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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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