# Homework Help: Coriolis Effect On A Projectile

1. Oct 3, 2011

### mj478

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

If you fire a projectile eastwards and upwards with a speed u=(uo,0,wo) at latitude theta, where will it land? You can neglect air resistance and (in the vertical momentum equation) the effects of rotation relative to a constant g. Sketch the solution(s). Find the landing place if g = 10m/s and u=(1000,0,1000)m/s.

2. Relevant equations

Du/Dt - fv = 0
Dv/Dt + fu = 0
Dw/Dt = -g

where f = 2*Omega sin (theta) = 10^-4 1/s

3. The attempt at a solution

I know that the projectile will move to the right due to coriolis effect, but I don't understand how to actually calculate the landing place.

2. Oct 3, 2011

### Cipherflak

Are those equations given? From the text it sounds like you should only neglect air resistance in the vertical.

Anyway, they must be integrated up so you have velocity as a function of time. Then I believe you must integrate again to get the position as a function of time. Is it a numerical (computer) problem or an analytical one?

3. Oct 3, 2011

### mj478

I don't think we are supposed to use a computer. The equations weren't given in the problem, but the prof said in class that we would need them to solve the problem.

4. Oct 3, 2011

### Cipherflak

Ok, so all air resistance could be ignored then. Such things are impossible without a computer anyway, and you don't got no equation for it either. Ok.. i think we first have to integrate all the equations to we get the actual velocities and not the acceleration.

∫Du/Dt dt= ∫fv dt
∫Dv/Dt dt= -∫fu dt
∫Dw/Dt dt= -∫g dt

integrating with respect to time yields,

u(t) = fvt+C
v(t) = -fut+C
w(t) = gt+C

Now you insert your initial conditions, which is u(t=0) = u0 , v(t=0) = 0 and w(t=0) = w0
This means that when you actually set t to equal 0 in the integrated equations, they should be equal to these conditions. They are 1000m/s, but let's keep the symbols for now.

initial conditions gives simply the values for C in this case:
u(t) = fvt + u0
v(t) = -fut
w(t) = gt + w0

These equations seem to make very much sense regarding the velocity. But then...
I was about to integrate these again to gain an expression for distance versus time, but I realized that it's not that easy, because v and u are functions of time as well.. hmm... I gotta think this one over later.. perhaps inserting the value v(t) into the value for v in the first equation and vice versa. Other ideas out there?

any other insightful ideas?

Last edited: Oct 3, 2011
5. Oct 3, 2011

### Cipherflak

vertical equation integrated with time gives: S(t) = 0.5gt^2 + w0*t

solve this one and for t and set S = 0 to find how long the projectile is in the air. then you know a whole lot more.

6. Oct 3, 2011

### davo789

I haven't done a problem like this for a long time (so take this with a pinch of salt), but could you use the equation for the (apparent) Coriolis Force? This way you could calculate an acceleration experienced by the particle, and given the time in the air, you would know its displacement.

7. Oct 3, 2011

### mj478

I got the time the projectile is in the air from basic kinematics. I am stuck trying to solve a second degree DE for u. I took the derivative of the first eqn and substituted in from the second one to get:

u'' -(f^2)v' = 0.

The characteristic equation doesn't have real roots. So I am having trouble getting an actual number for u at a given time.

8. Oct 4, 2011

### Cipherflak

I'm starting to have doubts whether this is possible to do analytically. If the problem could be broken down to many small iterations, one could assume that v and u was constant within each iteration, and it would be no problem solving this... but that would require a computer.

I hope someone can answer this, or I hope you will if you eventually get the solution from your professor. I'm studying meteorology and should know this! :uhh: