Question 1165015
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Resultant velocity magnitude:  *[tex \Large \sqrt{(-9)^2\,+\,16^2}]


Resultant velocity direction: *[tex \Large \arctan\(\frac{-9}{16}\)]  you can represent this as an angle in the range *[tex \Large 270^\circ\ <\ \theta\ <\ 360^\circ] or *[tex \Large -90^\circ\ <\ \theta\ <\ 0^\circ]


Time to reach the other side:  Divide the distance across by the magnitude of horizontal component of the velocity.


Distance downstream:  Multiply the time value from the previous part by the magnitude of the downstream component of the velocity

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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