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Professor Line
PHYS 163 A
Albert Square
April 1, 2024
Experiment #0
Free Fall
My Partners: Emmy Noether
Billy Cobham
Activity Goals:
The goal of this activity is to determine if free-fall motion correctly describes the motion of a
500-g steel ball falling vertically through air.
Procedure:
The procedure used for this activity can be found on pp. 300-305 of the Phys 163 Fall 2024 Lab
Manual.
1
Data:
A 500-g steel ball was dropped from four different measured heights, and the time to reach the ground
was measured.
Ball mass: 500 ± 1 g
Initial
Height 1
y (m)
3.5
±
0.1
Measure
±
1
2
3
Fall
time t
(s)
0.82
0.88
0.86
Avg t
Fall Time
0.85
0.85
0.02
0.03
Initial
Height 3
y (m)
10.5
±
0.1
Measure
±
1
2
3
Fall
time t
(s)
1.48
1.50
1.45
Avg t
Fall Time
1.48
1.48
0.02
0.03
Trial
Trial
0.02
0.02
0.02
0.02
0.02
0.02
Initial
Height 2
y (m)
7
Deviation
from
average
0.03
0.03
0.01
Initial
Height 4
Deviation
from
average
0.00
0.02
0.03
±
0.1
Measure
±
1
2
3
Fall
time t
(s)
1.19
1.20
1.19
Avg t
Fall Time
1.19
1.19
0.02
0.02
Measure
±
1
2
3
Fall
time t
(s)
1.64
1.66
1.63
Avg t
Fall Time
1.64
1.64
0.02
0.02
Trial
y (m)
13.5
0.02
0.02
0.02
Deviation
from
average
0.00
0.01
0.00
±
0.1
Trial
0.02
0.02
0.02
Deviation
from
average
0.00
0.02
0.01
Measurement Uncertainties:
The following quantities were measured:
• Initial height, H (m)
• Time of fall, t (s)
2
Uncertainty in initial height: 10 cm
The ball was dropped by hand from an open window located on each floor of Kirkbride Hall
and its roof. The distance from the location of release to the ground was determined from the
length of a string, weighted at one end and lowered until it reached the ground. Using a 3meter ruler, with reading uncertainty of 0.001 m, the reading uncertainty associated with the
largest distance was estimated at 0.005 m. However, because the ball was dropped by hand
reaching out the open window, an initial height uncertainty of 0.1 m was chosen to account for
a lack of precision in locating the hand at the same height as the bottom of the window.
Uncertainty in fall time: 0.02 s
A digital stopwatch, which displayed time to 0.01 seconds, was used. It was assumed that the
stopwatch worked correctly, so the reading uncertainty was taken to be 0.005 s. However, to
account for reaction times (starting and stopping the stop watch just at the beginning and end of
the fall) an uncertainty of 0.02 s was used. This was determined by making several trials of
timing a second hand on the lab’s wall clock as it ticked-off 2 seconds.
Analysis of Data:
Sample Calculations:
The following calculations were made:

Initial height, using a 3-meter ruler. As an example, the following are the calculations used in
finding height 4, along with its uncertainty:
H4 = (3±.001) + (3±.001) + (3±.001) + (3±.001) + (1.5±.001) =13.5± 0.005 m
δH4 = 0.001+ 0.001+ 0.001+ 0.001+ 0.001= 0.005 m

Absolute difference from the average – the absolute deviation
For example, the deviation associated with gcalc for the height 4 (H4) data:
10.0 −9.7 = 0.3 ms 2

Average value
3
The average value of N measured or calculated quantities, as well as the average uncertainty
for the average value, is determined using the formula
xavg
1 N xi
For example, to determine the average measured value of t (for Initial Height 1 data):
tavg = (0.82 + 0.88+ 0.86) s = 0.85 s
3
For repeated measurements or calculations, the reported uncertainty for the average is
determined by the larger of (the average of the individual uncertainties) and (the largest
absolute difference between the individual quantities and the average of those quantities). The
absolute difference from the average is given by xi − xavg .

Uncertainty for repeated measurements
The largest of the deviation values was compared to the average reading uncertainty. The
larger of these two was the reported uncertainty:
For the Initial Height 1 data, 0.03 > 0.02, and the reported uncertainty is 0.03 s.

δ tavg = 0.03 s
Calculation of the value for g
The value of g was calculated using the constant acceleration ∆y = vo,y∆t − g(∆t)2
For example, using the height 4 data:
(
)

g =−(2∆ t)y2 =− 2×(1 .−6413s.)52m =10.0ms2
δg = g
δ∆(∆yy)+δ(∆∆tt)+δ(∆∆tt)
10..6402
=10.0
130..15 + 10..6402 +
=10.0(0.0074 + 0.012 + 0.012)
4
=10.0(0.031)= 0.3 ms 2
• Comparison of gcalc to gtheory by % difference
9.7 −9.8
×100% = 0.8%
9.8

Comparison of gcalc to gtheory by agreement within uncertainty
For Initial Height 3 data, g = 9.6 ± 0.4 m/s2. This means the experimental value falls
somewhere within the range
9.2ms 2 ≤ gexperimental ≤10.2ms 2
The accepted value for g is 9.81 ± 0.01 m/s2, which falls within the experimentally-determined
range.
Results:

