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236
Chapter 2
EVALUATE:
165 m/s
369 mi/h
=
, so only Superman could jump downward with this initial speed.
Figure 2.90
2.91.
IDENTIFY:
Apply constant acceleration equations to the motion of the rocket and to the motion of the canister
after it is released. Find the time it takes the canister to reach the ground after it is released and find the height of
the rocket after this time has elapsed. The canister travels up to its maximum height and then returns to the ground.
SET UP:
Let
y
+
be upward. At the instant that the canister is released, it has the same velocity as the rocket.
After it is released, the canister has
2
9.80 m/s
y
a
=−
. At its maximum height the canister has
0
y
v
=
.
EXECUTE:
(a)
Find the speed of the rocket when the canister is released:
0
0
y
v
=
,
2
3.30 m/s
y
a
=
,
0
235 m
yy
−=
.
22
00
2(
)
y
vv a
=+
−
gives
2
0
2
(
)
2(3.30 m/s )(235 m)
39.4 m/s
va
y
y
=
=
. For the
motion of the canister after it is released,
0
39.4 m/s
y
v
,
2
9.80 m/s
y
a
,
0
235 m
−
.
2
1
2
y
y
yy v
t a
t
−= +
gives
235 m
(39.4 m/s)
(4.90 m/s )
tt
−
. The quadratic formula gives
12.0 s
t
=
as the
positive solution. Then for the motion of the rocket during this 12.0 s,
2
11
235 m
(39.4 m/s)(12.0 s)
(3.30 m/s )(12.0 s)
945 m
t
=
+
+
=
.
(b)
Find the maximum height of the canister above its release point:
0
39.4 m/s
y
v
,
0
y
v
=
,
2
9.80 m/s
y
a
.
)
y
−
gives
2
0
0
2
0 (39.4 m/s)
79.2 m
(
9
.
8
0
m
/
s
)
y
vv
a
−
−
=
=
−
. After its release the canister travels
upward 79.2 m to its maximum height and then back down 79.2 m
235 m
+
to the ground. The total distance it
travels is 393 m.
EVALUATE:
The speed of the rocket at the instant that the canister returns to the launch pad is
2
0
39.4 m/s
(3.30 m/s )(12.0 s)
79.0 m/s
y
t
=+=
+
=
. We can calculate its height at this instant by
)
y
−
with
0
0
y
v
=
and
79.0 m/s
y
v
=
.
2
0
0
2
(79.0 m/s)
946 m
(
3
.
3
0
m
/
s
)
y
a
−
=
=
, which agrees
with our previous calculation.
2.92.
IDENTIFY:
Both objects are in freefall and move with constant acceleration
2
9.80 m/s , downward. The two
balls collide when they are at the same height at the same time.
SET UP:
Let
y
+
be upward, so
2
9.80 m/s
y
a
for each ball. Let
0
y
=
at the ground. Let ball
A
be the one
thrown straight up and ball
B
be the one dropped from rest at height
H
.
0
0
A
y
=
,
0
B
yH
=
.
EXECUTE:
(a)
2
1
2
y
y
t
applied to each ball gives
2
1
0
2
A
yv
tg
t
and
2
1
2
B
yHg
t
.
AB
=
gives
0
vt
g
t
H
g
t
−
and
0
H
t
v
=
.
(b)
For ball
A
at its highest point,
0
yA
v
=
and
0
y
t
gives
0
v
t
g
=
. Setting this equal to the time in
part (a) gives
0
0
H
v
vg
=
and
2
0
v
H
g
=
.
EVALUATE:
In part (a), using
0
H
t
v
=
in the expressions for
A
y
and
B
y
gives
2
0
1
2
gH
yyH
v
⎛⎞
== −
⎜⎟
⎝⎠
.
H
must be
less than
2
0
2
v
g
in order for the balls to collide before ball
A
returns to the ground. This is because it takes ball
A
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237
time
0
2
v
t
g
=
to return to the ground and ball
B
falls a distance
2
2
0
1
2
2
v
gt
g
=
during this time. When
2
0
2
v
H
g
=
the
two balls collide just as ball
A
reaches the ground and for
H
greater than this ball
A
reaches the ground before
they collide.
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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, Acceleration

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