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Motion in Two or Three Dimensions
321
SET UP:
Take the origin of
coordinates at the point
where the canister is
released. Take
y
+
to be
upward. The initial
velocity of the canister is
the velocity of the plane,
64.0 m/s in the
direction.
x
+
Figure 3.53
Use the vertical motion to find the time of fall:
?,
t
=
0
0,
y
v
=
2
9.80 m/s ,
y
a
=−
0
90.0 m
yy
−=
−
(When the canister reaches the ground it is 90.0 m below
the
origin.)
2
1
00
2
y
y
yy v
t a
t
−= +
EXECUTE:
Since
0
0,
y
v
=
0
2
2(
)
2( 90.0 m)
4.286 s.
9.80 m/s
y
t
a
−−
==
=
−
SET UP:
Then use the horizontal component of the motion to calculate how far the canister falls in this time:
0
?,
xx
0,
x
a
−
0
64.0 m/s,
x
v
=
EXECUTE:
2
1
2
(64.0 m/s)(4.286 s)
0
274 m.
xx v
t
=
+=
EVALUATE:
The time it takes the cannister to fall 90.0 m, starting from rest, is the time it travels horizontally at
constant speed.
3.54.
IDENTIFY:
The equipment moves in projectile motion. The distance
D
is the horizontal range of the equipment
plus the distance the ship moves while the equipment is in the air.
SET UP:
For the motion of the equipment take
x
+
to be to the right and
y
+
to be upwards. Then
0
x
a
=
,
2
9.80 m/s
y
a
,
0
cos
7.50 m/s
x
vv
α
and
0
sin
13.0 m/s
y
. When the equipment lands in the front
of the ship,
0
8.75 m
−
.
EXECUTE:
Use the vertical motion of the equipment to find its time in the air:
2
1
2
y
y
t
gives
( )
2
1
13.0
( 13.0)
4(4.90)(8.75) s
9.80
t
=±
−
+
. The positive root is
3.21 s
t
=
. The horizontal range of the
equipment is
2
1
2
(7.50 m/s)(3.21 s)
24.1 m
t
=
=
. In 3.21 s the ship moves a horizontal distance
(0.450 m/s)(3.21 s)
1.44 m
=
, so
24.1 m 1.44 m
25.5 m
D
=+=
.
EVALUATE:
The equation
2
sin2
v
R
g
=
from Example 3.8 can’t be used because the starting and ending points
of the projectile motion are at different heights.
3.55.
IDENTIFY:
Projectile motion problem.
Take the origin of
coordinates at the point
where the ball leaves the
bat, and take
y
+
to be
upward.
0
cos
x
=
0
sin
,
y
=
but we don&t know
0
.
v
Figure 3.55
Write down the equation for the horizontal displacement when the ball hits the ground and the corresponding
equation for the vertical displacement. The time
t
is the same for both components, so this will give us two
equations in two unknowns (
0
v
and
t
).
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View Full Document322
Chapter 3
(a) SET UP:
y
component
:
2
9.80 m/s ,
y
a
=−
0
0.9 m,
yy
−=
−
00
sin45
y
vv
=
°
2
1
2
yy v ta
t
−= +
EXECUTE:
22
1
0
2
0.9 m
( sin45 )
( 9.80 m/s )
vt
t
°
+
−
SET UP:
x
component
:
0,
x
a
=
0
188 m,
xx
cos45
x
=°
2
1
2
x
x
x
xv
t a
t
EXECUTE:
0
188 m
cos45
x
t
−
==
°
Put the expression for
t
from the
x
component motion into the
y
component equation and solve for
0
.
v
(Note that
cos45 .
°=
°
)
2
2
0
188 m
188 m
0.9 m
(4.90 m/s )
cos45
cos45
v
⎛⎞
°
−
⎜⎟
°°
⎝⎠
2
2
0
188 m
4.90 m/s
188 m
0.9 m
188.9 m
cos45
v
=+=
°
2
2
0
cos45
4.90 m/s
,
188 m
188.9 m
v
°
=
2
0
188 m
4.90 m/s
42.8 m/s
cos45
188.9 m
v
°
(b)
Use the horizontal motion to find the time it takes the ball to reach the fence:
SET UP:
x
component
:
0
116 m,
0,
x
a
=
cos45
(42.8 m/s)cos45
30.3 m/s,
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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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