# Thread: Projectile's length of path

1. ## Projectile's length of path

Hey brainys,

I was wondering how to derive a formula for a length of projectiles path in the air (2 dimensions). I figured out the relationship of x and y, y=x*tg(a) - gx2/2v02cos2(a)
Then i combined some logic with the mean value theorem and got that lenght of path is integral from zero to x=v02sin(2a)/g of sqrt(1 + (tg(a) - xg/v02cos2(a))2) dx
When we raise (tg(a) - xg/v02*cos2(a))2 and simplify a little bit we get integral from zero to x=v02sin(2a)/g of sqrt((v04cos2(a) - 2v02sin2(a)*xg + x2g)/v04cos4(a)) dx . ​Any ideas how to solve it? Thank you!

2. ## Re: Projectile's length of path

Originally Posted by BlueBeast
Hey brainys,

I was wondering how to derive a formula for a length of projectiles path in the air (2 dimensions). I figured out the relationship of x and y, y=x*tg(a) - gx2/2v02cos2(a)
Then i combined some logic with the mean value theorem and got that lenght of path is integral from zero to x=v02sin(2a)/g of sqrt(1 + (tg(a) - xg/v02cos2(a))2) dx
When we raise (tg(a) - xg/v02*cos2(a))2 and simplify a little bit we get integral from zero to x=v02sin(2a)/g of sqrt((v04cos2(a) - 2v02sin2(a)*xg + x2g)/v04cos4(a)) dx . ​Any ideas how to solve it? Thank you!
Hi BlueBeast!

Let me first reformat your equation to something I can understand

$\displaystyle \displaystyle \int_0^{v_0^2\sin(2a)/g} \sqrt{\frac {v_0^4\cos^2(a) - 2v_0^2\sin^2(a) \cdot xg + x^2g} {v_0^4\cos^4(a)}} dx$

I haven't tried to verify the validity of that length, but to solve something like this, first simplify it to for instance

$\displaystyle \int \sqrt{A+Bx+Cx^2}$

with appropriate choices for A, B, and C.
Now you can feed it to for instance Wolfram|Alpha to do the difficult integration part.
You can see the result here.

Afterward you can substitute your values for A, B, and C again.

3. ## Re: Projectile's length of path

WOW.... I didn't think it is sooo complicated.. I realised that it is but i thought that it is possible to solve it with somekind of substitution. When i saw result in wolfram alfa, i was shocked :O . Thank you very much for this answer, i will try to derive that formula finally. When i'll do, i will post it here .

4. ## Re: Projectile's length of path

It may still be that a lot of that stuff cancels, but Wolfram did not accept the original formula.

Btw, I can already see that the formula cannot be correct.
The numerator should have units that are $\displaystyle m^4/s^4$, but $\displaystyle x^2g$ has the unit $\displaystyle m^3/s^2$

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