Openstax College Physics
Instructor Solutions Manual
Chapter 3
CHAPTER 3: TWO-DIMENSIONAL
KINEMATICS
3.2 VECTOR ADDITION AND SUBTRACTION: GRAPHICAL METHODS
1.
Find the following for path A in Figure 3.54: (a) the total distance traveled, and (b) the
magnitude and direction of the displacement from start to finish.
Solution
(a) d  (3 120 m)  (1120 m)  480 m
(b) s  s x  s y 
2
2
1  120 m 2  3  120 m 2
 379 m
 sx 
  tan 1  1  120 m   18.4, E of N

 3  120 m 
 sy 
  tan 1 
2.
Solution
Find the following for path B in Figure 3.54: (a) the total distance traveled, and (b) the
magnitude and direction of the displacement from start to finish.
(a) d  (4 120 m)  (3 120 m)  3 120 m  1.20 km
(b) s  s x 2  s y 2 
 sy
 sx
  tan 1 
1  120 m 2  3  120 m 2
 379 m

3  120 m 
  tan 1 
  71.6, N of E
 1  120 m 

3.
Find the north and east components of the displacement for the hikers shown in
Figure 3.52.
Solution
N-component: 5.00 kmsin 40.0  3.21 km
E-component: 5.00 kmcos 40.0  3.83 km
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4.
Solution
Instructor Solutions Manual
Chapter 3
Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are
you from your starting point, and what is the compass direction of a line connecting
your starting point to your final position? (If you represent the two legs of the walk as
vector displacements A and B , as in Figure 3.55, then this problem asks you to find
their sum R  A  B .)

B
N

R
25 m
W

A 
18 m

E
S
AB  R
A  B2  R2
2
R
A2  B 2 
18.02  25.02
 30.8 m
B
 25.0 m 
θ  tan 1    tan 1 
  54.25
 A
 18.0 m 
θ  φ  90  φ  90  54.25  35.75
Compass reading = 35.8 W of N
5.
Suppose you first walk 12.0 m in a direction 20.0 o west of north and then 20.0 m in a
direction 40.0 o south of west. How far are you from your starting point, and what is
the compass direction of a line connecting your starting point to your final position?
(If you represent the two legs of the walk as vector displacements A and B , as in
Figure 3.56, then this problem finds their sum R = A + B .)
Solution
R  AB
Openstax College Physics
Instructor Solutions Manual
Chapter 3
Rx  Ax  Bx  A sin 20  B cos 40  12.0 m sin 20  20.0 m  cos 40  19.425 m
R y  Ay  B y  A cos 20  B sin 40  12.0 m  cos 20  20.0 m sin 40  1.579 m
 19.425 m 2   1.579 m 2
R  Rx  R y 
2
2
 1.579 m 
 tan 1 
  4.65 S of W
Rx
 19.425 m 
Ry
θ  tan 1
6.
 19.5 m
Repeat the problem above, but reverse the order of the two legs of the walk; show
that you get the same final result. That is, you first walk leg B , which is 20.0 m in a
direction exactly 40 o south of west, and then leg A , which is 12.0 m in a direction
exactly 20 o west of north. (This problem shows that A  B  B  A .)
Solution
R x  20.0 m cos 40  12.0 m sin 20  19.425 m
R y  20.0 m sin 40  12.0 m cos 20  1.579 m
R  R x  R y  19.5 m, and θ  tan 1
2
7.
2
 1.579 
 tan 1 
  4.65 S of W
Rx
 19.425 
Ry
Repeat the problem two problems prior, but for the second leg you walk 20.0 m in a
direction 40.0 o north of east (which is equivalent to subtracting B from A —that is,
to finding R'  A  B ). (b) Repeat the problem two problems prior, but now you first
walk 20.0 m in a direction 40.0 o south of west and then 12.0 m in a direction 20.0 
east of south (which is equivalent to subtracting A from B —that is, to finding
R  B  A  R ). Show that this is the case.
Solution
(a) R   A   B 
R x  Ax  B x  Acos110  B cos40
 12.0 m  cos 110  20.0 m  cos 40  11.217 m
R y  Ay  B y  A sin 110  B  sin 40
 12.0 m sin 110  20.0 m sin 40  24.132 m

2
2
R  Rx  Ry

1/ 2
 26.6 m
 Ry ' 
24.132 
  tan 1 
  65.1 N of E
 11.217 
 Rx ' 
  tan 1 
Openstax College Physics
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Chapter 3
(b) R   B  A
R x  B x  Ax  20.0 m  cos 220  12.0 m  cos 110  11.217 m
R y  B y  Ay  20.0 m sin 220  12.0 m sin 110  24.132 m

R   R x  R y
2

2 1/ 2
 26.6 m
 Ry " 
  65.1 S of W
R
"
 x 
  tan 1 
This is consistent with part (a) because A  B  B  A.
8.
Show that the order of addition of three vectors does not affect their sum. Show this
property by choosing any three vectors A , B , and C , all having different lengths
and directions. Find the sum A + B + C then find their sum when added in a different
order and show the result is the same. (There are five other orders in which A , B ,
and C can be added; choose only one.)
Solution
Refer to the solution for Exercise 3.7 above.
9.
Show that the sum of the vectors discussed in the example Subtracting Vectors
Graphically: A Woman Sailing a Boat gives the result shown in Figure 3.24.
Solution
R x  Ax  B x  27.5 m  cos 66  30.0 m  cos112  0.0529 m
R y  Ay  B y  27.5 m sin 66  30.0 m sin 112  52.938 m
R  Rx  R y 
2
 Ry
θ  tan 1 
 Rx
2
 0.0529 m 2  52.938 m 2
 52.9 m

