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26. (a) Using the same coordinate system assumed in Eq. 422, we solve for
y
=
h
:
2
00
0
1
sin
2
hy v
t
g
t
θ
=+
−
which yields
h
= 51.8 m for
y
0
= 0,
v
0
= 42.0 m/s,
0
= 60.0° and
t
= 5.50 s.
(b) The horizontal motion is steady, so
v
x
=
v
0
x
=
v
0
cos
0
, but the vertical component of
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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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