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Problem 13.49 A sky diver jumps from a helicopter
and is falling straight down at 30 m/s when her parachute
opens. From then on, her downward acceleration is
approximately a=g−cv2, where g=9.81 m/s2and
cis a constant. After an initial “transient” period she
descends at a nearly constant velocity of 5 m/s.
(a) What is the value of c, and what are its SI units?
(b) What maximum deceleration is the sky diver
subjected to?
(c) What is her downward velocity when she has fallen
2 meters from the point at which her parachute
opens?
c=g
v2=9.81 m/s2
(5)2m2/s2=0.3924 m−1
(c) Choose coordinates such that distance is measured positive down-
ward. The velocity is related to position by the chain rule:
dv
dt =dv
ds
ds
dt =vdv
ds =a,
from which
vdv
g−cv2=ds.
2cln |g−cv2|=s+C.
When the parachute opens s=0 and v=30 m/s, from which
C=−1
v=14.4 m/s
Problem 13.50 The rocket sled starts from rest and
accelerates at a=30 +2tm/s2until its velocity is
400 m/s. It then hits a water brake and its acceleration
is a=−0.003v2m/s2until its velocity decreases to
100 m/s. What total distance does the sled travel?
v=30t+t2m/s
s1
(0.003)vf
v1
v2
Problem 13.51 In Problem 13.50, what is the sled’s
total time of travel?
0.003td=3
400
Problem 13.52 A car’s acceleration is related to its
position by a=0.01sm/s2. When s=100 m, the car
is moving at 12 m/s. How fast is the car moving when
s=420 m?
ds =0.01sm/s2
vf
12
vdv =0.01420
100
sds
Problem 13.53 Engineers analyzing the motion of a
linkage determine that the velocity of an attachment
point is given by v=A+4s2ft/s, where Ais a con-
stant. When s=2 ft, its acceleration is measured and
determined to be a=320 ft/s2. What is its velocity of
the point when s=2ft?
30
Problem 13.54 The acceleration of an object is given
as a function of its position in feet by a=2s
2(ft/s2).
When s=0, its velocity is v=1 ft/s. What is the veloc-
ity of the object when s=2 ft?
Solution: We are given
a=vdv
ds =2
ft-s2s2,
v=3.42 ft/s.
Problem 13.55 Gas guns are used to investigate
the properties of materials subjected to high-velocity
s=1.5 m and accelerates until it reaches the end of the
barrel at s=3 m. Determine the value of the constant
Problem 13.56 If the propelling gas in the gas gun
described in Problem 13.55 is air, a more accurate
Solution:
c
Problem 13.57 A spring-mass oscillator consists of a
mass and a spring connected as shown. The coordinate
smeasures the displacement of the mass relative to its
position when the spring is unstretched. If the spring
is linear, the mass is subjected to a deceleration pro-
portional to s. Suppose that a=−4sm/s2, and that you
give the mass a velocity v=1 m/s in the position s=0.
(a) How far will the mass move to the right before the
spring brings it to a stop?
(b) What will be the velocity of the mass when it has
returned to the position s=0?
s
Problem 13.58 In Problem 13.57, suppose that at
t=0 you release the mass from rest in the position
s=1 m. Determine the velocity of the mass as a func-
Solution: From the solution to Problem 13.57, the velocity as a
function of position is given by
32
Problem 13.59 A spring-mass oscillator consists of a
mass and a spring connected as shown. The coordinate
smeasures the displacement of the mass relative to its
position when the spring is unstretched. Suppose that
the nonlinear spring subjects the mass to an acceleration
a=−4s−2s3m/s2and that you give the mass a velocity
v=1 m/s in the position s=0.
(a) How far will the mass move to the right before the
springs brings it to a stop?
(b) What will be the velocity of the mass when it has
returned to the position s=0?
s
Solving for dwe find d=0.486 m.