52.938 m
  tan 1
 89.9 N of W  90.1w.r.t.  x - axis
 0.0529 m

(note Rx  0 )
10.
Find the magnitudes of velocities vA and v B in Figure 3.57.
Solution
Start with the information given: vtot  6.72 m/s; 49 N of E and we know that vA is
22.5 N of E. To calculate the angle of vB use the fact that the external angle of a
triangle,  equals the sum of the inner angles, so that   26.5  23.0  49.5 so
Openstax College Physics
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Chapter 3
that vB is at an angle of     72.0 N of E.
Now, getting the components of all vectors gives:
vtot, x  6.72 m/s cos 49.0  4.409 m/s; vtot, y  6.72 m/s sin 49.0  5.072 m/s;
vA, x  v A cos 22.5; vA, y  vA sin 22.5; vB, x  vB cos 72.0; vB, y  vB sin 72.0.
vtot x  4.409 m/s  vA cos 22.5  vB cos 72.0  vA 0.9239  vB 0.3090
(i)
vtot y  5.072 m/s  vA sin 22.5  vB sin 72.0  vA 0.3827   vB 0.9511
(ii)
Dividing equation (i) by (0.9239) gives: vA  4.772 m/s  0.3345vB . Substituting into
equation (ii) gives 4.772 m/s  0.3345vB 0.3827  vB 0.9511  5.072 m/s, or
vB = 3.94 m/s and vA = 3.45 m/s .
11.
Find the components of v tot along the x- and y-axes in Figure 3.57.
Solution
v tot x  6.72 m/s  cos( 49  )  4.41 m/s
v tot y  6.72 m/s  sin( 49  )  5.07 m/s
v tot x  4.41 m/s ; v tot y  5.07 m/s
12.
Find the components of vtot along a set of perpendicular axes rotated 30°
counterclockwise relative to those in Figure 3.57.
Solution
y'
vtot x  v tot cos19.0  6.72 m/s  cos19.0  6.35 m/s
v tot y  v tot sin 19.0  6.72 m/s sin 19.0  2.19 m/s
y


vtot = v'tot
49°
19°
30°
x'
x
3.3 VECTOR ADDITION AND SUBTRACTION: ANALYTICAL METHODS
Openstax College Physics
13.
Solution
Instructor Solutions Manual
Chapter 3
Find the following for path C in Figure 3.58: (a) the total distance traveled and (b) the
magnitude and direction of the displacement from start to finish. In this part of the
problem, explicitly show how you follow the steps of the analytical method of vector
addition.
(a) d  1  120 m   5  120 m   2  120 m   1  120 m   1  120 m   3  120 m 
 1.56  10 3 m
(b) s x  0  600  0  120  0  360 m  120 m
s y  120  0  240  0  120  0m  0 m
s  s 2 x  s 2 y  (120 m) 2  (0 m) 2  120 m

s

θ  tan 1  y   tan 1 0 m
120 m
 sx 

0  east , so S = 120 m, East
14.
Solution
Find the following for path D in Figure 3.58: (a) the total distance traveled and (b) the
magnitude and direction of the displacement from start to finish. In this part of the
problem, explicitly show how you follow the steps of the analytical method of vector
addition.
(a) d  2 120 m  6 120 m  4 120 m  1120 m  1560 m  1.56 km
(b) s x  0  720  0  120m  600 m
s y  0  720  0  120m  240 m
s  s 2 x  s 2 y  (600 m) 2  (240 m) 2  646 m
 sy
θ  tan 1 
 sx

240 m 
  tan 1 
  21.8 N of E
 600 m 

S = 646 m, 21.8 N of E
15.
Find the north and east components of the displacement from San Francisco to
Sacramento shown in Figure 3.59.
Openstax College Physics
Solution
Instructor Solutions Manual
Chapter 3
N-component: s x  123 kmsin 45  86.97 km  87.0 km
E-component: s y  123 km  cos 45  86.97 km  87.0 km
16.
Solution
Solve the following problem using analytical techniques: Suppose you walk 18.0 m
straight west and then 25.0 m straight north. How far are you from your starting
point, and what is the compass direction of a line connecting your starting point to
your final position? (If you represent the two legs of the walk as vector displacements
A and B , as in Figure 3.60, then this problem asks you to find their sum R  A  B .)

B
25 m
W
N

R

A 
18 m

E
AB  R
A  B2  R2
2
R
A2  B 2 
18.0 m 2  25.0 m 2
 30.8 m
B
 25.0 m 
θ  tan 1    tan 1 
  54.25
 A
 18.0 m 
θ  φ  90  φ  90  54.25  35.75
Compass reading = 35.8 W of N
17.
Repeat Problem 3.16 using analytical techniques, but reverse the order of the two legs
of the walk and show that you get the same final result. (This problem shows that
adding them in reverse order gives the same result—that is, B  A  A  B .) Discuss
how taking another path to reach the same point might help to overcome an obstacle
blocking your other path.
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Chapter 3

A
Solution
  
RAB

B
  
BAR

B


A
R  A  B  B  A by the commutative property of vector addition.
R  B 2  A2 
25.0 m 2  18.0 m 2
 30.8 m
 A
 18.0 m 
φ  tan 1    tan 1 
  35.8 W of N
B
 25.0 m 
18.
Solution
You drive 7.50 km in a straight line in a direction 15 east of north. (a) Find the
distances you would have to drive straight east and then straight north to arrive at the
same point. (This determination is equivalent to find the components of the
displacement along the east and north directions.) (b) Show that you still arrive at the
same point if the east and north legs are reversed in order.
(a) DE  Rcos  7.50 km  cos 75  1.94 km
DN  Rsin   7.50 km  sin 75  7.24 km
(b)

