# Thread: Need help with plane type of Questions

1. ## Need help with plane type of Questions

Hello, these two questions are from my Calc and Vectors book
1)A small airplane takes off at an airspeed of 180km/hr, at an angle of inclination of 14degrees, toward the east. A 15km/hr wind is blowing from the southwest. Determine the resultant ground velocity

2)An airplane takes off at a ground velocity of 200km/hr toward the east, climbing at an angle of 14 degrees. A 20km/hr wind is blowing from the north. Determine the resultant air and ground velocities and their magnnitudes.

2. Originally Posted by supersaiyan
Hello, these two questions are from my Calc and Vectors book
1)A small airplane takes off at an airspeed of 180km/hr, at an angle of inclination of 14degrees, toward the east. A 15km/hr wind is blowing from the southwest. Determine the resultant ground velocity
air vector parallel to the ground ... $\displaystyle 180\cos(14)$

east component ... $\displaystyle v_x = 180\cos(14) + 15\cos(45)$

north component ... $\displaystyle v_y = 15\sin(45)$

resultant ground speed = $\displaystyle \sqrt{v_x^2 + v_y^2}$

direction relative to east ... $\displaystyle \theta = \arctan\left(\frac{v_y}{v_x}\right)$

3. Originally Posted by skeeter
air vector parallel to the ground ... $\displaystyle 180\cos(14)$

east component ... $\displaystyle v_x = 180\cos(14) + 15\cos(45)$

north component ... $\displaystyle v_y = 15\sin(45)$

resultant ground speed = $\displaystyle \sqrt{v_x^2 + v_y^2}$

direction relative to east ... $\displaystyle \theta = \arctan\left(\frac{v_y}{v_x}\right)$
Thanks, but why didn't we also use $\displaystyle v_y = 15\sin(45) + 180\sin(14)$
Why did we only use $\displaystyle 15\sin(45)$

4. Originally Posted by supersaiyan
Thanks, but why didn't we also use $\displaystyle v_y = 15\sin(45) + 180\sin(14)$
Why did we only use $\displaystyle 15\sin(45)$
there is no component of the air vector in the y-direction.