# Projectile motion: variation of the equation of the trajectory and motion introducing air resistance

In this site I have seen several questions/answer of the projectile motion. For example very close question/answer is into this link: Add air resistance to projectile motion

My students of an high school (16 years old) know the simple laws of parabolic motion. For example:

$$y=(\tan \theta) \ x-\frac{g}{2(v_0\cos \theta)^2} \ x^2 \tag{1}$$ $$x=x_0+(v_0\cos\theta)t, \quad y=y_0+(v_0\sin\theta)t-\frac12 g t^2\tag{2}$$ and others.

In a very simple way if I should introduce air resistance these equations or others like flight time, how will these change or how will they change? This question is a further knowledge that I do not know with such sincerity.

• Depends on what you consider the air resistance term to be. Do you assume that it's proportional to $v$? Or to $v^2$ ? Or is it equal to $Av +Bv^2$? Whatever you choose it to be, just put it into F = ma and solve the differential equation, if it's solvable analytically Nov 16 '19 at 13:46
• More on projectile motion with air drag. Nov 16 '19 at 14:43
• @Qmechanic Thank you very much for your precious link. Nov 16 '19 at 16:06
• @BrainStrokePatient Thank you also for you for your comment. I've never treated or studied this argument considering air resistance. I thought that the equations were modified by some coefficient of resistance such as viscosity. In other words, I thought that the equations changed little like the equations of ideal gases with the real gases of van der Waals. Nov 16 '19 at 16:15

In general, the effects of air resistance are rather complicated, and not all that accessible to high-school students except in a quantitative way. But here's a brief run-down of how air resistance works and how it affects the solutions of the equations of motion.

For most everyday situations, the drag force from a fluid can be modeled in one of two ways: linear drag ($$\vec{F} = - a \vec{v}$$ for some constant $$a>0$$) and quadratic drag ($$\vec{F} = - b |\vec{v}|\vec{v}$$ for some constant $$b>0$$.) In general, linear drag applies if the object in question is rather small or the fluid it's moving through is rather viscous. For human-scale objects moving through air, neither condition holds, and the quadratic drag force is a better model.

### Linear drag

For the linear drag equations, Newton's laws become $$\ddot{x} = - \alpha \dot{x} \\ \ddot{y} = -g - \alpha \dot{y},$$ where $$\alpha = a/m$$. These equations can be solved exactly for $$x$$ and $$y$$ as a function of $$t$$: $$x = x_0 + \frac{v_{x0}}{\alpha} (1 - e^{-\alpha t}) \\ y = y_0 + \frac{1}{\alpha} \left[ \left( \frac{g}{\alpha} + v_{y0} \right) (1 - e^{-\alpha t}) - gt\right]$$ In principle, you could solve these equations to get a direct relationship between $$y$$ and $$x$$, but the result would rather complicated and not terribly enlightening.

A few points to note:

• As $$t \to \infty$$, the object approaches a distance $$v_{x0}/\alpha$$ from its starting coordinate $$x_0$$. This is in contrast to the case without air resistance, where the velocity remains constant.

• The velocities are $$v_x = v_{x0} e^{-\alpha t} \\ v_y = v_{y0} e^{-\alpha t} - \frac{g}{\alpha} (1 - e^{-\alpha t}).$$ We can see that the velocity in the $$x$$-direction approaches 0 as $$t \to \infty$$, while the velocity in the $$y$$-direction approaches $$-g/\alpha$$. It is not hard to show that when this is the case, the net force on the object is zero, with the weight of the object canceling out with the drag force.

While this is the more relevant case for most everyday objects, it's also much harder to address exactly. The equations of motion are now $$\ddot{x} = - \beta \sqrt{\dot{x}^2 + \dot{y}^2} \dot{x} \\ \ddot{y} = -g - \beta \sqrt{\dot{x}^2 + \dot{y}^2} \dot{y},$$ where $$\beta = b/m$$. These equations are non-linear and coupled, both of which features make it harder to solve. In fact, I'm unaware of any closed-form solution for these equations for the general case of arbitrary initial $$\vec{v}_0$$. It's possible to write down closed-form expressions for simple cases, such as a ball released from rest; in this case, the motion is 1-D and the solution for $$y(t)$$ is $$y = y_0 - \frac{1}{\beta} \ln (\cosh (\sqrt{g \beta} t)).$$ But in general, the best way to investigate the properties of such equations is through numerical modeling, as was done in the other question you linked to.