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(a) Different approaches are available, but since it will be useful (for the rest of the
problem) to first find the initial
y
velocity, that is how we will proceed. Using Eq. 216,
we have
22
2
10
0
2
(6.1 m/s)
2(9.8 m/s )(9.1 m)
yy
y
vv
g
y
v
=−
Δ
¡
which yields
v
0
y
= 14.7 m/s. Knowing that
v
2
y
must equal 0, we use Eq. 216 again but
now with
Δ
y
=
h
for the maximum height:
2
2
20
2
0 (14.7 m/s)
2(9.8 m/s )
g
h
h
¡
which yields
h
= 11 m.
(b) Recalling the derivation of Eq. 426, but using
v
0
y
for
v
0
sin
θ
0
and
v
0
x
for
v
0
cos
0
,
we have
2
00
1
0,
2
yx
vt
g
t R vt
=
which leads to
2/
.
xy
R
vv g
=
Noting that
v
0
x
=
v
1
x
= 7.6 m/s, we plug in values and
obtain
R
= 2(7.6 m/s)(14.7 m/s)/(9.8 m/s
2
) = 23 m.
(c) Since
v
3
x
=
v
1
x
= 7.6 m/s and
v
3
y
= –
v
0
y
= –14.7 m/s, we have
2
2
33
3
(7.6 m/s)
( 14.7 m/s)
17 m/s.
v
=+
=
+
−
=
(d) The angle (measured from horizontal) for
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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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