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(b) And we use Eq. 422 to solve for the initial height
y
0
:
2
22
00
0
0
11
(
sin
)
0
( 40.3 m/s)(10.0 s)
(9.80 m/s )(10.0 s)
yy v
t g
t
y
θ
−=
−
¡
−
−
which yields
y
0
= 897 m.
27. We adopt the positive direction choices used in the textbook so that equations such as
Eq. 422 are directly applicable. The coordinate origin is at ground level directly below
the release point. We write
0
= –30.0° since the angle shown in the figure is measured
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 Spring '08
 Any
 Physics

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