DE

DN

R

R

DE
North then east

DN
East then north
It is easily seen that D N  DE  DE  D N .
19.
Do Problem 3.16 again using analytical techniques and change the second leg of the
walk to 25.0 m straight south. (This is equivalent to subtracting B from A —that is,
finding R' = A - B ) (b) Repeat again, but now you first walk 25.0 m north and then
18.0 m east. (This is equivalent to subtract A from B —that is, to find A  B  C . Is
that consistent with your result?)
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Solution
Instructor Solutions Manual
(a)
Chapter 3
N

A

W


B' = – B
E

R'
S
R x  18.0 m, R y  25.0 m
R'  R x  R y  (18.0 m) 2  (25.0 m) 2  30.8 m
2
θ  tan 1
2
opp
 25.0 m 
 tan 1 
  54.2 S of W
adj
 18.0 m 


A' = – A
(b)
N

B

R''


W
E
S
R x  18.0 m, R y  25.0 m
R"  R x  R y  (18.0 m) 2  (25.0 m) 2  30.8 m
2
θ  tan 1
2
opp
 25.0 m 
 tan 1 
  54.2 N of E
adj
 18.0 m 
which is consistent with part (a).
20.
A new landowner has a triangular piece of flat land she wishes to fence. Starting at
the west corner, she measures the first side to be 80.0 m long and the next to be 105
m. These sides are represented as displacement vectors A from B in Figure 3.61. She
then correctly calculates the length and orientation of the third side C . What is her
result?
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Solution
Instructor Solutions Manual
Chapter 3
C  A  B
C y   Ay  B y  80.0 m sin  21  105 m sin 101  74.40 m
C x   Ax  B x
C x   A cos-21  B cos 101  80.0 m  cos 21  105 m cos101  54.65 m
C  C x  C y  (54.65 m) 2  (74.40 m) 2  92.3 m
2
θ  tan 1
21.
Solution
2
 74.40 m 
 tan 1 
  53.7 S of W
Cx
 54.65 m 
Cy
You fly 32.0 km in a straight line in still air in the direction 35.0 south of west. (a)
Find the distances you would have to fly straight south and then straight west to
arrive at the same point. (This determination is equivalent to finding the components
of the displacement along the south and west directions.) (b) Find the distances you
would have to fly first in a direction 45.0 south of west and then in a direction 45.0
west of north. These are the components of the displacement along a different set of
axes—one rotated 45 .
(a) DS  Rsin   32.0 km sin 35  18.4 km
DW  Rcos  32.0 km cos 35  26.2 km
(b)
Consider rotated axes, x and y :
DSW  Rcos   32.0 km cos10  31.5 km
DNW  Rsin    32.0 km sin 10  5.56 km
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22.
Instructor Solutions Manual
Chapter 3
A farmer wants to fence off his four-sided plot of flat land. He measures the first three
sides, shown as A, B, and C in Figure 3.62, and then correctly calculates the length
and orientation of the fourth side D . What is his result?
Solution
D  A  B  C
D x   Ax  B x  C x  4.70 km  cos 7.50  2.48 km  cos 106  3.02 km  cos 161
 1.12 km
D y   Ay  B y  C y  4.70 km sin  7.50  2.48 km sin 106  3.02 km sin 161
 2.75 km
D  D x  D y  (1.12 km) 2  (2.75 km) 2  2.97 km
2
θ  tan 1
23.
2
 1.12 km 
 tan 1 
  22.2 W of S
Dy
 2.75 km 
Dx
In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind
shifts a great deal during the day, and he is blown along the following straight lines:
2.50 km 45.0 north of west; then 4.70 km 60.0 south of east; then 1.30 km 25.0
south of west; then 5.10 km straight east; then 1.70 km 5.00 east of north; then
7.20 km 55.0 south of west; and finally 2.80 km 10.0 north of east. What is his
final position relative to the island?
Solution
Gilligan’s travels are composed of seven vectors. Their lengths in km and angles (ccw
from due east) are: A = 2.50 km @ 135 , B = 4.70 km @ 300 , C = 1.30 km @ 205 ,
D = 5.10 km @ 0 , E = 1.70 km @ 85 , F = 7.20 km @ 235 , G = 2.80 km @ 10
The resultant vector is: R  A  B  C  D  E  F  G
 2.50 cos 135  4.70 cos 300  1.30 cos 205  5.10 cos 0 
 km  3.280 km
R x  
  1.70 cos 85  7.20 cos 235  2.80 cos 10

 2.50 sin 135  4.70 sin 300  1.30 sin 205  5.10 sin 0 
 km  - 6.570 km
R y  
  1.70 sin 85  7.20 sin 235  2.80 sin 10

