is the initial speed and
θ
0
is the firing angle. The coordinates of the highest point on the
trajectory are
()
2
2
0
00
0
2
20 m/s
cos
sin
cos
sin 60 cos60
17.7 m
9.8 m/s
x
v
xv
tv
t
g
θθ
==
=
=
°
°
=
and
yv
t g
t
v
g
y
=−
=
=
=
°
0
2
0
2
2
0
2
2
1
2
1
2
1
2
20
98
60
153
sin
.
.
m / s
m/s
m.
2
b
g
Since no horizontal forces act, the horizontal component of the momentum is conserved.
Since one fragment has a velocity of zero after the explosion, the momentum of the other
equals the momentum of the shell before the explosion. At the highest point the velocity
of the shell is
v
0
cos
0
, in the positive
x
direction. Let
M
be the mass of the shell and let
V
0
be the velocity of the fragment. Then
Mv
0
cos
0
=
MV
0
/2, since the mass of the
fragment is
M
/2. This means
Vv
00 0
2
2 20
60
20
=
°
cos
cos
m/s.
bg
This information is used in the form of initial conditions for a projectile motion problem
to determine where the fragment lands. Resetting our clock, we now analyze a projectile
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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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