It is obvious that the motion of an object is resisted by air resistance both in the horizontal and vertical components. However what I fail to understand is that the projectile of an object with air resistance taken into account is not parabolic and instead is steeper on its way down than it is on its way up. Why is this?
[Physics] Why does air resistance affect projectile motion in the way it does
fluid dynamicsnewtonian-mechanics
Related Solutions
First be aware that, so far, you have been dealing with projectile velocity as a 2-dimensional phenomenon. This is going to have to change.
In order to conform to standard notation, you'll need to refer to altitude (vertical motion) as the z-coordinate, while horizontal will be handled by x and y coordinates. For ease of calculation, you can assume that you initial velocity has a zero component in one axis, let's say the y axis. Now the equations you're familiar with look the same, except that they are in terms of x and z rather than x and y. Let's start by looking at how to do the simulation in the absence of crosswind.
At any point, you can calculate the x and z velocities $V_x$ and $V_z$, and combine them to get the speed of the projectile, V: $$V = \sqrt{(V_x)^2+(V_y)^2}$$
Use this to calculate the drag on the projectile from the answer you linked to, then decompose the drag force into its x and z components. Apply the components to modify the x and z velocities and calculate the next step.
Now for crosswind. The simplest approach is to assume that the crosswind is much less than the projectile velocities. In this case, you can assume that velocities in the y axis don't affect the other 2 axes.
Start by calculating the relative crosswind - that is the speed of the wind minus the actual y-axis speed. In order to do this, you need to calculate the component of the crosswind which is actually perpendicular to the projectile's motion. For instance, if the wind is at 45 degrees to the projectile's direction, the crosswind sideways to the projectile will be .707 times the wind speed. Recognize that the coefficient of drag is different for crosswind than for the previous calculation, since the "regular" calculation looked at the projectile head-on, while crosswind applies sideways. Knowing the relative crosswind and the crosswind drag coefficient, use the linked answer to calculate the sideways force on the projectile, then apply this to calculate the new velocity and then the new y-axis position.
From the correction for wind direction, it should be clear that, for best accuracy, you ought to factor in the "head-on" component of the wind into the overall drag calculation, but you can always point out that the crosswind is assumed to be much lower than the projectile speed.
Hint: I think the issue here is that the motion here is not symmetric. If it starts out with a speed $u$, it is not necessary that it will have the same speed when it reaches the bottom - because the acceleration is not the same in both the cases.
In the case without air resistance, it is valid to write $t=\frac{u}{g}+\frac{u}{g}$, because the particle goes from $u$ to $0$ and then $0$ to $u$, which doesn't happen in this case.
Try studying the problem by taking into account the distance the projectile travels - because the distance it travels up is always the distance it'll travel down.
Best Answer
A projectile's trajectory is only parabolic in the first place because the force is constant in magnitude and direction.
Air resistance is not constant in magnitude or direction, so once you include air resistance trajectories can't be parabolic any more.
As for why it's steeper on the way down, a good way to visualize this is to imagine something where air resistance completely dominates: a feather, for instance. If you throw a feather at a high speed, it very quickly loses virtually all of its momentum to air resistance, after which it begins to fall at terminal velocity. As a result, it falls straight down, whatever its initial trajectory was.
You can imagine making a projectile smaller and smaller. For a large projectile, it has a parabolic arc. A very small projectile has effectively a linear rise and a fall straight downwards. A projectile like a baseball hit off a bat is somewhere in the middle: the fall is steeper than the rise, but not straight down.