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A puck is released from the top of curved, frictionless track. The puck descends, then rises again at the end, such that it leaves the track and continues in free fall until hitting the ground. The ground is assumed to be a plane on the same level as the exit point from the curve.

Question

Does a brachistochrone such as the one shown in the diagram, produce greatest range from the exit point? If not, how can one characterise and calculate the curve that would?

Assumptions

  1. No friction from the slope or the air
  2. The x-axis shown, represents ground level for the purposes of determining maximum range.

Please ask for any necessary clarifications before answering - thanks.

enter image description here

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The speed of the puck at the exit point is fixed: it can be found from energy conservation as $$mgh = \frac{1}{2}mv^2$$ $$v = \sqrt{2gh}.$$ The only way we can control the exit velocity is by adjusting the slope of the track at the exit point. It is a well known result that a launch angle of $45^\circ$ from the horizontal maximizes the projectile range, so as long as the track smoothly curves up to a $45^\circ$ angle at the exit point, the range should be maximum.

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  • $\begingroup$ Thank you. However that does not explain how to get the maximum exit speed and best angle from a curve that descends from height h and ends at height zero. I want to know the curve that will produce the maximum range. This does not necessarily end at 45 degrees. For example, if the exit speed at the end of the track is greater following the descent along a particular path, then this extra speed may compensate for the less than ideal angle. On the other hand, a track that is constrained to end at 45 degrees may not produce the greatest exit velocity. This is not a simple problem. $\endgroup$
    – chasly - supports Monica
    Commented Mar 25, 2023 at 19:34
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    $\begingroup$ @chasly-supportsMonica, as the answer says, the curve doesn't matter. All curves from A to B that are smooth enough to not cause drag leave the ball with the same energy at B. Only the exit angle matters. $\endgroup$
    – BowlOfRed
    Commented Mar 25, 2023 at 19:34
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    $\begingroup$ @chasly-supportsMonica The exit speed (not velocity) is independent of the path, as required by conservation of energy. Yes, this is the claim. $\endgroup$
    – Puk
    Commented Mar 25, 2023 at 19:37
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    $\begingroup$ The Brachistochrone is special in that the ball reaches B in the earliest possible time. But it has the same speed at B as other curves. $\endgroup$
    – BowlOfRed
    Commented Mar 25, 2023 at 19:38
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    $\begingroup$ @chasly-supportsMonica Sure, I assumed that was fixed since it's specified in your diagram. $\endgroup$
    – Puk
    Commented Mar 25, 2023 at 20:02

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