(a) Calculate the impulse experienced when a 60 kg person lands on frm ground aFter jumping From a
height oF 2.7 m. (b) Estimate the average Force exerted on the person’s Feet by the ground iF the landing
is sti±legged.
With sti± legs, assume the body moves 1.9 cm during impact.
(b) Estimate the average
Force exerted on the person’s Feet by the ground iF the landing is with bent legs. When the legs are bent,
assume the body moves about 44cm during impact.
Strategy.
²or Part (a), the ImpulseMomentum Theorem tells us that the impulse experienced by the
woman is equal to the change in her momentum. In order to calculate the change in momentum, we will
use conservation oF energy to fnd the woman’s momentum just beFore she hits the ground. ²or Parts (b)
and (c), The WorkEnergy Theorem tells us that the net work done on the woman is equal to the change
in her kinetic energy, which is zero. In order to calculate the average Force exerted on her by the ground,
we must evaluate the work done by both gravity and the ground.
Solution.
There are three “points oF interest” in this problem: (1) the instant at which the woman is at
a height
h
above the ground; (2) the instant immediately beFore the woman hits the ground; and, (3) the
instant immediately aFter the woman has landed and come to rest.
We will denote the woman’s energy
and momentum at each oF the points by
E
1
,
2
,
3
and
p
1
,
2
,
3
, respectively. Then
E
1
=
mgh,
E
2
=
p
2
2
/
(2
m
)
,
E
3
= 0
,
p
1
=
0
,
p
2
=

p
2
ˆ
j
,
p
3
=
0
,
(9.32.1)
where
m
= 60 kg is the mass oF the woman and
h
= 2
.
7 m is the height From which she Falls. Here
ˆ
i
,
ˆ
j
,
and
ˆ
k
are unit vectors that point in the
x
,
y
, and
z
 directions, respectively
Because there are no dissipative Forces acting on the woman while she Falls, energy is conserved between
points 1 and 2:
E
1
=
E
2
⇒
p
2
=
m
p
2
gh.
(9.32.2)
According to the ImpulseMomentum Theorem, when the woman lands, she experiences an impulse From
the ground given by
J
23
=
p
3

p
2
⇒
J
23
=
m
p
2
gh
ˆ
j
.
(9.32.3)
Using
g
= 9.8 m
/
s
2
, we fnd that
J
23
= 440 kg
·
m
/
s. Here
J
23
is the magnitude oF
J
23
.
The work done by gravity is given by
W
(gravity)
13
=

(
U
3

U
1
) =
mg
(
h
+
d
)
,
(9.32.4)
where
U
1
,
23
denotes the gravitational potential energy at points 1, 2, and 3.
Here
d
is the distance the
body moves during the impact. The work done by the ground is
W
(ground)
13
=
Z
3
1
F
(
y
)
·
d
x
=
Z
0
d
F
(
y
)
dy
=

Z
d
0
F
(
y
)
dy,
(9.32.5)
where
F
(
y
) =
F
(
y
)
ˆ
j
is the Force oF the ground on the woman and
d
x
=
dx
ˆ
i
+
dy
ˆ
j
+
dz
ˆ
k
. According to
the WorkEnergy Theorem, the net work done on the woman is zero. ThereFore,
W
(net)
13
=
W
(ground)
13
+
W
(gravity)
13
=
K
3

K
1
= 0
⇒
Z
d
0
F
(
y
)
dy
=
mg
(
h
+
d
)
.
(9.32.6)
²rom the defnition oF “average,” we know that the average Force is given by
h
F
i
=
1
d
Z
d
0
F
(
y
)
dy
=
mg
±
h
d
+ 1
²
.
(9.32.7)
The average Force is
h
F
i
= 8
.
4
×
10
4
N when
d
= 1.9 cm and