Answer:
The resultant velocity is 360.5 m/s and direction 79° north of east.
Explanation:
Given that,
Velocity of airplane = 300 m/s
Velocity of wind = 100 m/s
Angle θ₁ = 25°
Angle θ₂ =35°
The horizontal velocity component
Using formula of velocity
![v_{x}=v_{1}\cos\theta-v_{2}\cos\theta](https://tex.z-dn.net/?f=v_%7Bx%7D%3Dv_%7B1%7D%5Ccos%5Ctheta-v_%7B2%7D%5Ccos%5Ctheta)
Put the value into the formula
![v_{x}=300\cos65-100\cos55](https://tex.z-dn.net/?f=v_%7Bx%7D%3D300%5Ccos65-100%5Ccos55)
![v_{x}=69.42\ m/s](https://tex.z-dn.net/?f=v_%7Bx%7D%3D69.42%5C%20m%2Fs)
The vertical velocity component
Using formula of velocity
![v_{y}=v_{1}\sin\theta+v_{2}\sin\theta](https://tex.z-dn.net/?f=v_%7By%7D%3Dv_%7B1%7D%5Csin%5Ctheta%2Bv_%7B2%7D%5Csin%5Ctheta)
Put the value into the formula
![v_{y}=300\sin65+100\sin55](https://tex.z-dn.net/?f=v_%7By%7D%3D300%5Csin65%2B100%5Csin55)
![v_{y}=353.8\ m/s](https://tex.z-dn.net/?f=v_%7By%7D%3D353.8%5C%20m%2Fs)
We need to calculate the resultant velocity
Using formula of resultant velocity
![v=\sqrt{v_{x}^2+v_{y}^2}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7Bv_%7Bx%7D%5E2%2Bv_%7By%7D%5E2%7D)
Put the value into the formula
![v=\sqrt{69.42^2+353.8^2}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B69.42%5E2%2B353.8%5E2%7D)
![v=360.5\ m/s](https://tex.z-dn.net/?f=v%3D360.5%5C%20m%2Fs)
We need to calculate the direction of the resultant velocity
Using formula of direction
![\tan\theta=\dfrac{v_{y}}{v_{x}}](https://tex.z-dn.net/?f=%5Ctan%5Ctheta%3D%5Cdfrac%7Bv_%7By%7D%7D%7Bv_%7Bx%7D%7D)
Put the value into the formula
![\theta=\tan^{-1}(\dfrac{353.8}{69.42})](https://tex.z-dn.net/?f=%5Ctheta%3D%5Ctan%5E%7B-1%7D%28%5Cdfrac%7B353.8%7D%7B69.42%7D%29)
![\theta=79^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D79%5E%7B%5Ccirc%7D)
Hence, The resultant velocity is 360.5 m/s and direction 79° north of east.