Question 1160718
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Depends on where the ball was launched. If the ball started at ground level, then the velocity is zero with no direction because it hit the ground at roughly 1.0204 seconds.  On the other hand, if it was launched from the edge of a structure that was in excess of -3.525 meters tall such that the ball when it was falling missed the roof and continued to the ground, then the velocity was downward and the magnitude found by taking the first derivative of the height function relative to the point of launch, *[tex \LARGE h(t)\ =\ 4.9t^2\ +\ 5t], evaluated at *[tex \LARGE t\ =\ 1.5]
								
								
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">
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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