hillbillybob
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waynejkruse10 said:or 9.8 meters/second/second.
no, it is 9.8 meters / second ^ 2
This is like he said, the acceleration due to gravity, or the acceleration at which a falling object would reach terminal velocity. That said you can do the math with the following equation
x = (-b +- sqrt(b^2-4ac)) / 2a
Now someone check me on this, but the following:
Where a = initial height
b = acceleration due to gravity
c = mass of object
TI-83 graphing calcs work wonders, easy to program too