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87. Using Eq. 216, we obtain
22
0
2
vv
g
h
=−
, or
0
()
/
2
.
hvv
g
(a) Since
0
v
=
at the maximum height of an upward motion, with
0
7.00 m/s
v
=
, we
have
(7.00m/s) /2(9.80m/s )
2.50 m.
h
==
(b) The relative speed is
0
7.00 m/s 3.00 m/s
4.00 m/s
rc
vvv
=−=
−
=
with respect to the
floor. Using the above equation we obtain
(4.00 m/s) / 2(9.80 m/s )
0.82 m.
h
(c) The acceleration, or the rate of change of speed of the ball with respect to the ground
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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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