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Chapter 3
The negative sign is present because air resistance acts in opposition to the motion of the
object. Therefore the net force acting on the shell is
F
=
F
1
+
F
2
=
−
mg
−
(0
.
1)
v
2
.
We now apply Newton’s second law to obtain
m
dv
dt
=
−
±
mg
+(0
.
1)
v
2
²
.
Because the initial velocity of the shell is 500 m/sec, a model for the velocity of the rising
shell is expressed as the initialvalue problem
m
dv
dt
=
−
±
mg
.
1)
v
2
²
,v
(
t
= 0) = 500
,
(3.11)
where
g
=9
.
81. Separating variables, we get
dv
10
mg
+
v
2
=
−
dt
10
m
and so
Z
dv
10
mg
+
v
2
=
−
Z
dt
10
m
⇒
1
√
10
mg
tan
−
1
³
v
√
10
mg
´
=
−
t
10
m
+
C.
Setting
m
=3,
g
.
81 and
v
= 500 when
t
= 0, we Fnd
C
=
1
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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