Question

An airplane of mass 10,000 kg is flying over level ground at a constant altitude of 4.30 km with a constant velocity of 165 m/s west. Its direction is currently straight toward the top of a mountain. At a particular instant, a house is directly below the plane at ground level. (a) At this instant, what is the magnitude of the plane's angular momentum relative to the house (in kg m-/s)? kg. m/s (b) As the plane continues its constant velocity motion, how does the angular momentum vector relative to the house change? The magnitude of the angular momentum vector stays the same, but the direction changes. Both the magnitude and direction of the angular momentum vector change. Both the magnitude and direction of the angular momentum vector stay the same. The direction of the angular momentum vector stays the same, but the magnitude changes. (c) At this same instant, what is the magnitude of the plane's angular momentum relative to the mountaintop (in kg. m/s)? kg .m2/5

Answer 1

165 m top of the mountain 4.3km - 4300m level ground a) Angular momentum of plane=8x mv =4300 X 10.000 x 165 og /s = 7.095 x 10 regruß As the perpendicular distance between the relorety vector and Cdirection) house stays the same and also plane velocity doesn't change (6) y "both direction and magnitude staye same." (c) Angular momentum relative to the mountains top = 0[:8=0] as dion of relocity vector passes through moustain top.