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3.23:
a) With
0
=
y
v
in Eq. (3.17), solving for
t
and substituting into Eq. (3.18) gives
m
6
.
13
)
m/s
80
.
9
(
2
0
.
33
sin
)
m/s
0
.
30
(
2
sin
2
)
(
2
2
2
0
2
2
0
2
0
0
=
°
=
=
=

g
v
g
v
y
y
y
α
b) Rather than solving a quadratic, the above height may be used to find the time the
rock takes to fall from its greatest height to the ground, and hence the vertical component
of velocity,
m/s
7
.
23
)
m/s
m)(9.80
6
.
28
(
2
2
2
=
=
=
yg
v
y
, and so the speed of the
rock is
m/s
6
.
34
))
33.0
m/s)(cos
0
.
30
((
m/s)
7
.
23
(
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Unformatted text preview: 2 2 = ° + . c) The time the rock is in the air is given by the change in the vertical component of velocity divided by the acceleration –g ; the distance is the constant horizontal component of velocity multiplied by this time, or m. 103 ) m/s 80 . 9 ( )) m/s)sin33. ((30.0 m/s 23.7 ( m/s)cos33. . 30 ( 2 =°° = x d)...
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