# Need help with plane type of Questions

• Apr 27th 2009, 11:58 AM
supersaiyan
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.
• Apr 27th 2009, 04:28 PM
skeeter
Quote:

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 ... $180\cos(14)$

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

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

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

direction relative to east ... $\theta = \arctan\left(\frac{v_y}{v_x}\right)$
• Apr 27th 2009, 05:02 PM
supersaiyan
Quote:

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

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

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

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

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

Thanks, but why didn't we also use $v_y = 15\sin(45) + 180\sin(14)$
Why did we only use $15\sin(45)$
• Apr 27th 2009, 06:17 PM
skeeter
Quote:

Originally Posted by supersaiyan
Thanks, but why didn't we also use $v_y = 15\sin(45) + 180\sin(14)$
Why did we only use $15\sin(45)$

there is no component of the air vector in the y-direction.