Problem 1.
Solution:
In order to solve this problem, it is simplest to determine the absolute velocity vector for each
projectile when it begins its fight. Once we have established what the initial velocities are, we can compute
the projectiles’ positions from
d
2
r
/dt
2
=
−
g
k
.
Let subscripts
h
and
t
denote the Humber and the tank, respectively. Also let subscripts
b
and
a
denote
the
shell buster
and the artillery shell, respectively. We are given the velocities of each projectile relative
to the vehicle from which it is launched, and the given velocities are
v
b/h
=
v
cos
φ
i
+
v
sin
φ
j
and
v
a/t
=
−
√
3
V
cos
φ
i
+
√
3
V
sin
φ
j
The vehicles’ velocity vectors are
v
h
=
V
i
and
v
t
=
−
1
2
V
i
Therefore, the initial values of the projectiles’ velocity vectors are
v
b
=
v
h
+
v
b/h
= (
V
+
v
cos
φ
)
i
+
v
sin
φ
j
v
a
=
v
t
+
v
a/t
=
−
w
1
2
V
+
√
3
V
cos
φ
W
i
+
√
3
V
sin
φ
j
The equations governing the motion of the
shell buster
are
d
2
x
b
dt
2
= 0;
x
b
(0) = 0
,
˙
x
b
(0) =
V
+
v
cos
φ
d
2
z
b
dt
2
=
−
g
;
z
b
(0) = 0
,
˙
z
b
(0) =
v
sin
φ
Solving, we find
x
b
(
t
) = (
V
+
v
cos
φ
)
t
and
z
b
(
t
) =
vt
sin
φ
−
1
2
gt
2
The equations governing the motion of the artillery shell are
d
2
x
a
dt
2
= 0;
x
a
(0) =
L,
˙
x
a
(0) =
−
w
1
2
V
+
√
3
V
cos
φ
W
d
2
z
a
dt
2
=
−
g
;
z
a
(0) = 0
,
˙
z
a
(0) =
√
3
V
sin
φ
Solving, we find
x
a
(
t
) =
L
−
w
1
2
V
+
√
3
V
cos
φ
W
t
and
z
a
(
t
) =
√
3
V
sin
φ
−
1
2
gt
2
(a)
The projectiles meet when
x
a
=
x
b
and
z
a
=
z
b
. Equating the vertical coordinates tells us that
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 Spring '06
 Shiflett
 Velocity, ... ...

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