Suppose we have a cannon at the origin. This cannon has a defined power, that translates to a defined speed of the projectile \(v_0\).
The envelope of all trajectories form a parabola that is called the “safety parabola” because it delimits the space in two regions, safe and not safe.
The equation of this parabola is \(s(x) = \frac{v_0^2}{2g} - \frac{g}{2v_0^2}x^2\).
In order to derive the parabola we are only going to need some basic physics & algebra.
Let’s have the initial speed a vector \(v_0\) with an angle \(\theta\) from the \(x\) axis.
We then have the following projections of \(v_0\):
From the second law of Newton we have:
Using the projections of \(v_0\) we arrive at:
Now we are going to need to find \(y_{max}\) the highest point reached. We can calculate \(v_y(t) = 0\) to find this.
\(v_y(t) = 0 \Leftrightarrow -gt + v_0\sin(\theta) = 0 \Leftrightarrow t_{max} = \frac{v_0\sin(\theta)}{g}\)
We can then subsitute in \(y(t)\), this gives \(y_{max}(\theta) = -\frac{1}{2}g\left(\frac{v_0\sin(\theta)}{g}\right)^2 + \frac{v_0^2\sin^2(\theta)}{g}\).
After some algebra we derive \(y_{max}(\theta)=\frac{v_0^2\sin^2(\theta)}{2g}\).
In order to find the range of the trajectory we can calculate \(t_r\) such that \(y(t_r) = 0\).
Because we are not interested in \(t = 0\) we can divide the equation by \(t\) to obtain: \(-\frac{1}{2}gt_r + v_0\sin(\theta)=0\).
By isolating \(t_r\) we get \(t_r=\frac{2v_0\sin(\theta)}{g}\). In order to find the range we need to check \(x(t_r) = \frac{2v_0^2\cos(\theta)\sin(\theta)}{g}\).
We can simplify a bit by noting that \(2\cos(\theta)\sin(\theta) = \sin(2\theta)\). We obtain: \(x_{max}(\theta) = \frac{v_0^2}{g}\sin(2\theta)\).
We have \(y(t)\) and \(x(t)\), but to find our trajectory we are going to need to find \(y(x)\).
This is easily done by setting \(t=\frac{x}{v_0\cos(\theta)}\), and substituting in \(y(t)\). We obtain \(y(x) = \frac{-g}{2v_0^2\cos^2(\theta)}x^2 + \tan(\theta)x\)
Using this formula with can plot a number of trajectories for different values of \(\theta\).
There are two main approaches to creating “spread out” parabolas:
We have \(x_{max}(\theta) = \frac{v_0^2}{g}\sin(2\theta)\), so \(\sin(2\theta)=\frac{gx_{max}}{v_0^2}\).
This leads to \(\theta = \frac{\arcsin\left(\frac{gx_{max}}{v_0^2}\right)}{2}\)
So if we take equally spaced out \(x\) intersections we get the corresponding \(\theta\) for our ballistic trajectory.
We know that the safety parabola is going to intersect the \(x\) axis at furthest point reachable by the cannon. We can find that point using \(x_{max}\).
Indeed we have \(\frac{\mathrm{d}{x_{max}}}{\mathrm{d}\theta} = \frac{2v_0^2}{g}\cos(2\theta)\). This derivative is zero when \(cos(2\theta) = 0\).
This means that \(2\theta = \frac{\pi}{2}\) or that \(\theta=\frac{\pi}{4}\).
By substituting in the formula for \(x_{max}\) we get \(x_{range} = \frac{2v_0^2}{g} \times \frac{\sqrt{2}}{2} \times \frac{\sqrt{2}}{2} = \frac{v_0^2}{g}\)
We could derive \(y(\theta)\) to find the angle giving the best hight, but we can see quite intuitively that we need all the energy to go in the \(y\) axis, giving \(\theta=\frac{\pi}{2}\) as the best angle for height.
By looking at our formula for the height of a parabola we can get the maximum height: \(y_{summit} = \frac{v_0^2\sin^2(\theta)}{2g} = \frac{v_0^2}{2g}\) because for our angle \(sin(\theta)=1\).
We know that our safety parabola is going to intersect the \(x\) axis at \(x_{range}\) and \(-x_{range}\). As such the safety parabola \(s(x)\) is of the form: \(a\left(x+\frac{v_0^2}{g}\right)\left(x-\frac{v_0^2}{g}\right)\). We can distribute a bit to have \(s(x) = a\left(x^2 - \frac{v_0^4}{g^2}\right)\).
We can find \(a\) using the fact that \(s(0) = y_{summit}\). This means \(-a\frac{v_0^4}{g^2}=\frac{v_0^2}{2g}\), solving for \(a\) we get \(a=\frac{-g}{2v_0^2}\). We can distribute in \(s(x)\) to find:
\(s(x) = \frac{v_0^2}{2g} - \frac{g}{2v_0^2}x^2\)