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alexander (jra2623) – homework 07 – Turner – (92510)
1
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001
(part 1 oF 4) 10.0 points
Consider the setup oF a gun aimed at a target
(such as a monkey) as shown in the fgure
below. The target is to be dropped From the
point
A
at
t
= 0, the same moment as the
gun is fred. The bullet hits the target at a
point
P
. Let the initial speed oF the bullet
be
v
0
= 107 m
/
s, let the angle between the
vector
V
v
0
and the horizontal (
x
) direction be
θ
= 44
.
8
◦
and let
AB
= 94
.
4 m. The distance
d
=
OB
is the
x
coordinate oF the target.
Denote the time taken to hit the target by
t
.
v
0
θ
ℓ
h
x
y
O
A
B
P
The acceleration oF gravity is 9
.
8 m
/
s
2
.
This time is given by
1.
t
=
d
v
0
cos
θ
.
correct
2.
t
=
d
v
0
.
3.
t
=
d
v
0
tan
θ
.
4.
t
=
d
v
0
sin
θ
.
5.
t
=
d
v
0
cot
θ
.
Explanation:
Basic Concepts:
Constant acceleration:
x

x
0
=
v
0
t
+
1
2
at
2
(1)
v
=
v
0
+
at
(2)
Solution:
Think oF this twodimensional motion as
two onedimensional trajectories, one in the
x
and one in the
y
direction. The horizontal (
x
)
initial velocity is
v
0
x
=
v
0
cos
θ .
The
x
motion is unaccelerated (gravity only
a±ects the
y
motion) so equation (1) gives
d
=
v
0
x
t
+ 0
.
Thus
t
=
d
v
0
x
=
d
v
0
cos
θ
.
002
(part 2 oF 4) 10.0 points
Assume:
Up is the positive direction.
With the time taken to hit the target given
by
t
, the vertical (
y
) component oF the veloc
ity oF the bullet is
1.
v
y
=

g t
2.
v
y
=
v
0
sin
θ
+
g t
3.
v
y
=
v
0
cos
θ
+
g t
4.
v
y
=

v
0
cos
θ
+
g t
5.
v
y
=
v
0
cos
θ

g t
6.
v
y
=
g t
7.
v
y
=
v
0
sin
θ

g t
correct
8.
v
y
=

v
0
sin
θ

g t
9.
v
y
=

v
0
cos
θ

g t
10.
v
y
=

v
0
sin
θ
+
g t
Explanation:
The vertical (
y
) initial velocity is given by
v
0
y
=
v
0
sin
θ
We take positive direction upward, so
a
=

g
and equation (2) gives us For the vertical
component oF the velocity at P
v
Py
=
v
0
y

g t
=
v
0
sin
θ

g t
.
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2
003
(part 3 of 4) 10.0 points
The height BP where the collision takes place
is
Correct answer: 86
.
7185 m.
Explanation:
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 Summer '08
 Kaplunovsky
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