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33. We first find the time it takes for the volleyball to hit the ground. Using Eq. 422, we
have
2
22
00
0
11
(
sin
)
0 2.30 m
( 20.0 m/s)sin(18.0 )
(9.80 m/s )
yy v
t g
t
t
t
θ
−=
−
¡
−
°
−
which gives
0.30 s
t
=
. Thus, the range of the volleyball is
()
cos
(20.0 m/s)cos18.0 (0.30 s)
5.71 m
Rv
t
==
°
=
On the other hand, when the angle is changed to
0
8.00
′
=°
, using the same procedure as
shown above, we find
2
0
(
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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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