It so happened that since I secured full marks in a test of projectile motion, my teacher challenged me to find the answer to a question. Here it is:-

\(\color{blue}{\text{What is the distance covered by a projectile thrown with initial velocity}} \ \color{red}{u} \ \color{blue}{\text{and an angle of}} \ \color{red}{\theta} \ \color{blue}{\text{with the horizontal?}}\)

The first thing that came to my mind was the formula \(\dfrac{u^2 \sin 2\theta}{g}\), but I realized that that was the formula for finding the horizontal range(displacement) by the projectile and not the distance.

Here is my approach for the problem:-

We break the trajectory into infinitely small bits such that it forms a right triangle of sides \(dy\) and \(dx\). Let the hypotenuse be \(dl\).

Using Pythagoras' theorem:-

\(\begin{align}

dl & = \sqrt{{(dx)}^2 + {(dy)}^2}\\ \\
dl & = \sqrt{{(dx)}^2\left(1 + {\left(\dfrac{dy}{dx}\right)}^2\right)}\\ \\
\text{Integrating on both sides}:-\\ \\
\int_{0}^{l} 1 \ dl & = \int_{0}^{R} \sqrt{\left(1 + {\left(\dfrac{dy}{dx}\right)}^2\right)} \ dx \end{align}\)

Note that \(R\) here is the horizontal range of the projectile.

But my method seems too long. Is there any shorter method? Please post a shorter solution if there is one. (Air resistance is neglected).

## Comments

Sort by:

TopNewesthttps://brilliant.org/discussions/thread/arc-length-of-projectile-2/

I think u might be looking for this...

Log in to reply

Thanks!!!! For sharing, tht is of cool help.

Log in to reply

i was supposed to find the same but was not asked by the teacher, i thought of it on my own and my answer is right but just one factor needed a little bit correction, thanks anyways........

Log in to reply

Nice thought then! :P

Log in to reply

pleased to read that

Log in to reply

@Ashish Siva also see this: https://www.physics.harvard.edu/uploads/files/undergrad/probweek/sol71.pdf

Log in to reply

Thanks for the tip!

Log in to reply

i was supposed to find the same but was not asked by the teacher, i thought of it on my own and my answer is right but just one factor needed a little bit correction, thanks anyways........

Log in to reply

@Hung Woei Neoh @Swapnil Das @Rishabh Tiwari please do comment.

Log in to reply

If I remember correctly the arc length of a curve is calculated this way

Log in to reply

Yes it is, but i tried womething with eccentricity and something "stupid" :P it was a total mess.

Log in to reply

This is the only method.

However, here's a slight simplification: Substitute \(x\) and \(y\) in terms of the parameter \(t\).

Log in to reply

Hmm, ok i tried an alternative method too by using the formula of arc length of parabola, but that came out too tedious than this.

Log in to reply

I guess u are asked to find the range of the projectile since it is in horizontal distance. I suggest you make use of R=u^2sin2@/g

Log in to reply

Nah, teacher explained the question first. Its not the horizontal range, its the length of the trajectory.

Log in to reply