Problem 13.60 The mass is released from rest with
the springs unstretched. Its downward acceleration is
a=32.2−50sft/s2, where sis the position of the mass
measured from the position in which it is released.
(a) How far does the mass fall? (b) What is the
maximum velocity of the mass as it falls?
s
0
0
2=32.2s−25s2.
ds =0.
(32.2−50s) ft/s2. Since we want maximum velocity, we can
assume that v= 0 at this point. Thus, 0 =(32.2−50s),ors=
(32.2/50)ft when v=vMAX. Substituting this value for sinto
the equation for v,weget
13.60 is in the position s=0 and is given a downward
velocity of 10 ft/s.
(a) How far does the mass fall?
(b) What is the maximum velocity of the mass as
it falls?
v2=(10 ft/s)2+(64.4 ft/s2)s −(50 s−2)s2
v=10.99 ft/s
Problem 13.62 If a spacecraft is 100 mi above the sur-
face of the earth, what initial velocity v0straight away
from the earth would be required for the vehicle to
reach the moon’s orbit 238,000 mi from the center of
the earth? The radius of the earth is 3960 mi. Neglect
0
100 mi
vdv =−gR2
E
s2.
Integrate:
1h 5280 ft
1mi 1h
3600 s =35,913.1 ft/s .
Check: Use the result of Example 13.5
34
Problem 13.63 The moon’s radius is 1738 km. The
4.89 ×1012
s2m/s2.
Suppose that a spacecraft is launched straight up from
the moon’s surface with a velocity of 2000 m/s.
(a) What will the magnitude of its velocity be when it
is 1000 km above the surface of the moon?
(b) What maximum height above the moon’s surface
will it reach?
Solution: Set G=4.89 ×1012 m3/s2,r
0=1.738 ×106m,
ds =−G
s2⇒v
v0
r0
s2ds ⇒v2
2−v02
2=G1
r−1
r0
v2=v02+2Gr0−r
rr0
(a) v(r0+1.0×106m)=1395 m/s
(b) The maximum velocity occurs when v=0
Problem 13.64* The velocity of an object subjected
only to the earth’s gravitational field is
Solution:
v=v2
0+2gR2
E1
Take the derivative with respect to s.
E
Problem 13.65 Suppose that a tunnel could be drilled
straight through the earth from the North Pole to the
South Pole and the air was evacuated. An object dropped
from the surface would fall with the acceleration a=
−gs/RE, where gis the acceleration of gravity at sea
level, REis radius of the earth, and sis the distance of the
object from the center of the earth. (The acceleration due
to gravitation is equal to zero at the center of the earth
and increases linearly with the distance from the center.)
What is the magnitude of the velocity of the dropped
object when it reaches the center of the earth? S
N
Tunnel
RE
s
Problem 13.66 Determine the time in seconds required
for the object in Problem 13.65 to fall from the surface
of the earth to the center. The earth’s radius is 6370 km.
Solution: From Problem 13.65, the acceleration is
a=vdv
s
v=ds
dt =±
RER2
E−s2
0
ds
R2
g
dt
Problem 13.67 In a second test, the coordinates of the
position (in m) of the helicopter in Active Example 13.6
are given as functions of time by
x=4+2t,
y=4+4t+t2.
(a) What is the magnitude of the helicopter’s velocity
at t=3s?
(b) What is the magnitude of the helicopter’s acceler-
ation at t=3s?
y
x
36
Problem 13.68 In terms of a particular reference
frame, the position of the center of mass of the F-14
at the time shown (t =0)is r=10i+6j+22k(m).
The velocity from t=0tot=4sisv=(52 +6t)i+
(12 +t2)j−(4+2t2)k(m/s). What is the position of
the center of mass of the plane at t=4s?
Problem 13.69 In Example 13.7, suppose that the angle
between the horizontal and the slope on which the skier
45⬚
20⬚
2(9.81 m/s2)t2−(10 m/s)sin 20◦t
Problem 13.70 A projectile is launched from ground
level with initial velocity v0=20 m/s. Determine its
range Rif (a) θ0=30◦; (b) θ0=45◦(c) θ0=60◦.
