I did perform averaging over sets of 15 frames (averaging
y and t before calculating v and a) and the conclusion
is essentially the same, it was not a nearly free fall. It
seems that accelerations were actually decreasing in the
second half of the recorded sequence.
t=3.36 s (actually from ~3.12 to 3.59 s) a=6.2 m/s^2
t=3.86 s (actually from ~3.62 to 4.10 s) a=4.6 m/s^2
t=4.36 s (actually from ~4.13 to 4.60 s) a=3.6 m/s^2
If the differences between these three accelerations were
significant then one would say that everage accelerations
(over the 0.5 second intervals) were decreasing during
the fall. Are the differences significant? To answer this
question the timing boundaries were shifted. This gave
the following results (still avaraging over 15 frames).
t=2.99 s (actually from ~2.75 to 3.22 s) a=7.0 m/s^2
t=3.49 s (actually from ~3.26 to 3.73 s) a=6.7 m/s^2
t=3.99 s (actually from ~3.76 to 4.23 s) a=5.0 m/s^2
t=4.50 s (actually from ~4.26 to 4.73 s) a=1.4 m/s^2
The numbers are different, as one would expect, but
the trend remains. It is likely to be a very mon-smooth
trend but I am averging. Let me try another shift.
t=3.16 s (actually from ~2.92 to 3.39 s) a=5.6 m/s^2
t=3.66 s (actually from ~3.42 to 3.89 s) a=6.6 m/s^2
t=4.16 s (actually from ~3.93 to 4.40 s) a=4.4 m/s^2
t=4.66 s (actually from ~4.43 to 4.90 s) a=0.4 m/s^2
I tend to believe that the error in each a is close to 1 m/s^2.
But this is only a guess. It looks that toward the end,
shortly before the visibility was lost, the motion was at
nearly constant speed rather than at nearly constant close
to 9.8 m/s^2. I have no idea why there is a dramatic
discrepancy between the two conclusions. Something
must be wrong somewhere; we are using the same row
data and the results should be similar.
The frames at t<2.0 s are more difficult because velocities
were smaller. That is why I decided to stay above the
low t region. Let us focus on the second half of the
recording in order to address the discrepancy between
the two conclusions. Note that the traditional textbook
method, which I use, does not depend on knowing
where the object was at time zero.
The chi-sqr method could not be used because I was not
trying to justify a particular hypothesis. Oh yes, I could
have used this method to show that my accelerations are
not consistent with the nearly free fall. But why do I need
chi-sqr to justify this obvious conclusion?
P.S.
The attached text file may be opened and used in a
program, for example, True Basic or Fortran. Each
line of that file has only two numbers, t in seconds and
|y| in meters. The numbers are from the first and last
columns of the spreadsheet. Three "duplicate t" frames
were removed, as suggested by Glenn; in his spreadsheet
these three frames can easilly be identified by t=1.913,
3.659 and 4.263 seconds.
Ludwik Kowalski
.067 , -.689079
.1 , .689079
.134 , 0
.167 , .689079
.201 , 0
.235 , .689079
.268 , 0
.302 , .689079
.335 , -.689079
.369 , .689079
.402 , 0
.436 , 0
.47 , 0
.503 , -.689079
.537 , .689079
.57 , 0
.604 , -.689079
.637 , -.689079
.671 , 0
.705 , -.689079
.738 , 0
.772 , 0
.805 , -.689079
.839 , 0
.872 , -2.06724
.94 , -.689079
.973 , -1.37816
1.007 , -1.37816
1.04 , -2.06724
1.074 , -1.37816
1.107 , -1.37816
1.141 , -1.37816
1.175 , -2.06724
1.208 , -1.37816
1.242 , -1.37816
1.275 , -2.75632
1.309 , -2.06724
1.342 , -2.75632
1.376 , -2.75632
1.41 , -3.4454
1.443 , -3.4454
1.477 , -2.06724
1.51 , -2.75632
1.544 , -2.75632
1.577 , -2.75632
1.611 , -2.75632
1.645 , -3.4454
1.678 , -3.4454
1.712 , -3.4454
1.745 , -3.4454
1.779 , -3.4454
1.812 , -4.13448
1.846 , -4.13448
1.88 , -4.82356
1.913 , -4.82356
1.947 , -4.82356
1.98 , -4.82356
2.014 , -5.51264
2.047 , -6.20171
2.081 , -6.20171
2.115 , -6.20171
2.148 , -6.89079
2.182 , -6.89079
2.215 , -7.57987
2.249 , -8.26895
2.282 , -8.26895
2.316 , -8.95803
2.35 , -9.64711
2.383 , -10.3362
2.417 , -10.3362
2.45 , -11.0253
2.484 , -11.7144
2.517 , -12.4034
2.551 , -12.4034
2.585 , -13.0925
2.618 , -13.7816
2.652 , -14.4707
2.685 , -15.1597
2.719 , -15.1597
2.752 , -15.8488
2.786 , -16.5379
2.82 , -16.5379
2.853 , -17.227
2.887 , -17.9161
2.92 , -18.6051
2.954 , -18.6051
2.987 , -19.2942
3.021 , -19.9833
3.055 , -20.6724
3.088 , -22.0505
3.122 , -22.0505
3.155 , -23.4287
3.189 , -23.4287
3.222 , -24.1178
3.256 , -24.8069
3.29 , -25.4959
3.323 , -25.4959
3.357 , -26.8741
3.39 , -27.5632
3.424 , -28.2523
3.457 , -28.9413
3.491 , -29.6304
3.525 , -30.3195
3.558 , -31.0086
3.592 , -31.6977
3.625 , -32.3867
3.659 , -33.0758
3.692 , -33.0758
3.726 , -34.454
3.76 , -35.1431
3.793 , -35.8321
3.827 , -36.5212
3.86 , -37.8994
3.894 , -37.8994
3.927 , -39.2775
3.961 , -39.9666
3.994 , -40.6557
4.028 , -41.3448
4.062 , -42.0338
4.095 , -42.7229
4.129 , -43.412
4.162 , -44.1011
4.196 , -45.4792
4.229 , -46.1683
4.263 , -46.8574
4.297 , -46.8574
4.33 , -48.2356
4.364 , -48.9246
4.397 , -50.3028
4.431 , -50.9919
4.464 , -51.681
4.498 , -51.681
4.532 , -52.37
4.565 , -53.0591
4.599 , -54.4373
4.632 , -55.1264
4.666 , -55.8154
4.699 , -56.5045
4.733 , -57.1936
4.767 , -58.5718
4.8 , -59.2608
4.834 , -59.9499
4.867 , -60.639
4.901 , -61.3281
4.934 , -62.7062
4.968 , -63.3953