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Is there an elegant or easy way to derive the maximum distance from the origin to a point of the parabola created by projectile motion (assume initial position is the origin)? Other than differentiating with respect to $x$ and equate to zero the distance given by $\sqrt{x^{2}+y^{2}}$?

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  • $\begingroup$ You want the maximum distance of all the parabolas with fixed initial speed, or the point of intersect of a parabola to the ground? It is not clear from the question. $\endgroup$ – John Alexiou Jan 20 '14 at 15:24
  • $\begingroup$ The $\theta$ is given but otherwise arbitrary. And yes as @JohnRennie says, the maximum distance from the origin is not necessarily the maximum horizontal distance, it can be other distance from the origin in the course of the flight. In specific what I want is given an initial speed and an angle find the maximum distance from the origin to a point of the projectile's trajectory. $\endgroup$ – ace7047 Jan 20 '14 at 17:18
  • $\begingroup$ There is no maximum distance from the origin to a point. Each point as one distance. There is a maximum distance to any point on the path. $\endgroup$ – John Alexiou Jan 20 '14 at 18:04
  • $\begingroup$ For ONE dimension, you have-$H_{max} = \frac{u^2}{2g}$ $\endgroup$ – GRrocks Jun 28 '15 at 17:08
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    $\begingroup$ Wouldn't this question belong better at Math Stack exchange? It seems like you just want the maximum distance from the origin to a point on an arbitrary parabola, which doesn't really involve much physics does it $\endgroup$ – Joshua Lin Dec 12 '15 at 21:58
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You want the velocity vector to be perpendicular to the radius vector. So calculate the position and trajectory as a function of $t$. When the dot product of the two vectors is zero, that is the maximum distance. This will only work if the trajectory is high enough that the maximum is not a the impact on the ground.

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Actually distance is the complete length traced by the Particle So if you ask about distance for a given angle Theta ( which is Not mentioned in the question Too) Then It is the Point on Same Level wrt to initial Point (if It is projected on Ground) and if theta is arbitary for a particular speed Then Also max distance is at Same Point but for which theta is the max distance Covered can be the question. The distance is perimeter of the Parabola which can be caluclated only by Integration (as far as i know) and after trying It i end up with an implicit equation in theta which by finding got its Maxima nearly at 57.2 degree

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Yes - calculate the time it takes for the object to hit the ground (using the y-component of its motion) and then calculate the total distance travelled (using that time and the x-component of its speed).


If you can't change the path of the parabola, then yes that's what you'l have to do.

Remember that you can just remove the root before differentiating! The max of $\sqrt{x(t)}$ will be at the same t as the max of $x(t)$. - an useful trick you can sometimes use.

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    $\begingroup$ This calculates the horizontal distance travelled, but that might not be the maximum distance from the origin during the flight. Suppose the projectile was sent almost vertically. The maximum distance will be near the top of its trajectory not when it hits the ground. $\endgroup$ – John Rennie Jan 20 '14 at 10:56
  • $\begingroup$ It will be if the user is allowed to choose $\theta$; that's not clear - edited. $\endgroup$ – Kvothe Jan 20 '14 at 11:03

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