Because this is a blog and I make up what happens, I'm going to write about projectiles. Specifically, some object moving under constant gravity and a resistance that depends on velocity.

## No Resistance

The simplest projectile problem is one that is moving in a vacuum.

At a time t after launch, the projectile has a position $ \vec{r} $. This position is a vector quantity, with two components - $ x $, the distance along the ground, and $ y $, the height above the ground.

This means that the position can be written as $ \vec{r} = x \hat{\imath} + y \hat{\jmath} $. Here, $ \hat{\imath} $ and $ \hat{\jmath} $ are unit vectors in the x and y directions; they have a length of 1, and point in their respective directions.

For a particle moving in a vacuum with only gravity, you can write the equations of motion of the particle as

$ \ddot{x} = 0 $

$ \ddot{y} = -g $

The dots represent derivatives with respect to time. $ x $ is the distance along the ground, $ \dot{x} $ is the speed at which the particle is moving along the ground, and $ \ddot{x} $ is the acceleration of the particle in the direction along the ground.

Those two equations of motion mean that there is no acceleration in the direction along the ground, but there is an acceleration of $ g $ pointing downwards.

These can be solved to find that

$ x = At + B $

$ y = -\frac{1}{2}gt^2 + Ct + D $

There's one problem here: solving those second order differential equations gives us four unknown quantities, which I have nicely called A, B, C, D.

### Boundary Conditions

This is because those equations specify very little about the situation; we know that a particle has been thrown somewhere, at some speed, but we don't know where, and we don't know how fast.

To do this, we can set some boundary conditions. These are descriptions of the state of the system at a specific time. Because there are four unknowns, we will need four pieces of information.

The most useful set of information is probably the x location at the start, the y location of the start, the x speed at the start, and the y speed at that start.

At time zero, the values of $ x $ and $ y $ are

$ x = B $

$ y = D $

If we launch the projectile from $ \vec{0} $, which is usually the best option (because we can pick any point to be (0,0)), we hence get

$ B = D = 0 $

At time zero, the velocities of the particle are

$ \dot{x} = A $

$ \dot{y} = C $

Usually, the velocity of launch is more interesting that just being zero. But, starting off with velocity being zero (corresponding to the particle just being dropped),

$ A = C = 0 $

and $ \vec{r} = -\frac{1}{2} g t^2 \hat{\jmath} $.

If the projectile begins at a velocity of $ v_x \hat{\imath} + v_y \hat{\jmath} $, then $ A = v_x $, $ B = v_y $, and the position as a function of time is

$ \vec{r} = \left( v_y t - \frac{1}{2} g t^2 \right) \hat{\jmath} + v_x t \hat{\imath} $.

Now, there are all kinds of things you can do with this. One interesting thing is to find the velocity, which is obtained from differentiating this:

$ \dot{ \vec{r} } = \left( v_y - g t \right) \hat{\jmath} + v_x \hat{\imath} $

The acceleration, which is

$ \ddot{ \vec{r} } = -g \hat{\jmath} $

(This is what we started with, which is promising)

And the jerk, which is

$ \overset{...}{ \vec{r} } = 0 $

### The Trajectory

Or the trajectory, which gives y in terms of x. This lets the path that the particle takes be plotted on a graph. Since $ x = v_x t $, that means that $ t = \frac{x}{v_x} $, and hence the path that the particle takes can be written by substituting this into position as

$ \vec{r} = \left( \frac{ v_y }{ v_x } x - \frac{g}{2 v_{x}^{2} } x^2 \right) \hat{\jmath} + x \hat{\imath} $

You can also find out where and when the particle lands! The position at which the particle lands can be found using the trajectory. When the particle hits the ground, $ y = 0 $, so

$ \frac{ v_y }{ v_x } x - \frac{g}{2 v_{x}^{2} } x^2 = 0 $

and

$ x = 0 $ or $ x = \frac{2 v_x v_y}{g} $

The time at which it lands can be found using the position vector. As before, $ y=0 $, so

$ v_y t - \frac{1}{2} g t^2 = 0 $

$ t = 0 $ or $ t = \frac{2 v_y}{g} $

Either one of these can be calculated from the other by noticing that $ x = v_x t $, too.

What about when, and where, the projectile reaches its peak? When can be found using the position; differentiating it with respect to time, and then finding where the velocity in the y direction is equal to zero. Where can be found by then substituting this time back into the position.

