Dropped Particle with Rebound Problem(a)If we neglect the very small “centrifugal force” term in GPS Eq.(4.91),the equation of motion for a particle of massmin the rotating frame of theearth ismdvdt=mg+ 2mv×ω,(1)wherevis the particle velocity,ωis the earth’s angular velocity vector, andgis the acceleration of gravity at the earth’s surface,g=−gez.All the vectorsand derivatives are measured in a bodyfixed frame in which thezaxis is normalto the earth’s surface and thex-yplane is tangent to the earth’s surface and islocated atz= 0. Assume that we are in the Northern hemisphere, and let theyaxis point north and thexaxis point east. As we saw in class, the solutionsto Eq.(1) forvandrtofirst order inωarev=v0+gt+ 2(v0×ω)t+ (g×ω)t2,(2)r=r0+v0t+12gt2+ (v0×ω)t2+13(g×ω)t3,(3)where the subscript0denotes an initial value.Falling particle:This problem is solved in the textbook and in the classnotes, and the answer is that the defl
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