0
y
Solution: Set g=9.81 m/s2,v
0=20 m/s
ay=−g, vy=−gt +v0sin θ0,s
y=−1
Problem 13.71 Immediately after the bouncing golf
ball leaves the floor, its components of velocity are vx=
0.662 m/s and vy=3.66 m/s.
(a) Determine the horizontal distance from the point
where the ball left the floor to the point where it
hits the floor again.
y
Problem 13.72 Suppose that you are designing a
mortar to launch a rescue line from coast guard vessel
to ships in distress. The light line is attached to a weight
fired by the mortar. Neglect aerodynamic drag and the
weight of the line for your preliminary analysis. If you
want the line to be able to reach a ship 300 ft away
when the mortar is fired at 45◦above the horizontal,
what muzzle velocity is required?
y
45⬚
x
Problem 13.73 In Problem 13.72, what maximum
height above the point where it was fired is reached by
the weight?
2(32.2 ft/s2)=75 ft
is 1.82 m above the ground and its initial velocity is v0=
13.6 m/s. Determine the horizontal distance the shot trav-
els from the point of release to the point where it hits
the ground. 30⬚
v
0
sx=(13.6 m/s)cos 30◦t,
Problem 13.75 A pilot wants to drop survey markers
at remote locations in the Australian outback. If he flies
at a constant velocity v0=40 m/s at altitude h=30 m
and the marker is released with zero velocity relative to
the plane, at what horizontal dfrom the desired impact
point should the marker be released?
h
0
40
Problem 13.76 If the pitching wedge the golfer is
using gives the ball an initial angle θ0=50◦, what range
of velocities v0will cause the ball to land within 3 ft of
the hole? (Assume the hole lies in the plane of the ball’s
trajectory).
30 ft
v0
θ
0
2.
From the xequation, we can find the time at which the ball reaches
Problem 13.77 A batter strikes a baseball 3 ft above
home plate and pops it up. The second baseman catches
it 6 ft above second base 3.68 s after it was hit.
What was the ball’s initial velocity, and what was
the angle between the ball’s initial velocity vector and
the horizontal?
90 ft
Second
base
Home plate
Solution: The equations of motion g=32.2 ft/s2
ax=0ay=−g
vx=v0cos θ0vy=−gt +v0sin θ0
sx=v0cos θ0ts
y=−1
2gt2+v0sin θ0t+3ft
When the second baseman catches the ball we have
127.3 ft =v0cos θ0(3.68 s)
Problem 13.78 A baseball pitcher releases a fastball
with an initial velocity v0=90 mi/h. Let θbe the initial
angle of the ball’s velocity vector above the horizontal.
When it is released, the ball is 6 ft above the ground
(2) dvy
(3) dx
(4) dy
dt =−gt +v0sin θ, from which
y(t) =−g
2t2+v0sin θt +6,
since the initial position is y(0)=6 ft. At a distance d=58 ft,
the height is h. The time of passage across the home plate is
x(tp)=d=v0cos θt
p, from which
tp=d
v0cos θ.
For θ=1◦,h=3.91 ft ,Yes, the pitcher hits the strike zone.
For θ=2◦,h=4.92 ft No, the pitcher misses the strike zone.
Problem 13.79 In Problem 13.78, assume that the
pitcher releases the ball at an angle θ=1◦above the
horizontal, and determine the range of velocities v0(in
ft/s) within which he must release the ball to hit the
42
Problem 13.80 A zoology student is provided with a
bow and an arrow tipped with a syringe of sedative
and is assigned to measure the temperature of a black
rhinoceros (Diceros bicornis). The range of his bow
when it is fully drawn and aimed 45◦above the horizon-
tal is 100 m. A truculent rhino charges straight toward
him at 30 km/h. If he fully draws his bow and aims 20◦
above the horizontal, how far away should the rhino be
when the student releases the arrow?