R  (3.280 km) 2  (6.570 km) 2  7.34 km
  6.570 km 
α  tan 1 
  63.5  63.5 S of E
 3.280 km 
Openstax College Physics
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Chapter 3
24.
Suppose a pilot flies 40.0 km in a direction 60 north of east and then flies 30.0 km
in a direction 15 north of east as shown in Figure 3.63. Find her total distance R
from the starting point and the direction  of the straight-line path to the final
position. Discuss qualitatively how this flight would be altered by a wind from the
north and how the effect of the wind would depend on both wind speed and the speed
of the plane relative to the air mass.
Solution
R x  (40.0 km)cos60   (30.0 km)cos15   48.978 km
R y  (40.0 km)sin60   (30.0 km)sin15   42.406 km
R  R x  R y  64.8 km;
2
2
 42.406 km 
θ  tan 1 
  40.9 N of E
 48.978 km 
If the wind speed is less than the speed of the plane, it is possible to travel to the
northeast, but she will travel more to the east than without the wind. If the wind
speed is greater than the speed of the plane, then it is no longer possible for the
plane to travel to the northeast, it will end up travelling southeast.
3.4 PROJECTILE MOTION
25.
A projectile is launched at ground level with an initial speed of 50.0 m/s at an angle of
30.0° above the horizontal. It strikes a target above the ground 3.00 seconds later.
What are the x and y distances from where the projectile was launched to where it
lands?
Solution
Range of projectile on level ground: R 
v02
(50.0 m/s) 2
sin 2 
sin 60.0  221 m
g
9.80 m/s 2
The time in air is given as 3.00 s, so projectile landed above level ground. Find the
position relative to the launching point:
x  v0 x t  50.0 m/s cos 30.03.00 s   1.30  10 2 m
1
1
2
y  v0 y t  at 2  50.0 m/s sin 30.03.00 s   9.80 m/s 2 3.00 s   30.9 m
2
2
Therefore, the projectile landed 1.30 10 2 m horizontally and 30.9 m vertically from
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the launching point.
26.
A ball is kicked with an initial velocity of 16 m/s in the horizontal direction and 12 m/s
in the vertical direction. (a) At what speed does the ball hit the ground? (b) For how
long does the ball remain in the air? (c)What maximum height is attained by the ball?
Solution (a) vf2  v02x  v02y
vf  16 2  12 2 m/s   20 m/s
(b)
vf y  v0 y  at
0 m/s  12 m/s  9.8 m/s 2  t s 
t  1.224 s  t total  2  1.224 s  2.45 s
(c) vf2  v02  2ay


0 2  12 m/s   2 9.8 m/s 2 y m 
2
y  7.35 m
27.
Solution
A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from
the base of the building. Ignore air resistance. (a) How long is the ball in the air? (b)
What must have been the initial horizontal component of the velocity? (c) What is the
vertical component of the velocity just before the ball hits the ground? (d) What is the
velocity (including both the horizontal and vertical components) of the ball just before
it hits the ground?
(a)
1 2
at
2
y  v0 t 


1
 9.8 m/s 2 t 2
2
t  3.50 s
60 m  0 
(b)
x  v0 x t
100 m  v0 x 3.50 s 
v0 x  28.6 m/s
(c) v y  v0 y  at  0   9.8 m/s 2 3.50 s   34.3 m/s
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(d) v 2  v x2  v y2  28.6 m/s 2   34.3 m/s 2
v  44.7 m/s
 vy
 vx
  tan 1 
28.

 34.3 
  tan 1 
  50.2 below x - axis
 28.6 

(a) A daredevil is attempting to jump his motorcycle over a line of buses parked end to
end by driving up a 32° ramp at a speed of 40.0 m/s ( 144 km/h) . How many buses
can he clear if the top of the takeoff ramp is at the same height as the bus tops and
the buses are 20.0 m long? (b) Discuss what your answer implies about the margin of
error in this act—that is, consider how much greater the range is than the horizontal
distance he must travel to miss the end of the last bus. (Neglect air resistance.)
Solution
v0 sin 2θ0 (40.0 m/s) 2 sin(2  32)

 146.7 m
(a) R 
g
9.80 m/s 2
R
146.7 m
n

 7.34 buses
bus length
20.0 m
2
So, he can only make it over 7 buses.
(b) He clears the last bus by 6.7 m, which seems to be a large margin of error, but
since we neglected air resistance, it really isn’t that much room for error.
29.
An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at
same height as the release height of the arrow. (a) At what angle must the arrow be
released to hit the bull’s-eye if its initial speed is 35.0 m/s? In this part of the problem,
explicitly show how you follow the steps involved in solving projectile motion
problems. (b) There is a large tree halfway between the archer and the target with an
overhanging horizontal branch 3.50 m above the release height of the arrow. Will the
arrow go over or under the branch?
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Solution (a) R  75.0 m, v0  35.0 m/s,   ?
2
v sin 2θ0
Use the equation for a projectile on level ground: R  0
g
θ0 
1 1  gR  1 1  (9.80 m/s 2 )(75.0 m) 
sin  2   sin 
  18.4
2
2
2
(35.0
m/s)
v


 0 
(b) The arrow will be at the tree when the vertical velocity is zero:
v y  v0 sin   gt  t  v0 sin  / g  1.127 s
y  v0 sin θt 
1 2
gt
2
1
 (35.0 m/s)sin18. 4(1.13 s)  (9.80 m/s 2 )(1.13 s) 2  6.23 m
2
The arrow goes over the branch!
30.
A rugby player passes the ball 7.00 m across the field, where it is caught at the same
height as it left his hand. (a) At what angle was the ball thrown if its initial speed was
12.0 m/s, assuming that the smaller of the two possible angles was used? (b) What
other angle gives the same range, and why would it not be used? (c) How long did this
pass take?
Solution
2
 gR 
v0 sin 2θ 0
1
, θ 0  sin 1  2 
(a) R 
g
2
 v0 
1 1  (9.8 m/s 2 )(7.0 m) 
  14.2
 0  sin 
2
2
(
12
m/s)


