Hello,(adsbygoogle = window.adsbygoogle || []).push({});

New here. I have designed a segment of a track that I am building that I need to pull some data out of. It is a high-to-low transition (down hill) that takes the form of cosine over the domain [0,pi]. Thea graph of the transition's profile is y=5.5cos(xpi/18)+5.5 It is 11 inches high and 18 inches long.

I want to find out its exit velocity (x=18) for a ~100g ball that starts form rest at the top. The graph of the formula I used above is position-vs-position. I know that velocity is the graph of position-vs-time's first derivative.

How do I get from the position-vs-position graph/formula to the position-vs-time graph/formula? I think I could do it if the slope was constant, but I'm not sure about the curve and I don't want to average the slope. Shouldn't it be the same graph, just compressed along the x-axis?

Thanks,

Chopper

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# Kinematics of a ball and track

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