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A particle is thrown up vertically with initial speed, reaches a maximum height and falls back to ground. Show that the Coriolis deflection when it again reaches the ground is opposite in direction, and four times greater in magnitude, than the Coriolis deflection when it is dropped at rest from the same maximum height.

I worked out this problem as 2 parts : going up and going down. I see that while going up, I get a deflection of factor 2/3 times $\omega\sin\theta v_{0}^3/g^2$ and while going down, I get -1/3 factor which is the same when dropped down from the same height. How is this wrong ? To get the correct answer, I have to integrate over twice the time required to reach the maximum height and then I get 4/3 for going up and coming down and -1/3 for being dropped down. How is the first method not correct since coming down part of the trajectory is same as being dropped down from the maximum height it reaches ? Should the deflection while coming down be not opposite to the deflection while going up ? I think I can derive it rigorously but it will be good to get an intuitive idea why the things work out by taking the total time of fall (ignoring the horizontal motion) but they don't if up and down motion are treated separately- Does the effect of horizontal motion cancel over going up and down together ?

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After the particle reaches the top of its trajectory, it is still moving due to the Coriolis force on the way up. So it continues being deflected in that direction on the way down, even as it picks up force in the opposite direction.

You can get the factor of $\frac{4}{3}$ if you add $vt$ to your downward deflection, where $v$ is the maximum speed of upward deflection, and $t$ is the time of descent. This gives you a factor of 1, from which you subtract your $\frac{1}{3}$, getting $\frac{2}{3}$ going down plus $\frac{2}{3}$ going up = $\frac{4}{3}$.

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