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October 30, 2014

October 30, 2014

Posted by **walter hughes** on Sunday, October 20, 2013 at 8:19am.

- math -
**Steve**, Sunday, October 20, 2013 at 12:27pmso, plug in your numbers. You don't say what m is, but I assume it's the slope of the trajectory at kickoff

y = (1/2)x - 32/(2*80^2)(1+1/4)x^2

= 1/320 (160x-x^2)

this is a parabola with vertex at (80,20)

so, knowing that, what's the max height?

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