ([tex]\vec v_{H/W}[/tex]), also call velocity at still air, measured in meters per second. That is:
[tex]\vec v_{H} = \vec v_{W}+\vec v_{H/W}[/tex] (1)
In addition, vectors in rectangular form are defined by the following expression:
[tex]\vec v = \|\vec v\| \cdot (\cos \alpha, \sin \alpha)[/tex] (2)
Where:
[tex]\|\vec v\|[/tex] - Magnitude, measured in meters per second.
[tex]\alpha[/tex] - Direction angle, measured in sexagesimal degrees.
Then, (1) is expanded by applying (2):
[tex]\vec v_{H} = \|\vec v_{W}\| \cdot (\cos \alpha_{W},\sin \alpha_{W}) +\|\vec v_{H/W}\| \cdot (\cos \alpha_{H/W},\sin \alpha_{H/W})[/tex] (3)
[tex]\vec v_{H} = \left(\|\vec v_{W}\|\cdot \cos \alpha_{W}+\|\vec v_{H/W}\|\cdot \cos \alpha_{H/W}, \|\vec v_{W}\|\cdot \sin \alpha_{W}+\|\vec v_{H/W}\|\cdot \sin \alpha_{H/W} \right)[/tex]
If we know that [tex]\|\vec v_{W}\| = 25\,\frac{m}{s}[/tex], [tex]\|\vec v_{H/W}\| = 125\,\frac{m}{s}[/tex], [tex]\alpha_{W} = 240^{\circ}[/tex] and [tex]\alpha_{H/W} = 325^{\circ}[/tex], then the resulting velocity of the helicopter is:
[tex]\vec v_{H} = \left(\left(25\,\frac{m}{s} \right)\cdot \cos 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \cos 325^{\circ}, \left(25\,\frac{m}{s} \right)\cdot \sin 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \sin 325^{\circ}\right)[/tex][tex]\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)[/tex]
The resultant velocity of the helicopter is [tex]\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)[/tex].
