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:-

What is the distance covered by a projectile thrown with initial velocity u and an angle of θ with the horizontal?\color{#3D99F6}{\text{What is the distance covered by a projectile thrown with initial velocity}} \ \color{#D61F06}{u} \ \color{#3D99F6}{\text{and an angle of}} \ \color{#D61F06}{\theta} \ \color{#3D99F6}{\text{with the horizontal?}}

The first thing that came to my mind was the formula u2sin2θg\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 dydy and dxdx. Let the hypotenuse be dldl.
Using Pythagoras' theorem:-
dl=(dx)2+(dy)2dl=(dx)2(1+(dydx)2)Integrating on both sides:0l1 dl=0R(1+(dydx)2) dx\begin{aligned} 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{aligned}

Note that RR 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
3 years, 4 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

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

I think u might be looking for this...

abc xyz - 3 years, 4 months ago

Log in to reply

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

Ashish Siva - 3 years, 4 months ago

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........

A Former Brilliant Member - 3 years, 3 months ago

Log in to reply

Nice thought then! :P

Ashish Siva - 3 years, 3 months ago

Log in to reply

pleased to read that

A Former Brilliant Member - 3 years, 3 months ago

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 - 3 years, 4 months ago

Log in to reply

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

Ashish Siva - 3 years, 4 months ago

Log in to reply

This is the only method.

However, here's a slight simplification: Substitute xx and yy in terms of the parameter tt.

Deeparaj Bhat - 3 years, 4 months ago

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.

Ashish Siva - 3 years, 4 months ago

Log in to reply

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

Ashish Siva - 3 years, 4 months ago

Log in to reply

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

Hung Woei Neoh - 3 years, 4 months ago

Log in to reply

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

Ashish Siva - 3 years, 4 months ago

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........

A Former Brilliant Member - 3 years, 3 months ago

Log in to reply

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

Harry Jones - 2 years, 10 months ago

Log in to reply

Thanks for the tip!

Ashish Siva - 2 years, 10 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...