Once we know the vertical displacement of the bullet as it passes between the buildings, we
can determine the time
1
t
required for the bullet to reach the window using Equation 3.4b.
Since the motion in the
x
direction is not accelerated, the distance
D
can then be found from
0
1
x
D
v
t
=
.
SOLUTION
Assuming that the direction to the right is positive, we find that the time that
the bullet spends in the building is (according to Equation 3.5a)
0
6.9 m
0.0203 s
340 m/s
x
x
t
v
=
=
=
The vertical displacement of the bullet after it enters the building is, taking down as the
negative direction, equal to –0.50 m.
Therefore, the vertical component of the velocity of
the bullet as it passes through the window is, from Equation 3.5b,
1
2
1
1
2
2
0 (window)
2
2
–0.50 m
(
9.80 m/s
)(0.0203 s) = –24.5 m/s
0.0203 s
y
y
y
y
a t
y
v
a t
t
t

=
=

=


The vertical displacement of the bullet as it travels between the buildings is (according to
Equation 3.6b with
0
0
y
v
=
m/s)
2
2
2
(
24.5 m/s)
30.6 m
2
2(–9.80 m/s
)
y
y
v
y
a

=
=
= 
Therefore, the distance
H
is
30.6 m
0.50 m
31 m
H
=
+
=
The time for the bullet to reach the window, according to Equation 3.4b, is
134
KINEMATICS IN TWO DIMENSIONS
1
0
2
2
2(–30.6 m)
2.50 s
(–24.5 m/s)
y
y
y
y
y
t
v
v
v
=
=
=
=
+
Hence, the distance
D
is given by
0
1
(340 m/s)(2.50 s)
850 m
x
D
v
t
=
=
=
50.
REASONING
AND
SOLUTION
Let
H
be the initial height of the can above the muzzle of
the rifle. Relative to its initial position the vertical coordinate of the bullet at time,
t
, is
2
1
2
0
y
y
v
t
gt
=

Relative to the can’s initial position the vertical coordinate of the can at the
same
time is
y
' = (1/2)
gt
2
.
NOTE:
y
=
H
–
y
' if the bullet hits the can. Then
y
=
v
0
y
t
–
y
'. The horizontal distance
traveled by the bullet in time
t
is
x
=
v
0
x
t
.
Solving for
t
and substituting gives
y
= (
v
0
y
/
v
0
x
)
x
–
y
'
Now
v
0
y
/
v
0
x
= tan
θ
and it is seen from the figure in the text that
H
=
x
tan
θ
if the bullet is
to hit the can so
y
=
H
–
y
'.
Hence, both objects are at the same place at the same time, and
the bullet will always strike the can.
51.
REASONING
The drawing shows the
trajectory of the ball, along with its
initial speed
v
0
, horizontal displacement
x
, and vertical displacement
y
. The angle
that the initial velocity of the ball makes
with the horizontal is
θ
. The known data
are shown in the tables below:
xDirection Data
x
a
x
v
x
v
0
x
t
+26.9 m
0 m/s
2
+(19.8 m/s) cos
θ
θ
v
0
y
x
+
x
+
y
Chapter 3
Problems
135
yDirection Data
y
a
y
v
y
v
0
y
t
+2.74 m

9.80 m/s
2
+(19.8 m/s) sin
θ
There are only two known variables in each table, so we cannot directly use the equations of
kinematics to find the angle
θ
. Our approach will be to first use the
x
direction data and
obtain an expression for the time of flight
t
in terms of
x
and
v
0
x
. We will then enter this
expression for
t
into the
y
direction data table. The four variables in this table,
y
,
a
y
,
v
0
y
, and
t
, can be related by using the appropriate equation of kinematics. This equation can then be
solved for the angle
θ
.
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 Spring '11
 rollino
 Physics, Acceleration, Velocity, m/s