I'm told (by a questionable authority) that the exact formula for the lateral deflection of a vertically fired projectile is 
.
There is an analysis of this in Deflection of a Projectile due to the Earth’s Rotation, it's result is in a slightly different form, with the projectile landing (4ωV3cos λ)/(3g2)) due West of the launch point.
I haven't checked if they are equivalent, but I'd hazard a guess that unless you can launch yourself at least a kilometre or so precisely vertically you won't learn anything.
Yep. Assuming V is muzzle velocity, λ is latitude, and ω (omega) is the rotation rate of earth (2 pi / T
E) (full rotation / time to make full rotation), they resolve to the same thing.
The key is knowing that t
f in the first formula is the total time in flight (upward
and downward), that is:
t
f = t
u + t
dSince, ideally, t
u = t
d, then t
u = t
f / 2.
The general form for velocity of an object under constant acceleration, a, for time t, is
V
t = V
0 + a t
where V
t is the velocity after being accelerated for time t, and V
0 is the velocity it had when t = 0.
Since we're using upward velocity, but the acceleration is the downward acceleration of gravity, then a = -g. So,
V
t = V
0 - g t
At the end of the upward flight t = t
u, and V
t is zero (it's stopped going up and is about to start falling, so its upward velocity is 0)
0 = V
0 - g t
uV
0 = g t
uRecall the second form of the formula is
(4 ω V
3 cos λ) / (3 g
2)
Substituting 2 pi / T
E for ω (rotation rate), and our V
0 for V (muzzle velocity), we get
4 (2 pi / T
E) (g t
u)
3 cos λ / (3g
2)
= 8 pi (g t
u)
3 cos λ / (3 g
2 T
E)
= 8 pi g
3 t
u3 cos λ / (3 g
2 T
E)
= 8 pi g t
u3 cos λ / (3 T
E)
Since t
u = t
f / 2,
= 8 pi g (t
f / 2)
3 cos λ / (3 T
E)
(1/2)
3 = 1/8, so
= 8 pi g (t
f3 / 8 ) cos λ / (3 T
E)
= pi g t
f3 cos λ / (3 T
E)
Q.E.D.
They're the same.