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Falling EastwardProblem: Solution: and is directed eastward. The term v is the velocity of the falling test object as a function of time. If the eastward deflection is assumed sufficiently small compared to h, we may approximate v and h as Substituting 2. into 1. gives for the Coriolis acceleration
Integrating once give the eastward velocity as a function of time
Integrating a second time gives the required deflection
Using 3., t3 = (2h/g)3/2, so that for the entire fall from height h,
Let us
now assume that the initial height h = 100 m, and l = 42 deg. With t =
4.52 sec
v = 44.3 m/sec vE = 1.08 x 10-4 m/sec xE = 1.63 x 10-4 m = .16 mm Comparing vE and v, we also see that the deviation of the velocity from the vertical is arctan
(1.08 x 10-4/44.3) = 1.4 x 10-4 deg.
With h = 1000 m, l = 42 deg, we find t =
14.3 sec
v = 140 m/sec vE = 1.08 x 10-3 m/sec xE = 5.16 x 10-3 m = 5.16 mm with arctan (1.08 x 10-3/140) = 4.4 x 10-4 deg.
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