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98. (a) From Ch. 2, we have
22
0
2
vv
a
x
=+Δ
, where
a
may be interpreted as an average
acceleration in cases where the acceleration is not uniform. With
v
0
= 0,
v
= 11000 m/s
and
Δ
x
= 220 m, we find
a
= 2.75
×
10
5
m/s
2
. Therefore,
52
4
2
2.75
10 m/s
2.8
10
9.8 m/s
ag
g
§·
×
==
×
¨¸
©¹
.
(b) The acceleration is certainly deadly enough to kill the passengers.
(c) Again using
0
2
a
x
, we find
2
2
(7000 m/s)
7000 m/s
714 .
2(3500 m)
=
(d) Energy conservation gives the craft’s speed
v
(in the absence of friction and other
dissipative effects) at altitude
h
= 700 km after being launched from
R
= 6.37
×
10
6
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Acceleration

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