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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 v

_{vert}=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 = v

_{vert}-gt , g = const. => t

_{1}= 6.1s => t

_{total}= 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 v

_{B}= const.:

r(t) = r + ∫v(t)dt

2∏r(0)/d = v

_{B}, d...day

v

_{B}is constant => ω can not be constant:

ω(t) = v

_{B1}/v

_{B2}*ω

_{1}= u

_{1}/u

_{2}*ω

_{1}= r/r(t)*ω

_{1}(R

_{1}/R

_{2}= r

_{1}/r

_{2})

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

_{1}

Δω = ω

_{1}∫ -(r*v(t))/(r+∫v(t)dt)² dt

Δs = Δω*R*t = r*cos(α)*t*ω

_{1}∫ -(r*v(t))/(r+∫v(t)dt)² dt

Thank you for your troubles.