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v
0
y
–
gt
again but now “starting the clock” at the highest point so that
v
0
y
= 0 (and
1.0 s
t
=
). This leads to
v
y
= –9.8 m/s and
()
2
2
(10 m/s)
9.8 m/s
14 m/s
+−
=
.
(c) The
x
0
value may be obtained from
x
= 0 =
x
0
+ (10 m/s)(1.0s), which yields
0
10m.
x
=−
(d) With
v
0
y
= 9.8 m/s denoting the
y
component of velocity one second before the top of
the trajectory, then we have
yy
v
t
g
t
y
== +
−
0
00
1
2
2
where
t
= 1.0 s. This yields
0
4.9 m.
y
(e) By using
x
–
x
0
= (10 m/s)(1.0 s) where
x
0
= 0, we obtain
x
= 10 m.
(f) Let
t
= 0 at the top with
0
y
yv
==
. From
2
1
2
y
y
t
g
t
−=
−
, we have, for
t
= 1.0 s,
22
(9.8 m/s )(1.0 s) / 2
4.9 m.
y
125. At maximum height, the
y
component of a projectile’s velocity vanishes, so the
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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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