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position for this part of the problem) and take +
x
leftward and +
y
downward. The result
of part (a) implies
v
0
=
R
ω
= 6.3 m/s, and we see from the figure that (with these positive
direction choices) its components are
00
cos30
5.4 m s
sin30
3.1 m s.
x
y
vv
=°
=
=
The projectile motion equations become
xv
t
yv
t g
t
xy
==
+
2
1
2
and
.
We first find the time when
y
=
H
= 5.0 m from the second equation (using the quadratic
formula, choosing the positive root):
2
2
0.74s.
yy
vvg
H
t
g
−+
+
Then we substitute this into the
x
equation and obtain
x
54
074
40
...
ms
s
m
.
bg
b
g
9. (a) We find its angular speed as it leaves the roof using conservation of energy. Its
initial kinetic energy is
K
i
= 0 and its initial potential energy is
U
i
= Mgh
where
6.0sin30
3.0 m
h
=
(we are using the edge of the roof as our reference level for
computing
U
). Its final kinetic energy (as it leaves the roof) is (Eq. 11-5)

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