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ground is
v
o
→
= (
v
ox
–
v
s
) i
^
+
v
oy
j
^
.
The horizontal and vertical displacement (relative to
the ground) are therefore
x
land
–
x
launch
=
Δ
x
bg
= (
v
ox
–
v
s
)
t
flight
y
land
–
y
launch
=
0
= v
oy
t
flight
+
1
2
(
−
g
)(
t
flight
)
2
.
Combining these equations leads to
Δ
x
bg
=
2 v
ox
v
oy
g
–
©
¨
§
¹
¸
·
2v
oy
g
v
s.
The first term corresponds to the “
y
intercept” on the graph, and the second term (in
parentheses) corresponds to the magnitude of the “slope.” From Figure 454, we have
40 4 .
bg
s
x
v
Δ=−
This implies
v
oy
= (4.0 s)(9.8 m/s
2
)/2 = 19.6 m/s, and that furnishes enough information to
determine
v
ox
.
(a)
v
ox
= 40
g
/2
v
oy
= (40 m)(9.8 m/s
2
)/(39.2 m/s) = 10 m/s.
(b) As noted above,
v
oy
= 19.6 m/s.
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

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