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109. We make use of Eq. 425.
(a) By rearranging Eq. 425, we obtain the initial speed:
v
xg
xy
0
00
2
=
−
cos
( tan
)
θθ
which yields
v
0
= 255.5
≈
2.6
×
10
2
m/s for
x
= 9400 m,
y
= –3300 m, and
θ
0
= 35°.
(b) From Eq. 421, we obtain the time of flight:
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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, Light

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