If you were driving in a car with a constant velocity, and then you see a target through the window of the car. You fire an arrow exactly perpendicular so that the target and the car are perfectly at a 90 degree angle. So I tried this experiment in real life, but the arrow always moved with the wind that the velocity of the car was creating, so it never made it to the target, why, and is there an equation for this? Thank you so much for reading this. It has been a question on my mind for a long time now.

If you fire the arrow perpendicularly from the car it will move travel across the ground at an angle because the arrow has velocity both in the perpendicular direction (equal to the speed that the arrow is fired from the bow) plus velocity in the parallel direction (equal to the velocity of the car). So if the target is exactly perpendicular to the car at the instant you fire the bow the arrow will miss the target in the direction the car is moving. Is that what you're trying to describe? Note that since an arrow has feathers, which cause it to rotate into the oncoming wind, after you fire it in a perpendicular direction it should rotate to face slightly forward (in the direction of the car) so that it is aligned with its direction of travel. But for the arrow or any other object that is thrown or launched from the car the direction of travel is dictated by the relative velocity of the car to the launch velocity. The angle that the object travels determined by:

\tan \theta = \frac {V_c}{ V_o}

where V_c is the velocity of the car and V_o is the launch velocity of the object.

Why would the arrow go the same direction as the car, when a substantial amount of wind and air density is drifting the arrow off to the other direction?

Why would the arrow go the same direction as the car, when a substantial amount of wind and air density is drifting the arrow off to the other direction?

Assuming the air is calm, the "wind" that the arrow feels is always in the direction it is traveling - there is no side wind relative to the direction of travel of the arrow. This is because the feathers of the arrow cause it to rotate into the direction of the prevailing wind, and ensures that the arrow points in that direction. This angle is where the "wind" due to V_c and the "wind" due to V_o add up to beibng a direct head wind on the arrow. So the arrow does not deviate from that path.

This is a bit different than the case of throwing a ball or similar obhject out the car window - a ball will feel the side wind and will tend to arc backwards from the straight path that I illustrated in the previous post.

Oh okay so I'm guessing an arrow has a lot different flight characteristics than a ball right? And thank you so much for answering my question all the other people said that its Newtons 1st Law, and nothing else... but because it is launched out of a car with a velocity, there has to be more variables, so thanks for doing that.