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September 19, 2014

September 19, 2014

Posted by **Cody** on Friday, February 3, 2012 at 5:28pm.

- physics -
**Steve**, Friday, February 3, 2012 at 5:44pmThe height of the rocket fired at speed v at angle a is given by

y(x) = -g*sec^2(a)/2v^2 x^2 + x*tan(a)

= -.0065x^2 + 2.55x

y(26) = 61.9m so it clears the 11-m wall by 50.9m

Makes sense, since if the rocket went in a straight line at an angle of 68.6°, at 26m out it would be 66.3m high.

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