(b)  0   0 '  90 or  0 '  90  14.22  75.8
This angle is not used as often, because the time of flight will be longer. In rugby
that means the defense would have a greater time to get into position to knock
down or intercept the pass that has the larger angle of release.
(c) To find the time of the pass we use:
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x  x0  v x t , where v x  v cos   12 m/s cos(14.2)  11.63 m/s
t
31.
Solution
32.
Solution
x  x0
7.0 m

 0.60 s
vx
11.63 m/s
Verify the ranges for the projectiles in Figure 3.41(a) for   45.0 and the given initial
velocities.
2
2
v0 sin 2θ 0 v0

for   45 :
g
g
R  91.8 m for v0  30 m/s; R  163 m for v0  40 m/s; R  255 m for v0  50 m/s
R
Verify the ranges shown for the projectiles in Figure 3.41 (b) for an initial velocity of
50 m/s at the given initial angles.
R
50.0 m/s 2 sin 2θ 0
g
R = 255 m for   45.
33.
for v0  50 m/s : R  128 m for   15 and 75
The cannon on a battleship can fire a shell a maximum distance of 32.0 km. (a)
Calculate the initial velocity of the shell. (b) What maximum height does it reach? (At
its highest, the shell is above 60% of the atmosphere—but air resistance is not really
negligible as assumed to make this problem easier.) (c) The ocean is not flat, because
the Earth is curved. Assume that the radius of the Earth is 6.37  10 3 km . How many
meters lower will its surface be 32.0 km from the ship along a horizontal line parallel
to the surface at the ship? Does your answer imply that error introduced by the
assumption of a flat Earth in projectile motion is significant here?
Solution (a) The range is a maximum when   45 :
2
Rmax
v
 0 solving for the initial velocity
g
v0  gRmax  (9.80 m/s 2 )(32.0  10 3 m)  560 m/s
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(v0 sin  0 ) 2
(b) The height can be found from the equation h 

2g
2g
v0 y
2
(560 m/s)( sin 45)2  8.00 103 m
2(9.80 m/s 2 )
(c) Let R = radius of the earth
α
32.0 km
32.0 km
 360 
 360  0.2878
2π R
2  (6.37  10 3 km)
R
6.37  10 3 km

 6.37008  10 3 km
cos α
cos 0.2878
3
d = x - R = 0.00008  10 km  80 m.
R  x cos α  x 
This error is not significant because it is only 1% of the answer in part (b).
34.
An arrow is shot from a height of 1.5 m toward a cliff of height H . It is shot with a
velocity of 30 m/s at an angle of 60° above the horizontal. It lands on the top edge of
the cliff 4.0 s later. (a) What is the height of the cliff? (b) What is the maximum height
reached by the arrow along its trajectory? (c) What is the arrow’s impact speed just
before hitting the cliff?
Solution
1 2
1
2
2
(a) y  v0 y t  2 at  30 m/s sin 604.0 s   2  9.8 m/s 4 s   25.5 m

H  1.5 m  y  27.0 m

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(b) v 2  v 02  2ay

Chapter 3

0  30.0 m/s  sin 60  2  9.8 m/s 2 y
2
y  34.44 m
H  34.44 m  1.5 m  36.0 m


(c) vf y  v0 y  at  30 m/s sin 60   9.8 m/s 2 4.0 s   13.2 m/s
v 2  vf2x  vf2y
v
35.
30 cos 602  13.22 m/s   20 m/s
In the standing broad jump, one squats and then pushes off with the legs to see how
far one can jump. Suppose the extension of the legs from the crouch position is 0.600
m and the acceleration achieved from this position is 1.25 times the acceleration due
to gravity, g . How far can they jump? State your assumptions. (Increased range can
be achieved by swinging the arms in the direction of the jump.)
Solution


vf2  v02  2ax  0  2 1.25  9.8 m/s 2 0.600 m   3.83 m/s
Assume the person leaves at 45 and on level ground, thus vf y  v0 y  at


0  3.83 m/s sin 45  9.80 m/s 2 t
t  0.276 s to the top of the jump, tTOTAL  2  0.276  0.553 s
x  v0 x t  3.83 m/s  cos 450.553 s   1.5 m
36.
The world long jump record is 8.95 m (Mike Powell, USA, 1991). Treated as a
projectile, what is the maximum range obtainable by a person if he has a take-off
speed of 9.5 m/s? State your assumptions.
Solution Assume motion is on level ground (the person leaves at the ground height) and the
angle is 45 .
R  9.5 m 
2
sin 90  9.21 m
9.8 m/s 2
Or, assume the person leaves the ground with their center of mass 1.0 m above the
ground. To find the time in the air:
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y  1.0 m  9.5 m/s sin 45t 

Chapter 3

1
9.80 m/s 2 t 2  t  1.51 s
2
So, R  9.5 m/s cos 451.51 s  10.44 m
37.
Solution
Serving at a speed of 170 km/h a tennis player hits the ball at a height of 2.5 m and an
angle  below the horizontal. The baseline from which the ball is served is 11.9 m
from the net, which is 0.91 m high. What is the angle  such that the ball just crosses
the net? Will the ball land in the service box, which has an outermost service line 6.40
m from the net?
v0  170 km/h  47.2 m/s
1
y  2.50 m  0.91 m  1.59 m and y  v0 y t  at 2  47.2 m/s sin  t  4.9 m/s 2 t 2
2
In x direction x  11.9 m  47.2 m/s cos t , so t 
0.252
. Inserting this value into
cos θ
the first equation,


