Re: free fall data

[Raacc@aol.com]

Ludwik,

I plotted your position versus time data and saw a stray point at t = 7.33
seconds.  Throwing out every other data point, and using the finite
derivative I posted earlier, I arrive at this table:

t(s)	y(m)	v(m/s)	a(m/s/s)
0.367	0.000		
0.433	0.047	1.050	
0.500	0.140	1.680	9.732
0.567	0.271	2.348	10.236
0.633	0.453	3.045	10.237
0.700	0.677	3.713	10.240
0.767	0.948	4.410	10.516
0.833	1.265	5.115	
0.900	1.630		

I don't recommend you use the one-sided-two-point finite derivative method.
 It finds a slope somewhere between the two times you are using.  For
example, in your table you calculate the velocity at time t, using the
positions at time t and the previous time.  This will give a velocity at a
time somewhere between these two times, not at time t.  I recommend you use
positions and times on either side of the time where you want to calculate
the velocity (see my previous post).  The same is true for acceleration
calculations.  If you proceed with your one-sided-two-point finite
derivative, find the slope and intercept from the velocity plot, and generate
equations, then you will find your equations will not match the data since
the initial velocity will be incorrect.  This will be confusing to students.

Bob Carlson