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).

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I think u might be looking for this... – Abc Xyz · 1 year, 1 month ago

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– Ashish Siva · 1 year, 1 month ago

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........ – Brilliant Member · 1 year, 1 month ago

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– Ashish Siva · 1 year, 1 month ago

Nice thought then! :PLog in to reply

– Brilliant Member · 1 year, 1 month ago

pleased to read thatLog in to reply

@Ashish Siva also see this: https://www.physics.harvard.edu/uploads/files/undergrad/probweek/sol71.pdf – Harry Jones · 7 months, 3 weeks ago

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– Ashish Siva · 7 months, 3 weeks ago

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........ – Brilliant Member · 1 year, 1 month ago

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@Hung Woei Neoh @Swapnil Das @Rishabh Tiwari please do comment. – Ashish Siva · 1 year, 1 month ago

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– Hung Woei Neoh · 1 year, 1 month ago

If I remember correctly the arc length of a curve is calculated this wayLog in to reply

– Ashish Siva · 1 year, 1 month ago

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 \(t\). – Deeparaj Bhat · 1 year, 1 month ago

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– Ashish Siva · 1 year, 1 month ago

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 – Ayanlaja Adebola · 1 year, 1 month ago

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– Ashish Siva · 1 year, 1 month ago

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