1.59 m  11.9 m tan   0.311 m 1  tan 2   tan θ  0.107 or θ  6.1
Using this value, the time for the ball to fall the full 2.5 m can be calculated, and is
t  0.366 s . The range of the ball in this time is R  v x t  17.2 m , so yes, the ball lands
at 5.3 m from the net.
38.
A football quarterback is moving straight backward at a speed of 2.00 m/s when he
throws a pass to a player 18.0 m straight downfield. (a) If the ball is thrown at an
angle of 25° relative to the ground and is caught at the same height as it is released,
what is its initial speed relative to the ground? (b) How long does it take to get to the
receiver? (c) What is its maximum height above its point of release?
Solution (a) Note: the player’s backward motion will not be a factor in this problem.
α  angle relative to ground
2
R
v0 sin 2α
 v0 
g
Rg

sin 2α
(18.0 m)(9.80 m/s 2 )
 15.2 m/s
sin 50.0
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(b) v x  v0 cos α  (15.2 m/s) cos 25.0  13.8 m/s
t
(c) h 
39.
R
18.0 m

 1.31 s
v x 13.8 m/s
(v0 sin  ) 2 (15.2 m/s )(sin 25.0)2

 2.11 m
2g
2(9.80 m/s 2 )
Gun sights are adjusted to aim high to compensate for the effect of gravity, effectively
making the gun accurate only for a specific range. (a) If a gun is sighted to hit targets
that are at the same height as the gun and 100.0 m away, how low will the bullet hit if
aimed directly at a target 150.0 m away? The muzzle velocity of the bullet is 275 m/s.
(b) Discuss qualitatively how a larger muzzle velocity would affect this problem and
what would be the effect of air resistance.
Solution (a)
bullet path
ideal target
real target
v0
0
R
d
v0 xt '
Use the 100 m data to calculate the release angle for the bullet.
2
v sin 2θ0
R 0
so that
g
1 1  gR  1 1  (9.80 m/s 2 )(100 m) 
θ0  sin  2   sin 
  0.3712
2
(275 m/s) 2


 v0  2
Next, calculate the time to travel.
x  x0  150 m  v0 x t  v0 cos θ 0 t
t
x  x0
150 m

 0.5455 s
v0 cos θ 0 (275 m/s) cos 0.3712
Lastly, calculate the change in vertical position during the 150 m flight:
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Chapter 3
1
y  y0  v0 y t  gt 2
2
1
 (275 m/s) sin 0.3712(0.5455 s)  (9.80 m/s 2 )(0.5455 s) 2   0.486 m
2
(b) The larger the muzzle velocity, the smaller the deviation in the vertical direction,
because the time of flight would be smaller. Air resistance would have the effect
of decreasing the time of flight, therefore increasing the vertical deviation.
40.
An eagle is flying horizontally at a speed of 3.00 m/s when the fish in her talons
wiggles loose and falls into the lake 5.00 m below. Calculate the velocity of the fish
relative to the water when it hits the water.
Solution
x -direction (horizontal); given:
v0 x  3.00 m/s, a x  0 m/s 2 , v x  v0 x  constant  3.00 m/s
y -direction (vertical); given:
v0 y  0.00 m/s, a y   g  9.80 m/s 2 ,  y  y 0   5.00 m
v y2  v02y  2 g  y  y 0 
v y  (0 m/s) 2  2(9.80 m/s 2 )( 5.00 m)  9.90 m/s
v  v x  v y  (3.00 m/s) 2  (9.90 m/s ) 2  10.3 m/s
2
2
 vy 
  9.90 m/s 
θ  tan 1    tan 1 
  73.1
 3.00 m/s 
 vx 
v  10.3 m/s, 73.1 below the horizontal
41.
An owl is carrying a mouse to the chicks in its nest. Its position at that time is 4.00 m
west and 12.0 m above the center of the 30.0 cm diameter nest. The owl is flying east
at 3.50 m/s at an angle 30.0 below the horizontal when it accidentally drops the
mouse. Is the owl lucky enough to have the mouse hit the nest? To answer this
question, calculate the horizontal position of the mouse when it has fallen 12.0 m.
Solution The mouse will land in the nest if its horizontal, x , is > 3.85 m and < 4.15 m ( 4.0 m 
the radius of the nest). y  y0  12.0 m , and v 0 ,y  (3.50 m/s)sin30   1.75 m/s.
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To calculate the horizontal position, we need to first calculate the time it takes to fall
the distance of 12.0 m.
1 2
gt , so solve for t :
2
y  y 0  v0, y t 
 12 m  (1.75 m/s) t 
t
9.80 m/s 2 2
t , or t 2  (0.3571 s)t  2.449 s 2  0 so that
2
 (0.3571 s)  (0.3571 s) 2  4(2.449 s 2 )
 1.40 s
2
Next, we need to calculate the horizontal displacement.
x  x0  v0 x t , so that x  x0  (3.50 m/s) cos 30(1.40 s)  4.23 m
No, the owl is not lucky. The mouse just misses the nest.
42.
Suppose a soccer player kicks the ball from a distance 30 m toward the goal. Find the
initial speed of the ball if it just passes over the goal, 2.4 m above the ground, given
the initial direction to be 40 o above the horizontal.
Solution Given  0  40 , x -direction
x  x0  30.0 m
v0 x  v0 cos  0
x  x0  v0 x t  v0 cos  0 t
y -direction, y  y 0  2.4 m
v0 y  v0 sin  0
y  y 0  v0 y t 
1 2
gt , find v0 .
2
Solve the x -equation for t , then substitute into the y -equation:
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t
x
;
v0 cos  0

