- #1
mcgooskie
- 4
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In this problem I need to figure out how far I am from the airport.
The pilot tells us that we have to circle the aiport before we can land. We will maintain a speed of 450mph at an altitude of 20,000ft. while traveling in a horizontal circle around the airport. I notice that the pilot banks the plane so that the wings are oriented at 10deg to the horizontal. An article in the in-flight magazine says that planes can fly because the air exerts a force "lift" on the wings which is perpendicular to the wing surface.
So:
V=660 ft/s
h=20,000 ft
bank = 10deg to the horizontal
So I drew a FBD with my normal force perpendicular to my wings, which were banked to the horizontal.
Sum Fx=Wx=ma where a=v^2/r
sin10 (mg)=mv^2/r (mass drops out)
sin10 (32.2ft/s^2)=(660ft/s)/r
solving for r=118.03ft
add that to the current altitude of 20,000ft gives a distance of 20,118ft from the airport?
It doesn't seem like this problem could be this "easy". I feel like I am totally missing a point here.
Thanks!
The pilot tells us that we have to circle the aiport before we can land. We will maintain a speed of 450mph at an altitude of 20,000ft. while traveling in a horizontal circle around the airport. I notice that the pilot banks the plane so that the wings are oriented at 10deg to the horizontal. An article in the in-flight magazine says that planes can fly because the air exerts a force "lift" on the wings which is perpendicular to the wing surface.
So:
V=660 ft/s
h=20,000 ft
bank = 10deg to the horizontal
So I drew a FBD with my normal force perpendicular to my wings, which were banked to the horizontal.
Sum Fx=Wx=ma where a=v^2/r
sin10 (mg)=mv^2/r (mass drops out)
sin10 (32.2ft/s^2)=(660ft/s)/r
solving for r=118.03ft
add that to the current altitude of 20,000ft gives a distance of 20,118ft from the airport?
It doesn't seem like this problem could be this "easy". I feel like I am totally missing a point here.
Thanks!