Rotational earth motion...I am extremely close to what the answer is supposed to be..

[jb646]

1. The problem statement, all variables and given/known data

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)

2. Relevant 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

3. 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

[Mindscrape]

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.

[jb646]

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