$ v_y - gt = 0 $

$ t = \frac{v_y}{g} $

And substituting this time into the position gives

$ \vec{r} = \frac{v_{y}^{2}}{2g} \hat{\jmath} + \frac{v_x v_y}{g} \hat{\imath} $

## Resistance Proportional to Velocity: Laminar Drag

The next situation is where there is some resisting force, propertional to the velocity of the particle. This is known as laminar drag, and is present when a particle is moving slowly and with no turbulence.

For this, the equations of motion are more complicated, because drag applies in both the x and y directions, and adds an extra term to the calculations.

$ \ddot{x} = -\frac{\zeta}{m} \dot{x} $

$ \ddot{y} = -\frac{\zeta}{m} \dot{y} - g $

Solving these two second order differential equations gives you a more complicated expression for position than before

$ \vec{r} = \left( \frac{gm}{\zeta} t + C e^{- \frac{\zeta}{m} t} + D \right) \hat{\jmath} + \left( A e^{-\frac{\zeta}{m} t} + B \right) \hat{\imath} $

Looking at the equation, you can see that the vertical position has two components: an exponentially decaying one, based on initial conditions, and a linear one. The exponential term represents the initial velocity, which vanishes exponentially because of the drag. The linear term represents the terminal velocity; an object dropped will eventually end up falling at a constant velocity $ \frac{gm}{\zeta} $ when the exponential term vanishes.

The horizontal position only has one component, which is the initial horizontal velocity, although now falling off exponentially.

Like before, let's put in some boundary conditions for further analysis - these will tell us how our projectile started off.

### Boundary Conditions

Let's fire it off from the location $ \vec{0} $ at a velocity $ v_x \hat{\imath} + v_y \hat{\jmath} $, like before.

At time 0, the position vector of the projectile is $ \vec{r} = \left(C + D \right) \hat{\jmath} + \left( A + B \right) \hat{\imath} $

So we know $ C = -D $ and $ A = -B $. Progress! Number of constants has been cut down from four to two.

The velocity at time 0 is

$ \dot{ \vec{r} } = \left( \frac{gm}{\zeta} - C \right) \hat{\jmath} - \left( A \frac{m}{\zeta} \right) \hat{\imath} $

Taking the x and y components seperately,

$ \frac{gm}{\zeta} - C = v_y $

$ A \frac{m}{\zeta} = v_x $

This gives our four constants as:

- $ A = v_x \frac{\zeta}{m} $
- $ B = - v_x \frac{\zeta}{m} $
- $ C = \frac{gm}{\zeta} - v_y $
- $ D = v_y -\frac{gm}{\zeta} $

And hence the position of the projectilve as a function of time is $ \vec{r} = \left( \frac{gm}{\zeta} t + \left( \frac{gm}{\zeta} - v_y \right) \left( e^{- \frac{\zeta}{m} t} -1 \right) \right) \hat{\jmath} + \left( \left( v_x \frac{\zeta}{m} \right) \left( e^{-\frac{\zeta}{m} t} - 1 \right) \right) \hat{\imath} $

Very fresh.

## Resistance Proportional to Velocity Squared: Turbulent Drag

Another situation that you might find with a particle fired out of a cannon is a resisting force proportional to the velocity squared. This is known as turbulent drag, and occurs when the particle is moving at a high velocity and turbulence is occuring. Most real drag is of this form.

Naturally, of course, this lends itself to the least palatable equations.

The equations of motion for this form of drag are

$ \ddot{x} = -\frac{\beta}{m} \dot{x}^2 $

$ \ddot{y} = -\frac{\beta}{m} \dot{y}^2 - g $

And solving these two equations gives the horrendous

$ \vec{r} = \left( \frac{\sqrt[3]{6} \wp \left(\frac{\sqrt[3]{-\frac{\beta}{m}} \left( t + C \right)}{\sqrt[3]{6}} ; \frac{2 \sqrt[3]{6} g}{\sqrt[3]{-\frac{\beta}{m}}}, D \right) }{\sqrt[3]{-\frac{\beta}{m}}} \right) \hat{\jmath} + \left( \frac{\sqrt[3]{6} \wp \left(\frac{\sqrt[3]{-\frac{\beta}{m}} \left( t + A \right)}{\sqrt[3]{6}} ; 0, B \right) }{\sqrt[3]{-\frac{\beta}{m}}} \right) \hat{\imath} $

The covenant must be sealed. The pact must be fulfilled.