which yields
F
≈
2.8
×
10
4
kg. Since we are not at all certain of his mass, we express this
as a guessedat range (in kN) 25 <
F
< 30.
Since
F
mg
>>
, the impulse
G
J
due to the net force (while he is in contact with the water)
is overwhelmingly caused by the upward force of the water:
Fdt
J
=
z
G
to a good
approximation. Thus, by Eq. 929,
Fdt
p
p
m
gh
fi
=−
=
−
−
z
G
G
02
di
(the minus sign with the initial velocity is due to the fact that downward is the negative
direction) which yields
()
23
(70 kg) 2 9.8 m/s
12 m
1.1 10 kg m s.
=×
⋅
Expressing this as
a range we estimate
33
1.0 10 kg m s
1.2 10 kg m s.
×⋅
<
<
³
25. We estimate his mass in the neighborhood of 70 kg and compute the upward force
F
of the water from Newton’s second law:
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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, Force, Mass

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