Posted by Anonymous on Tuesday, October 12, 2010 at 1:39am.
The equations are (with distance Y and velocity V measured positive downward)
Y = 6t + 4.9 t^2
V = 6 + 9.8 t
Y=0 corresponds to the point at which it is dropped.
You do the graphing and plugging in of numbers, such as 2.00s for t in parts (a) and (b).
ay is 9.8 m/s^2 at all times it is falling
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