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Motion in Two or Three Dimensions
311
(c)
0
4.0 m/s
v
=
gives
2.29 m
y
=−
. In this case, the dart was fired with so slow a speed that it hit the ground before
traveling the 3meter horizontal distance.
EVALUATE:
For (a) and (d) the trajectory of the dart has the shape shown in Figure 3.26 in the textbook. For (c)
the dart moves in a parabola and returns to the ground before it reaches the
x
coordinate of the monkey.
3.23.
IDENTIFY:
Take the origin of coordinates at the roof and let the
direction
y
+
be upward. The rock moves in
projectile motion, with
0
x
a
=
and
.
y
ag
= −
Apply constant acceleration equations for the
x
and
y
components of
the motion.
SET UP:
00
0
cos
25.2 m/s
x
vv
α
==
0
sin
16.3 m/s
y
Figure 3.23a
(a)
At the maximum height
0.
y
v
=
2
9.80 m/s ,
y
a
0,
y
v
=
0
16.3 m/s,
y
v
=+
0
?
yy
−
=
22
2(
)
y
vv a
−
EXECUTE:
2
0
0
2
0
(16.3 m/s)
13.6 m
(
9
.
8
0
m
/
s
)
y
a
−
−
−=
=
=
+
−
(b) SET UP:
Find the velocity by solving for its
x
and
y
components.
0
25.2 m/s
xx
(since
0
x
a
=
)
?,
y
v
=
2
9.80 m/s ,
y
a
0
15.0 m
−
(negative because at the ground the rock is below its initial position),
0
16.3 m/s
y
v
=
)
y
−
2
)
y
a
y
y
+
−
(
y
v
is negative because at the ground the rock is traveling downward.)
EXECUTE:
(16.3 m/s)
2( 9.80 m/s )( 15.0 m)
23.7 m/s
y
v
+ −
−
Then
2
2
(25.2 m/s)
( 23.7 m/s)
34.6 m/s.
xy
v
=
+
−
=
(c) SET UP:
Use the vertical motion (
y
component) to find the time the rock is in the air:
t
=
23.7 m/s
y
v
(from part (b)),
2
9.80 m/s ,
y
a
0
16.3 m/s
y
v
EXECUTE:
0
2
23.7 m/s 16.3 m/s
4.08 s
9.80 m/s
y
t
a
−
−−
=
+
−
SET UP:
Can use this
t
to calculate the horizontal range:
4.08 s,
t
=
0
25.2 m/s,
x
v
=
0,
x
a
=
0
?
−
=
EXECUTE:
2
1
2
(25.2 m/s)(4.08 s)
0 103 m
xx v
t a
t
−= +
=
+=
(d)
Graphs of
x
versus
t
,
y
versus
t
,
x
v
versus
t
, and
y
v
versus
t
:
Figure 3.23b
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View Full Document312
Chapter 3
EVALUATE:
The time it takes the rock to travel vertically to the ground is the time it has to travel horizontally.
With
0
16.3 m/s
y
v
=+
the time it takes the rock to return to the level of the roof (
0)
y
=
is
0
2
/
3.33 s.
y
tvg
==
The
time in the air is greater than this because the rock travels an additional 15.0 m to the ground.
3.24.
IDENTIFY:
Consider the horizontal and vertical components of the projectile motion. The water travels 45.0 m
horizontally in 3.00 s.
SET UP:
Let
y
+
be upward.
0
x
a
=
,
2
9.80 m/s
y
a
=−
.
00
0
cos
x
vv
θ
=
,
000
sin
y
=
.
EXECUTE:
(a)
2
1
2
x
x
x
xv
t a
t
−= +
gives
0
(cos
)
x
t
−=
and
0
45.0 m
cos
0.600
(25.0 m/s)(3.00 s)
;
0
53.1
=
°
(b)
At the highest point
0
(25.0 m/s)cos53.1
15.0 m/s
xx
°
=
,
0
y
v
=
and
22
15.0 m/s
xy
v
=
. At all points
in the motion,
2
9.80 m/s
a
=
downward.
(c)
Find
0
yy
−
when
3 00s
t
=.
:
2
11
(25.0 m/s)(sin53.1 )(3.00 s)
( 9.80 m/s )(3.00 s)
15.9 m
yy v
t
=
°
+−
=
0
15.0 m/s
,
2
0
(25.0 m/s)(sin53.1 )
(9.80m/s )(3.00 s)
9.41 m/s
y
vv a
t
=+=
°
−
=
−
, and
2
2
(15.0 m/s)
( 9.41 m/s)
17.7 m/s
vvv
=
+
−
=
EVALUATE:
The acceleration is the same at all points of the motion. It takes the water
0
2
20.0 m/s
2.04 s
9.80 m/s
y
y
v
t
a
=
−
to reach its maximum height. When the water reaches the building it has passed
its maximum height and its vertical component of velocity is downward.
3.25.
IDENTIFY
and
SET UP:
The stone moves in projectile motion. Its initial velocity is the same as that of the balloon.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.
 Spring '06
 Buchler
 Physics

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