We will show that:

The arc length of any projectile subject to no forces other than gravity can be expressed by the formula:

\(\ell = \frac{(v_0\cos(\phi))^2}{2g} \left[ 2\sec(\phi)\tan(\phi) + \ln \lvert\frac{1+\sin(\phi)}{1-\sin(\phi)} \rvert \right]\)

Where:

\(\phi\) is the angle the projectile initially makes with the horizontal axis

\(v_0\) is the initial velocity

\(g\) is the force of gravity

I suppose before we start, I should explain why I decided to put such a disgusting looking formula up here. Arc length integrals are not particularly difficult (usually), but they are often times very annoying to actually compute, at least in my opinion. With this formula, arc length can now be considered a function of any of the above variables, even \(g\). For instance, suppose we wished to study the effect of gravitational fields with differing magnitude (but always constant direction) on the arc length of a projectile. We would simply make the above formula a function of the variable \(g\) and hold everything else constant (as long as all other factors are the same). This is why I felt this formula has its advantages and is thus worthy of sharing.

Proof:

Consider a massive particle projected from the origin of a Cartesian Coordinate system with initial velocity \(v_0\) and angle \(\phi\). There is negligible air resistance and the only force acting on said particle is gravity. Then, by Newton, we have:

\(\displaystyle \sum \vec{F} = m\vec{a} = \vec{W}\)

\(\vec{W}\) is just the weight of the particle:

\(\vec{W} = -mg \hat{j} \)

Then:

\(\vec{a} = -g \hat{j}\)

Acceleration is clearly constant, hence we may proceed to derive vector-valued functions for our velocity and position:

\(\vec{v} = \displaystyle \int \vec{a} \,dt = (v_0\cos(\phi))\hat{i} + (v_0\sin(\phi) - gt)\hat{j}\)

\(\vec{r} = \displaystyle \int \vec{v}\,dt = (v_0\cos(\phi)t)\hat{i} +(v_0\sin(\phi)t - \frac{gt^2}{2})\hat{j}\)

(This assumes that we zero-reference our position from the initial position)

Now, keep these in mind as we derive an integral formula for arc length:

By the Pythagorean Theorem, if we consider a triangle with infinitesimal hypotenuse on the trajectory of the particle, we have:

\(d\ell^2 = dx^2 +dy^2\)

Now, our position equations are parameterized in time. Then:

\(\frac{d\ell^2}{dt^2} = \frac{dx^2}{dt^2} + \frac{dy^2}{dt^2}\)

\(\Rightarrow\) \(\frac{d\ell}{dt} = \sqrt{\frac{dx^2}{dt^2} + \frac{dy^2}{dt^2}}\)

\(\Rightarrow\) \(\ell = \displaystyle \int_{0}^{t_0} \sqrt{\frac{dx^2}{dt^2} + \frac{dy^2}{dt^2}}\,dt\)

Where \(t_0\) is the time at which the particle has exactly zero height.

(As a quick side note, this integral should make sense. The integrand is simply the magnitude of the velocity as a function of time, hence we are summing all infinitesimal velocity elements along the curve at all times \(0\leq t \leq t_0\) to obtain our length)

Now we have our position functions for \(x\) and \(y\), these are just the components of our position vector. Then:

\(x(t) = v_0\cos(\phi)t\) \(\Rightarrow\) \(\frac{dx}{dt} = v_0\cos(\phi)\)

\(y(t) = v_0\sin(\phi)t - \frac{gt^2}{2}\) \(\Rightarrow\) \(\frac{dy}{dt} = v_0\sin(\phi) - gt\)

Then our integral is:

\(\ell = \displaystyle \int_{0}^{t_0} \sqrt{ v_0^2\cos^2(\phi) + \frac{dy^2}{dt^2}}\,dt\)

Let: \(u= \frac{dy}{dt} = v_0\sin(\phi) - gt\) \(\Rightarrow\) \(-\frac{du}{g} = dt\)

Then:

\(\ell =-\frac{1}{g} \displaystyle \int_{v_0\sin(\phi)}^{v_0\sin(\phi) -gt_0} \sqrt{ v_0^2\cos^2(\phi) + u^2}\,du\)

Recall that: \(\sin^2(\phi) + \cos^2(\phi) = 1\)

Dividing the entire equation by \(\cos^2(\phi)\) we obtain:

\(\tan^2(\phi) + 1 = \sec^2(\phi)\)

Let: \(u=v_0\cos(\phi)\tan(\theta)\) \(\Rightarrow\) \(du = v_0\cos(\phi) \sec^2(\theta) d\theta\)

Before plugging this in to our integral, we shall find an expression for \(t_0\). This will help us change our bounds:

\( 0 = v_0\sin(\phi) t_0 - \frac{gt_0^2}{2} \) \(\Rightarrow\) \(t_0 = 0\)

or

\(t_0 = \frac{2v_0\sin(\phi)}{g}\)

\(t_0=0\) is simply the initial time, hence we neglect this solution in favor of the latter.

