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Distance covered by projectile

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).

Note by Ashish Siva
2 months, 1 week ago

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

I think u might be looking for this... Abc Xyz · 2 months, 1 week ago

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@Abc Xyz Thanks!!!! For sharing, tht is of cool help. Ashish Siva · 2 months, 1 week ago

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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........ Shubham Dhull · 2 months ago

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@Shubham Dhull Nice thought then! :P Ashish Siva · 2 months ago

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@Ashish Siva pleased to read that Shubham Dhull · 2 months ago

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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........ Shubham Dhull · 2 months ago

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@Hung Woei Neoh @Swapnil Das @Rishabh Tiwari please do comment. Ashish Siva · 2 months, 1 week ago

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@Ashish Siva If I remember correctly the arc length of a curve is calculated this way Hung Woei Neoh · 2 months, 1 week ago

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@Hung Woei Neoh Yes it is, but i tried womething with eccentricity and something "stupid" :P it was a total mess. Ashish Siva · 2 months, 1 week ago

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This is the only method.

However, here's a slight simplification: Substitute \(x\) and \(y\) in terms of the parameter \(t\). Deeparaj Bhat · 2 months, 1 week ago

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@Deeparaj Bhat 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. Ashish Siva · 2 months, 1 week ago

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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 · 2 months, 1 week ago

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@Ayanlaja Adebola Nah, teacher explained the question first. Its not the horizontal range, its the length of the trajectory. Ashish Siva · 2 months, 1 week ago

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