Effect of earth's rotation in ballistics For this purpose, let's consider earth's rotations constant. Do earth rotation momentum get transfered to any object (a missile for example) that get's lauched? If so, why do we have to consider earth rotation when lauching the missiles? Wouldn't just follow earth rotation? (Btw, sorry for any grammar mistakes, I'm from a non-english speaking country).
 A: 
Do earth rotation momentum get transfered to any object (a missile for example) that get's lauched?

Yes
That is why they build rocket launch sites as close to the equator as possible, so that they can use that velocity to help reach orbital velocity. At the equator the Earth is moving roughly 1000 miles per hour, and low earth orbit is about 17000, so you get about 6% of the speed you need for free.
Getting right on the equator isn't always easy when you consider where the lower stages fall back to Earth, so for most cases its "as close as we can get". So you have launch sites in Florida for the US, French Guiana for Europe, and Kazakhstan for the USSR (which make more sense when the USSR still existed).
When launching into polar orbits, this may work against you, so you see launch sites for those satellites in more northern locations.
A: When an object is launched, it initially shares the earth's rotation. That is one reason why most spacecraft launching sites are situated fairly close to the equator: it gives the spacecraft a free initial velocity of 1600 km/h (at the equator).
The atmosphere also shares the earth's rotation. If it didn't, the equator would be subject to a 1600 km/h wind. So, while the object is in flight in the atmosphere, it keeps being affected by earth's rotation, and keeps the initial momentum. As a result, since its rotational momentum does not change, its course is effectively unaffected by the earths rotation. That is, if we ignore secondary effects like the Coriolis force, which will cause a deviation from the expected path.
However, once the missile leaves the atmosphere, that influence is no longer felt. Hence, when a spacecraft is in orbit, it is completely independent of the earth's rotation. The earth could suddenly stop, and the craft would continue on its path as if nothing had happened.
A: Earth's rotation velocity does get transferred to any object (a missile for example) that gets launched. The reason why we have to consider Earth's rotation when launching the missiles is the target rotates together with the Earth. Let us consider the following example: a missile is launched from point A to deliver some payload to point B. The typical flight time of an intercontinental ballistic missile is about half an hour. Due to Earth's rotation, point B travels up to 800 km in half an hour (depending on its latitude). So if one does not take Earth's rotation into account, the payload will be delivered to a point far away from B.
One can take Earth's rotation into account, for example, using a frame of reference rotating with the Earth. In this case, one needs to introduce fictitious centripetal and Coriolis forces. Littlewood gave the following example: allegedly, during a battle at Falkland Islands in 1914 (during World War I), German ships were destroyed from a maximum distance, but it required a lot of time, as initially the projectiles missed the targets by 100 yards, as corrections of about 50 yards for the Coriolis force were calculated for the latitude of 50 deg. North, whereas the Falkland Islands are at a latitude of about 50 deg. South. So ignoring or miscalculating the effect of Earth's rotation can be pretty risky. 
A: So the basic problem happens invloves moving closer or further away from the axis of rotation, which happens, everywhere other than the equator, in part when you move North or South. It can also happen if you travel East or West, if you go so fast that you start to notice the Earth curbing away under you. And of course everywhere other than the Poles it happens if you move up or down.
Why is moving away from the axis weird? Well, seen from an outer-space perspective, you are a distance $r=R\cos\alpha$ from the axis of rotation where $\alpha$ is your latitude and $R$ is the radius, and the Earth rotates with some angular velocity $\omega$ such that you have speed $v=R\omega\cos\phi.$ (In fact we have a convenient time unit to describe the Earth’s rotation: $\omega=2\pi/\text{day}$.)
Now let's suppose you were to grab a balloon and float to a height $h$. The problem is that you are still going at speed $v$ but your surroundings—what it means to stay above the same point on Earth—are now moving with speed $(R+h)\omega\cos\phi$. The wind will eventually drag you along and you will be moving again with the rotation of the Earth, but not before you have “fallen behind.”
So Earth rotates towards the East. Every time you jump, you land a tiny bit West of where you jumped, because you keep your Eastward velocity but you transfer into an altitude where staying in rotational equilibrium would require you to be moving faster than your were. This is the subject of the Coriolis effect, where in the Northern hemisphere something headed North wants to go a bit East and something headed South wants to go a bit West, because you're moving closer to the axis. You can see it in the curve that a hurricane or cyclone describes next time you see one on the news. 
It also causes hurricanes to have a certain rotation in the Northern hemisphere: to understand this, you have to understand that the eye of the storm is a very low pressure place, the air wants to come towards the center. So the air that's a little bit north of the eye of the storm travels Southwest and the air that's a little South of the eye travels Northeast, and if you just draw those two arrows on a piece of paper you can see a counter-clockwise pattern. It is truly a tiny force but it always breaks the clockwise/counter-clockwise symmetry in the laws in favor of counter-clockwise storms in the Northern hemisphere.
A: The earth's velocity doesn't need to be transferred to an object that is launched, because the object already has it. However, the object shares the Earth's linear velocity. Once it is launched, it continues to have that linear velocity. When it's on the ground, having the same linear velocity as the Earth means also having the same rotational velocity, but once it's in the air, the two can diverge. 
When an object rotates, the linear velocity of a point on the object is the rotational velocity times the radius. So points at a larger radius have a larger velocity for the same rotation, and vice versa for a smaller radius. If we calculate mechanics in the Earth's reference frame, we are dealing with a non-inertial reference frame, and have to take that into account.
If an object is launched directly upwards, it will continue to have the horizontal linear velocity that it had on the ground, which is smaller than what it "should" have for its new radius. If we compare it to the Earth's rotation, it's going too slow. So that means that, in the Earth's reference frame, it had an acceleration opposite to the Earth's rotation. 
Also, the equator is farther away from the axis of rotation than the poles are, and thus has a larger radius, so objects traveling towards the equator will also have an acceleration opposite to the Earth's rotation (again, in the Earth's reference frame), while objects traveling away from the equator will have an acceleration in the same direction as the Earth's rotation. This is known as the Coriolis force.
