Coriolis Effect - Vertical Shot, solution through integration?

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SUMMARY

The discussion focuses on calculating the distance Δs between the starting point and landing point of a vertically shot projectile, considering the Coriolis effect at latitudes 0° and 51°. The projectile's exit velocity is given as vvert = 60 m/s, and the total time of flight is calculated as ttotal = 12.2 seconds. The solution employs geometric principles and angular acceleration formulas, specifically using the Earth's radius of r = 6,370,000 meters and the relationship between angular velocity and latitude. The user seeks feedback on the correctness of their approach.

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Malor
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Hi, it's my first time on these forum, so I hope that I made this post correctly.

I have an example solution to this problem using tensors and matrices ^^ but I wanted to solve the problem in a different way and would like some feedback on whether or not this solution is correct.

Homework Statement



A gun shoots a projectile vertically into the air with an exit velocity of vvert=60ms-1 (assuming the nozzle is at sea level). Calculate the distance Δs, between the starting point and landing point of the projectile. Do this for the latitudes 0° and 51°. (neglect cascading effects)2. The attempt at a solution

Getting the time t for the whole process is a no-brainer:

v(t) = 0 = vvert-gt , g = const. => t1 = 6.1s => ttotal = 12.2s

Now to calculate Δs:

We assume that the Earth is an ideal globe, with a radius r = 6370000km. Basic geometry tells us that the distance from the axis of rotation is
R = sqrt(h(2r-h)) => R = sqrt(r²-r²sin²(α)) , α...latitude => R = r*cos(α)

The circumference of r(t) is the gives us the orbital velocity vB = const.:

r(t) = r + ∫v(t)dt

2∏r(0)/d = vB , d...day

vB is constant => ω can not be constant:

ω(t) = vB1/vB21 = u1/u21 = r/r(t)*ω1 (R1/R2 = r1/r2)

Angular acceleration: w'(t) = - (r*v(t))/(r+∫v(t)dt)² * ω1

Δω = ω1 ∫ -(r*v(t))/(r+∫v(t)dt)² dt

Δs = Δω*R*t = r*cos(α)*t*ω1 ∫ -(r*v(t))/(r+∫v(t)dt)² dt

Thank you for your troubles.
 
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