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2.46:
a) The vertical distance from the initial position is given by
;
2
1
2
0
gt
t
v
y
y

=
solving for
v
0
y
,
.
s
m
5
.
14
)
s
00
.
5
)(
s
m
80
.
9
(
2
1
)
s
00
.
5
(
)
m
0
.
50
(
2
1
2
0
=
+

=
+
=
gt
t
y
v
y
b)
The above result could be used in
(
29
,
2
0
2
0
2
y
y
g
v
v
y
y


=
with
v
= 0, to solve for
y
–
y
0
= 10.7 m (this requires retention of two extra significant figures in the calculation
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Unformatted text preview: for v y ). c) 0 d) 9.8 2 s m , down. e) Assume the top of the building is 50 m above the ground for purposes of graphing:...
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