20⬚
(1) dvx
dt =0, from which vx=Vx.Att=0,V
x=VAcos θ.
(2) dvy
y=
(3) dx
(4) dy
dt =vy=−gt +VAsin θ, from which
y=−g
2t2+VAsin θt +Cy.
g=2.18 s,
d=8.33(2.18)=18.2 m. The required range when the arrow is
Problem 13.81 The crossbar of the goalposts in
American football is yc=10 ft above the ground. To
kick a field goal, the ball must make the ball go between
the two uprights supporting the crossbar and be above
the crossbar when it does so. Suppose that the kicker
attempts a 40-yard field goal (xc=120 ft), and kicks
the ball with an initial velocity v0=70 ft/s and θ0=
40◦. By what vertical distance does the ball clear the
crossbar?
yc
xc
0
v0
θ
2. Set x=xc=120 ft and
find the time tcat which the ball crossed the plane of the goal posts.
Problem 13.82 An American football quarterback
stands at A. At the instant the quarterback throws the
football, the receiver is at Brunning at 20 ft/s toward
C, where he catches the ball. The ball is thrown at an
angle of 45◦above the horizontal, and it is thrown and
C
A
ax=0,v
x=v0cos θ0,s
x=v0cos θ0t
When the ball is caught we have
44
Problem 13.83 The cliff divers of Acapulco, Mexico
must time their dives that they enter the water at the
crest (high point) of a wave. The crests of the waves
are 1 m above the mean water depth h=4 m. The
horizontal velocity of the waves is equal to √gh. The
diver’s aiming point is 2 m out from the base of the cliff.
Assume that his velocity is horizontal when he begins
the dive.
(a) What is the magnitude of the driver’s velocity when
he enters the water?
(b) How far from his aiming point must a wave crest
be when he dives in order for him to enter the water
at the crest?
1 m
26 m
t1=tIMPACT =2.30 s
Vy(t1)=22.59 m/s
ax=0
Vx=8.4/t1=3.65 m/s =constant
The velocity at impact is
(a) |V|=(Vx)2+[Vy(t1)]2=22.9 m/s
The wave moves at √gh =6.26 m/s.
The wave crest travels 2.30 seconds while the diver is in their s=
√ght1=14.4m.
8.4 m
Problem 13.84 A projectile is launched at 10 m/s from
a sloping surface. The angle α=80◦. Determine the
10 m/s
Problem 13.85 A projectile is launched at 100 ft/s at
60◦above the horizontal. The surface on which it lands
is described by the equation shown. Determine the point
y
60°
100 ft/s
46
Problem 13.86 At t=0, a steel ball in a tank of oil
is given a horizontal velocity v=2i(m/s). The compo-
nents of the ball’s acceleration in m/s2are ax=−1.2vx,
ay=−8−1.2vy,az=−1.2vz. What is the velocity of
the ball at t=1s?
x
y
(1) dvx
dt =ax=−1.2vx. Separate variables and integrate:
dvx
(2) dvy
dt =ay=−8−1.2vy. Separate variables and integrate:
dvy
8
1.2(e−1.2t−1).
(3) dvz
vy(1)=−8
12 (1−e−1.2)=−4.66 m/s ,or
Problem 13.87 In Problem 13.86, what is the position
of the ball at t=1 s relative to its position at t=0?
lem 13.86. The equations for the coordinates:
dt =vx=2e−1.2t, from which
y(t) =−8
1.2e−1.2t
1.2+t+Cy.
y(t) =−8
(3) Since vz(0)=0 and z(0)=0, then z(t) =0. At t=1,
Problem 13.88 The point Pmoves along a circular
path with radius R. Show that the magnitude of its veloc-
ity is |v|=R|dθ/dt|.
Strategy: Use Eqs. (13.23).
P
y
Solution:
48
