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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
4 months, 3 weeks ago

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I think u might be looking for this... · 4 months, 2 weeks ago

Thanks!!!! For sharing, tht is of cool help. · 4 months, 2 weeks ago

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........ · 4 months, 2 weeks ago

Nice thought then! :P · 4 months, 2 weeks ago

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........ · 4 months, 2 weeks ago

@Hung Woei Neoh @Swapnil Das @Rishabh Tiwari please do comment. · 4 months, 2 weeks ago

If I remember correctly the arc length of a curve is calculated this way · 4 months, 2 weeks ago

Yes it is, but i tried womething with eccentricity and something "stupid" :P it was a total mess. · 4 months, 2 weeks ago

This is the only method.

However, here's a slight simplification: Substitute $$x$$ and $$y$$ in terms of the parameter $$t$$. · 4 months, 3 weeks 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. · 4 months, 3 weeks ago

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 · 4 months, 3 weeks ago