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Math 3F — HW #4 Solutions
Laney College, Fall 2017
Fred Bourgoin
Section 3.4
2. A 400-lb object is released from rest 500 ft above the ground and allowed to fall
under the influence of gravity. Assuming that the force in pounds due to air
resistance is
10v, where v is the velocity of the object in ft/sec, determine the
equation of motion of the object. When will the object hit the ground?
Solution. Using Newton’s Second Law of Motion, we have the equation
dv
12.5
= 400
10v
dt
which yields the IVP
dv
= 32
0.8 v,
v(0) = 0.
dt
Solving the equation (either as a separable ODE or as a linear ODE), we have
0.8 t
v = 40 + Ce
.
0.8 t
Plugging in the initial condition, we get v = 40 (1
e
). To find the position,
0.8 t
integrate v: x = 40 t + 50 e
+ D. With the initial condition x(0) = 0, the
position function is
0.8 t
x(t) = 40 t + 50 e
50.
The object hits the ground when
t ≈ 13.75 sec.
x(t) = 500
=⇒
7. A parachutist whose mass is 75 kg drops from a helicopter hovering 2000 m above
the ground and falls toward the ground under the influence of gravity. Assume that
the force due to air resistance is proportional to the velocity of the parachutist,
with the proportionality constant b
= 30 N-sec/m when the chute is closed and
1
b
= 90 N-sec/m when the chute is open. If the chute does not open until the
2
velocity of the parachutist reaches 20 m/sec, after how many seconds will she
reach the ground?
Solution. This problem is solved in two stages. First, when the parachutist is in
free-fall,
dv
1
= 9.81
0.4 v,
v
(0) = 0.
1
dt
0.4 t
The solution is v
(t) = 24.525
24.525 e
. Notice that v = 20 after approxi-
1
mately 4.23 seconds. Integrating v
and setting x
(0) = 0, we have the position
1
1
1

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