### Boyle’s Law Apparatus

```Acceleration Due to
Gravity: Free Fall
By: Rosa Marie Kinder and Kenisha
Marshall
SC441 Honors Class, Fall 2002, Dr. Roman Kezerashvili
Objective
To verify that the displacement of a freely
falling body released from rest is directly
proportional to the square of the elapsed
time.
To determine from the experiment the value
for g, the acceleration due to gravity.
To verify that the time of free fall does not
depend on the mass of the falling object.
Theory
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Objects acted upon only by the force of gravity are in
free fall. If we allow an object to fall in vacuum so that air
resistance does not affect its motion, we find a
remarkable fact, I all bodies, regardless of their size,
shape or composition, fall at the same acceleration the
same point near the Earth’s surface.
When an object is in free fall (whether up or down)
objects are accelerated (or decelerated) at a constant
rate by gravity, its magnitude, by the symbol g, and at or
near the Earth’s surface g is approximately 9.81m/s^2.
this is the standard for g.
Equations
Procedure
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Part One
The free fall apparatus was previously set up.
The free fall pad was positioned under the release mechanism.
The steel ball was placed in the release mechanism and secured, but not
yet released.
Connection to Digital Channel one as made on the Science Workshop
Interface.
The distance of the free falling object was measured at 0.3m.
The ball was released and the data was recorded.
The ball was dropped at the same distance for three times, enabling us to
calculate the mean.
These steps were carried out for several other heights: 0.5m, 0.7m, 0.9m,
1.1m, 1.3m, and 1.5m.
Part two
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The same steps were repeated from part
one except the height was changed to 1.0
meters and two balls of different weights
were used in order to verify that the time of
free falling objects do not depend on the
mass of the falling objects.
Excuse me,
the point, we
are running
out of time
No
problem
Professor
Data
Height
Y,m
Average t,s
Square of
average Time
Velocity from
eq. (4)
Height
y=v^2/2g,m
% Difference
for y
0.3
0.237
0.056
2.32
0.27
-10.5
0.5
0.319
0.101
3.13
0.56
11.3
0.7
0.376
0.141
3.69
0.69
1.43
0.9
0.428
0.183
4.20
0.90
0
1.1
0.465
0.216
4.56
1.06
-3.70
1.3
0.513
0.263
5.03
1.29
-0.77
1.5
0.511
0.303
5.40
1.49
-0.69
Verification that free fall does not depend
on the mass of the falling object.
Mass of
ball, g
Average t, s Acceleration
28.4
0.450
9.32
0.52
16.0
0.453
9.74
0.53
% error
due to g from
eq. (7) m/s^2
Height vs. Time
Graph(1) : Determination of g from the slope of the graph y
versus t
1.6
Height(m)
1.4
y = 4.8977x1.9957
1.2
1
0.8
0.6
0.4
0.2
0.2
0.25
0.3
0.35
0.4
Time(s)
0.45
0.5
0.55
0.6
Height vs. Time^2
Grap(2) : Determination of g from the slope of the graph y
versus t^2
1.5
Height( m)
1.3
y = 4.9032x + 0.0041
1.1
0.9
0.7
0.5
0.3
0.01
0.06
0.11
0.16
Time(s^2)
0.21
0.26
Graphs
Velocity vs. Time
Velocity(m/s)
Graph(3) : Instantaneous velocity v versus time t
6
5.5
5
4.5
4
3.5
3
2.5
2
1.5
1
y = 9.8033x + 0.0018
0.2
0.25
0.3
0.35
0.4
Time(s)
0.45
0.5
0.55
0.6
Questions
A ball is dropped from
the roof of a building.
1
The ball strikes the
y  gt
2
ground after 6 s. How
1
y  (9.8) (6 )
2
tall, in meters is the
y  176.4m
building?
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2
2
More Questions
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What is the magnitude
of the ball’s velocity
just before it reaches
the ground?
Yes, More Questions.
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A student throws a ball vertically
downward from the top of the
building. The ball leaves the
thrower’s hand with a speed of
10m/s.What is the speed after
falling for 1s, 2s, 3s?
How far does it fall in 1s, 2s, 3s?
y1 = 10 + 4.9(9)
y1 = 14.9m
y2 = 10 + 4.9(4)
y2 = 29.6m
y3 = 10 + 4.9(9)
y3 = 54.1m
v  10 9.8(1)
1
v1  19.8m / s
Last Question!
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What is the magnitude
of its velocity after
falling 10m?
Conclusion
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The experiment was conducted
successfully and we were able to obtain a
value for g (the acceleration due to gravity)
and compare it with the standard. This
experiment has shown that the acceleration
of free fall is independent of the mass of
the falling object.
The End
Thank You
```