How Does Earth's Rotation Affect a Vertically Projected Particle?

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SUMMARY

The discussion focuses on the effects of Earth's rotation on a vertically projected particle at a northern latitude θ. The key conclusion is that a particle projected vertically upward to a height h will strike the ground at a point given by the formula 4/3*ωcosθ*sqrt(8h^3/g) to the west, where ω represents the angular velocity of Earth and g is the acceleration due to gravity. The participants analyzed the equations of motion, including the Coriolis force, and clarified initial conditions for the problem. The final solution was reached by substituting time into the derived equations.

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Homework Statement



If a particle is projected vertically upward to a height h above the Earth's surface at a northern latitude θ, show that it strikes the ground at a point 4/3*ωcosθ*sqrt(8h^3/g) to the west. (Neglect air resistance and consider only small vertical heights.)

my initial conditions as I see them (...I have nothing more than I have already written):
I believe that (x0',y0',z0')=(0,0,h) and (x0dot',y0dot',z0dot')=(0,0,0)

the notation is '=in a spinning situation while dot=derivative so my initial position should be as I wrote (00h) and my initial velocity (000)

Homework Equations



x'(t)=1/3*w*g*t^3*cos(θ)-w*t^2(z0dot'*cos(θ)-y0dot'*sin(θ))+x0t+x0'

y'dot(t)=y0dot'(t)-w*xdot'*t^2*sin(θ)-2wtsinθ*x0'+y0'

z'(t)=-1/2gt^2+z0dot*t+wx0dot'*t^2cosθ+2wx0'tcosθ+z0

The Attempt at a Solution



reducing using my initial conditions, I believe that:

x'=1/3wgt^3cosθ and z'=-(gt^2)/2+h which setting z'=0 gives me t=sqrt(2h/g)

I then plugged t into x':
x'=1/3wsqrt(8h^3/g)cosθ but it is supposed to be 4/3, not 1/3 as is stated in the problem

Is one part of my initial equation for x'(t) supposed to be wgt^3cosθ or did I mess up my algebra somewhere
 
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Hey, I can't really understand all the mess is in the middle. I'm just going to use prime' for time derivative because I don't want to use latex. I made my coordinate system as z normal to the earth, y tangent to the earth, and x out of the page.

We know that the coriolis force is wxv, which gives only the force in the x direction
Fc=-2z'cosØ xhat
z''=-g
z'=v0-gt
z=v0t-1/2 gt^2
Since I hate mucking around with signs, move to the left for now
-x''=2z'wcosØ
-x'=2zwcosØ + x0
-x0=0 by IC
-x'=2(v0t-1/2 gt^2)

Take it from here and you ought to get it right.
 


hey thanks a lot, that really helped out, I have it now.
 

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