x
y  v0 sin  0 
 v0 cos  0

 1 
x
  g 
 2  v0 cos  0

y cos 2 θ 0 v0  x tan θ 0  cos 2 θ 0 v0 
2
Chapter 3
2



2
1 2
 1/ 2 gx 2
gx or v0 
2
cos 2 θ 0  y  x tan θ 0 
Substituting in the values gives:
v0 
43.
 0.5(9.80 m/s 2 )(30.0 m) 2
 18.2 m/s
(cos 2 40)[2.40 m  (30.0 m) tan 40]
Can a goalkeeper at her/ his goal kick a soccer ball into the opponent’s goal without
the ball touching the ground? The distance will be about 95 m. A goalkeeper can give
the ball a speed of 30 m/s.
Solution For the maximum range, we use   45.
R
v02 sin( 2 0 ) (30 m/s) 2 sin( 90)

 91.8 m
g
9.8 m/s 2
No, the maximum range is about 92 m (neglecting air resistance). The ball will not
travel 95 m before hitting the ground.
44.
Solution
The free throw line in basketball is 4.57 m (15 ft) from the basket, which is 3.05 m (10
ft) above the floor. A player standing on the free throw line throws the ball with an
initial speed of 8.15 m/s, releasing it at a height of 2.44 m (8 ft) above the floor. At
what angle above the horizontal must the ball be thrown to exactly hit the basket?
Note that most players will use a large initial angle rather than a flat shot because it
allows for a larger margin of error. Explicitly show how you follow the steps involved
in solving projectile motion problems.
Openstax College Physics
Instructor Solutions Manual
Chapter 3
x - direction, x  x0  4.57 m  L
vx  v0 cos θ0  (8.15 m/s) cos θ0
L   v0 cos θ0  t
t
L
v0 cos θ0
y -direction, y  3.05 m, y0  2.44 m
y  y0  0.61 m  h
v0 ,y  v0 sin θ0  (8.15 m/s)sin θ0
1
y  y0  h   v0 sin θ0  t  gt 2
2
Find θ 0 .
2

 1 

L
L
1  gL2 




h  v0 sin θ0 
  2 g  v cos θ   L tan θ0  2  v 2 cos 2 θ 
v
cos
θ
0 
0 
0 
 0
 0
 0
Using the identity 1  tan 2 α  sec 2 α 
1  gL2
h  L tan θ 0   2
2  v0
1  gL2
h  2
2  v0
1
,
cos 2 α

 2
 1  tan 2 θ 0   1  gL

2  v0 2



 2
  L tan θ 0  1  gL

2  v0 2



 2
  L tan θ 0  1  gL

2  v0 2

 2
 tan θ 0


 2
 tan θ 0  0  aX 2  bX  c where X  tan  0 ;


1  gL2  (9.80 m/s 2 )(4.57 m) 2
a  2 
 1.5407 m;b   L  4.57 m; and
2  v0 
2(8.15 m/s) 2
1  gL2 
c  h   2   0.61 m  1.5407 m  2.1507 m
2  v0 
X
so that
b  b 2  4ac
2a
4.57 m  (  4.57 m) 2  4(1.5407 m)(2.1507 m)

 0.5866 or 2.3796
2(1.5407 m)
Openstax College Physics
Instructor Solutions Manual
Chapter 3
 0  tan 1 0.5866  30.4
or
 0  tan 1 2.3796  67.2
45.
In 2007, Michael Carter (U.S.) set a world record in the shot put with a throw of 24.77
m. What was the initial speed of the shot if he released it at a height of 2.10 m and
threw it at an angle of 38.0 above the horizontal? (Although the maximum distance
for a projectile on level ground is achieved at 45 when air resistance is neglected, the
actual angle to achieve maximum range is smaller; thus, 38 will give a longer range
than 45 in the shot put.)
Solution Given:   38
x  x  x0  24.77 m, x  (v0 cos  )t , so that t 
y  y  y0  2.10 m , so that y  (v0 sin  )t 
x
,
v 0 cos 
1 2
gt , substituting for t gives:
2
 1  gx 2 
 x  1  x 
  g 
 or y  x tan    2 

y  (v0 sin  )
 v  2 cos 2   , so that


 v0 cos   2  v0 cos  
 0 
2
 gx 2 
1

v0   
2
 2 cos   ( y  x tan  )

46.
 (9.80 m/s 2 )(24.77 m) 2
 15.0 m/s
2cos 2 38[2.10 m  (24.77 m)tan38 ]
A basketball player is running at 5.00 m/s directly toward the basket when he jumps
into the air to dunk the ball. He maintains his horizontal velocity. (a) What vertical
velocity does he need to rise 0.750 m above the floor? (b) How far from the basket
(measured in the horizontal direction) must he start his jump to reach his maximum
height at the same time as he reaches the basket?
(a) Given: vx  5.00 m/s, y  y0  0.75 m, vy  0 m/s, a y   g  9.80 m/s 2 . Find v0, y .
Openstax College Physics
Instructor Solutions Manual
Chapter 3
v y2  v02y  2 g  y  y 0 
v0 ,y  v y  2 g  y  y 0   (0 m/s) 2  2(9.80 m/s 2 )(0.75 m)  3.83 m/s
2
(b) v y  v0, y  gt so that
v0 ,y  v y
(3.83 m/s)  (0 m/s)
 0.391 s
g
9.80 m/s 2
x  x0  v x t , so that  x  x0   v x t  5.00 m/s 0.391 s   1.96 m
t

