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54. We apply Eq. 421, Eq. 422 and Eq. 423.
(a) From
Δ
xv
t
x
=
0
, we find
0
40 m / 2 s
20 m/s.
x
v
==
(b) From
Δ
yv
t g
t
y
=−
0
1
2
2
, we find
()
22
1
0
2
53 m
(9.8 m/s )(2 s)
/ 2
36
y
v
=+
=
m/s.
(c) From
vv
g
t
y
y
′
0
with
v
y
= 0 as the condition for maximum height, we obtain
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 Spring '08
 Any
 Physics

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