Cannon

How far does a cannon shoot?

In the absence of air friction the distance $d$ only depends on height h, angle $\alpha$ and release speed $v$.

We will first calculate howe long it takes before the cannon ball hits the ground. To do this we take the familiar equation for the distance $x$ travelled by an object with inital vertical velocity $v_0$ and an acceleration $a$. \begin{equation} x = v_0 t + \frac12 a t^2 \end{equation}

If we use positive numbers for upward, then the initial velocity is $v_y$, the distance travelled by the cannon ball when it hits the ground is $-h$ and the acceleration is $-g$.

\begin{equation} -h = v_y t -\frac12 g t^2 \end{equation}

We can rewrite this the way we usually write down quadratic equations:

\begin{equation} -\frac12 g t^2 + v_y t + h = 0 \end{equation}

Solving this with the abc-equation gives

\begin{equation} t = \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \frac{-v_y \pm \sqrt{v_y^2-4(-\frac12g)h}}{2(-\frac12g)} = \frac{v_y + \sqrt{v_y^2+2gh}}{g} \end{equation}

Note that the last step is justified since we are looking for a value of $t$ after the ball leaves the cannon, not before.

Given the fact the the horizontal velocity $v_x=v\cos(\alpha)$ never changes, and the initial vertical velocity is $v_y = v\sin(\alpha)$ we can now calculate the total distance $d$ that the cannon ball travels.

\begin{equation} d = v_x t = v_x \frac{v_y + \sqrt{v_y^2+2gh}}{g} = v\cos(\alpha) \frac{v\sin(\alpha) + \sqrt{v^2\sin^2(\alpha)+2gh}}{g} \end{equation} \begin{equation} d = \frac{v^2}{g}\cos(\alpha)\sin(\alpha) \left(1 + \sqrt{1+2gh/\left(v^2\sin^2(\alpha)\right)}\right) \end{equation} \begin{equation} d(h, \alpha, v) = \frac{v^2}{2g}\sin(2\alpha) \left(1 + \sqrt{1+2gh/\left(v^2\sin^2(\alpha)\right)}\right) \end{equation}

For several values of $h$, with $g=9.81 m/s^2$, this equation is plotted below.

Note that for $h=0$ this equation simplifies to

\begin{equation} d(\alpha, v) = v^2\sin(2\alpha)/g \end{equation}

This is the bottom curve.

Angle for maximum distance

In order to simplify equation 7 we can rewrite it in terms of reduced height $c = 2 h g/v^2$:

\begin{equation} d(h, \alpha, c) = \frac{h}{c}\sin(2\alpha) \left(1 + \sqrt{1+c/\sin^2(\alpha)}\right); c = 2 h g/v^2 \end{equation}

For $h=0$ this equation has a maximum $\alpha = 45^\circ$. This is not the case in general though. To find this maximum, we equate the derivative to zero. We can ignore the constant factor $h/c$

\begin{equation} \frac{\mathrm{d}d(h, \alpha, c)}{\mathrm{d}\alpha } = 0 \end{equation} \begin{equation} \frac{\mathrm{d}}{\mathrm{d}\alpha } \sin(2\alpha) \left(1 + \sqrt{1+c/\sin^2(\alpha)}\right) = 0 \end{equation} \begin{equation} 2\cos(2\alpha)\left(\sqrt{c/\sin^2(\alpha)+1}+1)\right)-\frac{c\cos(\alpha)\sin(2\alpha)}{\sqrt{c/\sin^2(\alpha)+1}\sin^3(\alpha)}=0 \end{equation} This zero crossing, and therefore the angle at which the cannon ball goes furthest is \begin{equation} \alpha_{\mathrm{max}} = \arccos\left(\sqrt\frac{1+c}{2+c}\right) ; c = 2 h g/v^2 \end{equation}

where $c$ is the height in units of $v^2/(2g)$. The graph belows shows the optimal angle for shooting. The numbers on the upper x-axis give the height of the tip of the cannon barrel above the ground in meter, when $g=9.81\mathrm{m}/\mathrm{s}^2$ and $v=10\mathrm{m}/\mathrm{s}$