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()
2
00
0
1
sin
1.10m
2
yy v
t
g
t
θ
=+
−
=
which implies it does indeed clear the 0.90 m high fence.
(b) At
t
= 0.508 s, the center of the ball is (1.10 m – 0.90 m) = 0.20 m above the net.
(c) Repeating the computation in part (a) with
0
= –5.0° results in
t
= 0.510 s and
0.040 m
y
=
, which clearly indicates that it cannot clear the net.
(d) In the situation discussed in part (c), the distance between the top of the net and the
center of the ball at
t
= 0.510 s is 0.90 m – 0.040 m = 0.86 m.
32. We adopt the positive direction choices used in the textbook so that equations such as
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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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