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1
2
gt
1
,
l
1
¼
l
0
þ
V
0
t
1
þ
1
2
gt
2
, and
l
2
¼
l
0
þ
V
0
t
2
þ
1
2
gt
3
:
l
3
¼
l
0
þ
V
0
t
3
þ
Subtracting the first equation from the second and third equations, respectively,
the results are as follows:
9
=
;
1
2
gT
1
t
1
þ
L
1
¼
V
0
T
1
þ
ð
t
2
Þ
,
ð
2
:
26
Þ
1
2
gT
2
t
1
þ
L
2
¼
V
0
T
2
þ
ð
t
3
Þ
where L
1
¼
l
1
are the distances from the first position to the
second and third positions, respectively. T
1
¼
l
2
l
1
and L
2
¼
l
3
t
1
are the
times taken by the falling body in its motion from the first position to the second
and third positions. To eliminate V
0
, the two equations in (2.26) are divided by T
1
and T
2
, respectively, and the two results thus achieved subtracted from each other,
which reads as:
t
2
t
1
and T
2
¼
t
3
L
1
T
1
L
2
T
2
¼
1
2
gt
2
ð
t
3
Þ:
Since t
2
t
3
¼
T
1
T
2
, the formula of g can finally be written as:
2
T
2
L
2
T
2
L
1
T
1
g
¼
:
ð
2
:
27
Þ
T
1
Thus it can be seen that to determine gravity using the free-fall method requires
knowledge of the distances L
1
and L
2
traveled by the falling body within the time
periods T
1
and T
2
.
Rise-and-Fall Method
In this method, an object is thrown vertically upward and then allowed to fall freely.
To obtain the gravitational acceleration g, it is necessary to label two positions S
1
and S
2
in its course of motion. The time intervals T
1
and T
2
of the falling body past
each position are determined, cf. Fig.
2.20
. The transverse axis indicates time and
the ordinate axis indicates the vertical position of the falling body. Let H
1
and H
2
be
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