Now let us change our bounds:

\(\displaystyle \lim_{u \to v_0\sin(\phi)} \theta = \displaystyle \lim_{u \to v_0\sin(\phi)} \tan^{-1} \left(\frac{u}{v_0 \cos(\phi)} \right) = \phi\)

\(\displaystyle \lim_{u \to v_0\sin(\phi)-gt_0} \theta = \displaystyle \lim_{u \to v_0\sin(\phi)-gt_0} \tan^{-1} \left(\frac{u}{v_0 \cos(\phi)} \right) = -\phi\)

Then:

\(\ell = -\frac{1}{g} \displaystyle \int_{\phi}^{-\phi} \sqrt{ v_0^2\cos^2(\phi) +v_0^2\cos^2(\phi)\tan^2(\theta)}(v_0\cos(\phi)\sec^2(\theta))\,d\theta\)

\(\Rightarrow\) \(\ell = \frac{(v_0\cos(\phi))^2}{g} \displaystyle \int_{-\phi}^{\phi} \sec^3(\theta) \,d\theta = \frac{(v_0\cos(\phi))^2}{g} \displaystyle \int_{-\phi}^{\phi} \sec(\theta)\frac{d}{d\theta} \tan(\theta)\,d\theta\)

\(\Rightarrow\) \(\ell =\frac{(v_0\cos(\phi))^2}{g} \left[ \sec(\theta)\tan(\theta) |_{-\phi}^{\phi} - \displaystyle \int_{-\phi}^{\phi} \sec(\theta)\tan^2(\theta) \,d\theta \right] \)

\(\Rightarrow\) \(\frac{(v_0\cos(\phi))^2}{g} \displaystyle \int_{-\phi}^{\phi} \sec^3(\theta) \,d\theta =\frac{(v_0\cos(\phi))^2}{g} \left[\sec(\theta)\tan(\theta) |_{-\phi}^{\phi} - \displaystyle \int_{-\phi}^{\phi} \sec^3 (\theta) \,d\theta + \displaystyle \int_{-\phi}^{\phi} \sec(\theta) \,d\theta \right]\)

\(\Rightarrow\) \(\ell = \frac{(v_0\cos(\phi))^2}{2g}\left[\sec(\theta)\tan(\theta) |_{-\phi}^{\phi} + \displaystyle \int_{-\phi}^{\phi} \sec(\theta) \,d\theta \right]\)

Now, focus on \(\displaystyle \int_{-\phi}^{\phi} \sec(\theta) \,d\theta\). We'll deal with everything else later, we need to evaluate this integral first:

\(\displaystyle \int_{-\phi}^{\phi} \sec(\theta) \,d\theta = \displaystyle \int_{-\phi}^{\phi} \sec(\theta) \frac{\frac{1+\sin(\theta)}{\cos(\theta)}}{\frac{1+\sin(\theta)}{\cos(\theta)}} \,d\theta\)

If we let \(u= \frac{1+\sin(\theta)}{\cos(\theta)}\), then:

\(du = \sec(\theta) \frac{1+\sin(\theta)}{\cos(\theta)}d\theta\)

Then we have:

\(\displaystyle \int_{-\phi}^{\phi} \sec(\theta) \,d\theta = \displaystyle \int \frac{du}{u} |_{-\phi}^{\phi} = \ln \lvert \frac{1+\sin(\theta)}{\cos(\theta)} \rvert |_{-\phi}^{\phi} \)

\(\Rightarrow\) \(\ell = \frac{(v_0\cos(\phi))^2}{2g} \left [ \sec(\theta)\tan(\theta) + \ln \lvert \sec(\theta) +\tan(\theta) \rvert \right]_{-\phi}^{\phi}\)

\(\Rightarrow\) \(\ell = \frac{(v_0\cos(\phi))^2}{2g} \left [ 2 \sec(\phi) \tan(\phi) + \ln \lvert\frac{1+\sin(\phi)}{1-\sin(\phi)} \rvert \right]\)

Since \(\sec(-\phi) = \sec(\phi)\) and \( \tan(-\phi) = - \tan(\phi)\)

This is the intended result and we have, in the most un-elegant and brute-force way possible, proved our statement.

QED

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestI derived it before but I cant describe how more longwinded it was... In fact, at the last step, I just used Wolfram Alpha to integrate for me cause it was just too tedious to do it by hand

Log in to reply

Please assist as to if through this result we can find angle of projection for maximum arc length of projectile for a

fixed speed of projection.

i tried and it reduced to

please help.

I was trying out this question

https://brilliant.org/problems/the-arc-length-of-a-projectile/

Log in to reply

You could bash using newton's method

Log in to reply

@Ethan Robinett Can you add this to Arc Length - Basic and Intermediate? Thanks!

Log in to reply

Hey Calvin, I've been off the site for a while, just got back on (really busy with school and everything). I'll add this to both pages. Thanks!

Log in to reply

I'm not sure what is the use of finding the arclength of a projectile. I might be wrong though. Does the military care?

Log in to reply

Haha well suppose you have an artillery shell fired at long range and the shell is programmed to detonate after a certain amount of displacement. You couldn't use the horizontal or vertical displacement as these are not the actual distance traveled by the projectile, but you could use the arc length to find the horizontal rangeof the projectile before detonation.

My opinion though. why are people downvoting some of these comments?

Log in to reply

Wasn't me. I up-voted them.

Log in to reply

Log in to reply

I suppose that you can use this for calculating how much something (A spaceship or smth) corrodes when flying through Earth's atmosphere? Should be important...

Log in to reply

But then gravitational force will continuously change with distance from the earth.Then weight=mg won't hold

Log in to reply

Log in to reply

There is no use I suppose.

Log in to reply

Yes I too once derived this formula but not in the most unelegant way but instead using the result of arc length of a standard parabola.

Log in to reply

I originally did the same thing, but i figured the result would make a little more sense this way rather than deriving it for a parabola and relating it to a projectile. Though both methods will obviously work

Log in to reply

Agree to you.

Log in to reply

This problem was also there in Problems of the week by David morin.

Log in to reply