Vx=Vcos0; That's fast!

Vx=Vcos0 This equation is used to define a specific speed of an object traveling at a specific angle. So,for example, if you wanted to know the speed of a ball traveling in the "X" direction, you would be able to by accounting for the velocity (speed) multiplied by the cosine (degree of the angle). This information is useful for many reasons. I can understand how this speed would be important when trying to figure the speed needed for a rocket ship to leave earth's atmosphere. This simple equation could help identify how fast a ship would need to go in order to leave earth's atmosphere. Also, this equation could help to understand the quickest route to a destination. I tried to understand this physics problem from a unique perspective. I have to try to imagine what new angle could be used to help me utilize this equation. The result that I have concluded based on the calculation is applicable to travel as well. When we think about the quickest route to space, we think of a rocket ship shooting straight up to the sky. The velocity is assumed from the "Y" axis. The Y axis is a linear north/south direction the rocket ship travels. Since we are on planet earth, I have to understand that due to the curvature of the earth, our perception of travel to space, or air, is something that can be understood differently. Were it possible to take a linear path accross the surface of the earth, at say five feet off the ground, eventually we would head straight out to space. This being the case, if we were to try and understand how to get to a specific point in space from earth, understanding Vx=Vcos0 could become useful. Trying to find the speed of the rocket ship traveling along the "X" axis, we may be able to find that the quickest way to space is following a path along the X axis. For example, if you were to travel at 89 meters per second (198 mph) at an angle of 90 degrees, your speed would not change. However, as the angle changes to say fifty degrees, the speed slows. The angle slows. In fact, the speed of the rocket ship would be less until you reached a cosine of 2 degrees, at this point the speed would be equal to the 90 degree angle. Therefore, traveling straight accross the x axis is just as quick to get into outer space than if you were to take a 90 degree ascent into the heavens. Does this mean that regardless of how high off the ground you traveled, knowing the earth is round,would you be able to reach space as quickly if you traveled across the X axis? This of course is a questionable theory as the 90 degree angle to space is only 62 miles while taking the route on the X axis would take many more miles to reach space as the horizon is more than 62 miles. Though the 90 degree angle still seems like the quickest route, what if you were trying to reach a specific area in space? What if the area in space was more easily accesable taking a linear direction. Is it possible that an area in space is accessible more quickly following the X axis? Of course it is possible. Simply because you are traveling along the X axis, doesn't mean that you can't be, say, 1000 feet in the air. For that matter, it is not to say that you can not be 2 inches in the air from the surface of the earth. However, since the destination in space from earth, the curvature of the earth and your proximity is important. If you were 20 miles from the horizon or curvature of the earth, would you get to space quicker traveling on the X axis, rather than at any other angle, including a 90 degre angle? Not necessarily because this equation accounts for speed and not the time it would take to arrive at your destination. What this means is that your speed would be the same at the 90 degree angle as it would be at the 2 degree angle when determing how fast the object travels on the X axis. In other words, flat and verticle are the same in terms of speed. Although this is true, lets just say that the space ship leaving from the north pole wanted to get to a destination in space that is 90 degrees away on the X axis. It may be quicker to leave earth to that destination traveling along the X axis if the X axis is on that 90 degree angle that the space ship needs to reach. Though we would have to account for speed of the craft in outer space possibly being able to reach that place quicker, the fact is that spacially, this would be a quicker and more direct route on the X axis rather than the Y axis. That is to say, traveling at say a couple miles from the horizon to outer space on the X axis is closer in this case. In this scenario, it would take 64 miles to reach space rather than 62 miles. However, since the destination in space is closer, in this case on the X axis, would be faster. You would still be closer to space were you to take a 90 degree ascent into the sky say at three miles further up the path from the equater. This is because in this case, again, the rocket ship is traveling straight and would be at a 90 degree angle from the point in space to be arrived. However, once in space a control for this 90 degree angle would have to be taken. In essence, the only consideration that would be needed is to ascertain whether the 90 degree angle added up to more than the distance to the horizon plus 62 miles.