2H
Calculation of g, assuming free-fall analysis: g = t 2
Table R1 – Calculated values of g

H (m)
±H
t (s)
±t
3.5
0.1
0.85
0.03
9.6
1.0
7.0
10.5
13.5
0.1
0.1
0.1
1.19
1.48
1.64
0.02
0.03
0.02
9.8
9.6
10.0
0.5
0.4
0.3
gcalc (m/s2)
± gcalc
Summary of calculations of g from data: gtheory given in lab manual
Table R2 – Comparison of gexperimental to gtheoretical
5
9.6
±
gexp
1.0
9.8
9.6
10.0
0.5
0.4
0.3
gexp (m/s2)
9.81
±
gtheory
0.01
9.81
9.81
9.81
0.01
0.01
0.01
gtheory (m/s2)
% Diff
Agree?
2.0
Yes
0.2
1.8
1.9
Yes
Yes
Yes
Conclusions and Discussion
When a steel ball is dropped and falls through the air, both the gravitational and air drag forces will act
on the ball.
For free fall motion with constant gravitational force, the acceleration of the ball should be the
same value for all initial heights, and equal to 9.81 m/s2. If air resistance had a significant effect,
the acceleration should decrease with increasing initial height, and be less than 9.81 m/s2.
The results support the conclusion that free-fall motion describes the motion of the falling steel
ball over a range of heights up to 13.5 m.
This conclusion is supported by both graphical and numerical analysis.
For free fall motion with constant gravitational force, the value of g near Earth’s surface should be
9.81 ± 0.01 m/s2 ( source: Lab Manual, p. 302 ). Tables R1 and R2 in the Results section provide a
summary of the experimentally determined values of g for different heights.
Table R2 shows that the experimentally-determined acceleration of the steel ball is consistent with 9.81
m/s2 for all four heights investigated – each experimental value agrees with the accepted value of
9.81 ± 0.01 m/s2 within their uncertainties, with percentage differences between 0.2% and 2%.
In addition, Graph 1: Calculated values of g for different fall distances, gives results consistent with a
value for g which is independent of the initial height. This is seen from the fact that it is possible to
draw a horizontal line through the data + error bars; a horizontal line on a graph indicates that the
“dependent variable” is independent of the “independent variable.”
It can also be noted that, if air drag was significant, its effect would be to reduce the apparent
value of g with increasing height. On the graph Calculated values of g for different fall distances it
can be seen that a straight line of negative slope could also fit the results, which would be consistent
with a conclusion that air resistance plays a part in the motion. However, the data points themselves do
not appear to have any obvious negative slope trend, so a conclusion of significant air resistance
effects is not very well supported by the data. (See answer to Question below.)
In summary, the results of this activity supports the assumption that a 500-g steel ball, falling from a
height of 13.5 m or less, can be considered as having free fall motion.
6
Question:
1) How would the presence of air resistance show up in the data?
If air resistance were present, the net force acting on the falling ball would be decreased (Fg down,
but FR up, opposite the direction of travel), so the acceleration of the ball would be reduced from
its free fall value. Because the effects of air resistance increase with speed, the effects should also
increase with initial height (the greater the fall distance, the longer the fall time, and the longer the
time for the ball to accelerate). Therefore, the calculated value of g should decrease with
increasing initial height. A possible line that would support the presence of air resistance is shown
and labeled in Graph 1.
7
PHYS 164
LAB #21 The Speed of Sound in Air
DATA
f
(Hz)
1000
1500
2000
2500
3000
3500
4000
3rd
58,3
38,2
27,4
21,3
17,8
14,7
12,7
dL
(cm)
0,3
0,3
0,3
0,3
0,3
0,3
0,3
l/2
2nd-1st
16,9
11,4
8,9
7,1
5,8
4,9
4,3
l/2
3rd-2nd
17,1
11,8
8,1
6,7
5,9
4,9
4,3
d (l/2)
(cm)
0,6
0,6
0,6
0,6
0,6
0,6
0,6
Avg l
(m)
0,34
0,23
0,17
0,14
0,12
0,10
0,09
dl
(m)
0,01
0,01
0,01
0,01
0,01
0,01
0,01
v = fl
(m/s)
340
348
340
345
351
343
344
dv
(m/s)
15
21
27
33
40
45
51
vaverage
d(vaverage)
vexpected
d(vexpected)
(m/s)
344,4
(m/s)
33,4
(m/s)
345
(m/s)
1
% Diff
0,17
Agree?
yes
1st
24,3
15,0
10,4
7,5
6,1
4,9
4,1
L (cm)
2nd
41,2
26,4
19,3
14,6
11,9
9,8
8,4
Avg l/2
(cm)
17,0
11,6
8,5
6,9
5,85
4,9
4,3
ANALYSIS
f
(Hz)
1000
1500
2000
2500
3000
3500
4000
Dev from
Average
4,4
3,6
4,4
0,6
6,6
1,4
0,4
Speed Vs. frequency
450
400
Speed (m/s)
350
300
250
200
150
100
50
0
0
500
1000
1500
2000
2500
Frequency (Hz)
3000
3500
4000
4500

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