I would like to build a curve between two points A and B. A ball would roll down the curve in a gravitational uniform field (i.e., I'm actually going to build the thing here on Earth).

My question is, how can I ensure that my curve maximises the vector v at point B. enter image description here

I would like to chose v's orientation in advance (for example at 45º), and maximise |v|. There may be an obvious connection to the Brachistochrone, but I don't see it right now. Any ideas appreciated.

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    $\begingroup$ Conservation of energy. Choose any curve you like, and any final direction : the speed will be the same. $\endgroup$ Nov 25 '17 at 21:25

The speed of a ball starting from $A$ at rest and going to $B$ without friction is fixed by the difference in height between $A$ and $B$. In particular, if the ball has mass $m$, and we take $A$ to be at zero height, while $B$ is at height $h$, then by conservation of energy: $$0+mgh=\frac{1}{2}mv^2+0 \implies v=\sqrt{2gh}$$ If we take into account the fact that the ball has a size and is rolling, as long as friction is negligible (that is, the ball rolls without slipping), then the result is numerically slightly different, but still independent of the path taken (again, by conservation of energy).

Notice that the velocity is parallel to the path, so the direction of the final velocity is determined only by the shape of the final part of the path.

If you are trying to maximise velocity, then you should place $A$ and $B$ as far away as possible in height. At fixed $A$ and $B$, if we ignore friction the final speed, is determined as explained above. Therefore what you should be doing in practice is trying to minimise friction - but that's an engineering issue on which I can offer little help.

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    $\begingroup$ Sorry I don't understand your question. "given A, B and the angle of v", the final velocity is still fixed. A brachistocrone is a curve which minimises the time it takes to travel between two points. It has little to do with final velocity $\endgroup$
    – John Donne
    Nov 26 '17 at 0:19
  • $\begingroup$ I misunderstood your answer. You are absolutely right. The velocity is independent of the path followed due to conservation of energy. I was pretty thick today. $\endgroup$
    – Massagran
    Nov 26 '17 at 0:23
  • $\begingroup$ No problem! We all have our bad days $\endgroup$
    – John Donne
    Nov 26 '17 at 0:24

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