47.
A football player punts the ball at a 45.0 angle. Without an effect from the wind, the
ball would travel 60.0 m horizontally. (a) What is the initial speed of the ball? (b)
When the ball is near its maximum height it experiences a brief gust of wind that
reduces its horizontal velocity by 1.50 m/s. What distance does the ball travel
horizontally?
Solution
gR

(a) θ0  45,R  60.0 m, v0 
sin 2θ0
9.80 m/s 60.0 m  24.2 m/s
2
sin 2  45
(b) v0 ,x  v0 cos 45  (24.2 m/s)cos45   17.15 m/s
With the wind gust, the horizontal velocity is decreased by 1.50 m/s:
v x  v0, x  1.50 m/s  17.15 m/s  1.50 m/s  15.65 m/s
The time can be calculated for each half of the flight:
x  x0
30.0 m
t1 / 2 

 1.750 s
vx
17.1 m/s
(The time is not changed by wind gust, since it only acts in x-direction.) Finally, the
distance traveled after the wind gust is:
x  x0  v x t  15.65 m/s 1.750 s   27.38 m
So the ball travels a total of 57.4 m with the brief gust of wind.
Openstax College Physics
48.
Instructor Solutions Manual
Chapter 3
Prove that the trajectory of a projectile is parabolic, having the form y  ax  bx 2 . To
obtain this expression, solve the equation x  v0 x t for t and substitute it into the
expression for y  v0 y t  (1 / 2) gt 2 . (These equations describe the x and y positions
of a projectile that starts at the origin.) You should obtain an equation of the form
y  ax  bx 2 where a and b are constants.
Solution
x  v0 x t  t 
x
x

v0 x v0 cos 
Substitute into the y -equation: y  (v0 sin  )t 
 x  1  x 
  g 

y  (v0 sin  )
 v0 cos   2  v0 cos  
1 2
gt , giving
2
2

 2
g
 tan  x  
x  ax  bx 2
2 
 2v0 cos   
a  tan  , b 
49.
Derive R 
g
, which are constants.
2(v 0 cos  ) 2
v 02 sin 2 0
for the range of a projectile on level ground by finding the time
g
t at which y becomes zero and substituting this value of t into the expression for
x  x0 , noting that R  x  x0 .
Openstax College Physics
Solution
y  y 0  0  v0 y t 
Instructor Solutions Manual
Chapter 3
1 2
1
2(v 0 sin  )
gt  (v0 sin  )t  gt 2 so that t 
2
2
g
x  x0  v0 x t  (v0 cos  )t  R, and substituting for t gives:
 2v sin   2v0 sin  cos 
 
R  v0 cos   0
g
g


2
v sin 2
Since 2 sin  cos   sin 2 , the range is: R  0
g
2
3.5 ADDITION OF VELOCITIES
52.
Bryan Allen pedaled a human-powered aircraft across the English Channel from the
cliffs of Dover to Cap Gris-Nez on June 12, 1979. (a) He flew for 169 min at an average
velocity of 3.53 m/s in a direction 45 south of east. What was his total displacement?
(b) Allen encountered a headwind averaging 2.00 m/s almost precisely in the opposite
direction of his motion relative to the Earth. What was his average velocity relative to
the air? (c) What was his total displacement relative to the air mass?
Solution
60 s 

4
(a) d  v0 t  3.53 m/s 169 min 
  3.579  10 m  35.8 km, 45 S of E
1 min 

(b) vPG  vPA  vAG  vPA  vPG  vAG , so that
vPA  3.53 m/s   2.0 m/s   5.53 m/s, 45 S of E
60 s 

(c) x A  v PAt  5.53 m/s 169 min 
  56.1 km, 45 S of E
1 min 

Openstax College Physics
53.
Instructor Solutions Manual
Chapter 3
A seagull flies at a velocity of 9.00 m/s straight into the wind. (a) If it takes the bird
20.0 min to travel 6.00 km relative to the Earth, what is the velocity of the wind? (b) If
the bird turns around and flies with the wind, how long will he take to return 6.00 km?
(c) Discuss how the wind affects the total round-trip time compared to what it would
be with no wind.
Solution (a) A= air; S = seagull; G = ground, so v SG  v SA  v AG  v AG  v SG  v SA ;
xSG
6.00  103 m

 5.00 m/s
t
(20 min)(60 s/1 min)
 5.00 m/s  9.00 m/s   4.00 m/s
vSG 
v AG
(b) vSG  vSA  v AG  9.00 m/s   4.00 m/s   13.00 m/s
x
6.00  10 3 m
xSG  vSG t  t  SG 
 462 s  7 min and 42 s
vSG
13.00 m/s
(c) The wind will always slow down the round trip time, relative to having no wind
present.
54.
Near the end of a marathon race, the first two runners are separated by a distance of
45.0 m. The front runner has a velocity of 3.50 m/s, and the second a velocity of 4.20
m/s. (a) What is the velocity of the second runner relative to the first? (b) If the front
runner is 250 m from the finish line, who will win the race, assuming they run at
constant velocity? (c) What distance ahead will the winner be when she crosses the
finish line?
Openstax College Physics
Instructor Solutions Manual
Chapter 3
Solution (a) Label F for the front runner and S for the second runner:
vF  3.50 m/s, vS  4.20 m/s
vSF  vS  vF  4.20 m/s  3.50 m/s  0.70 m/s (faster th an the first runner)
(b) t F 
tS 
xF
250 m

 71.43 s
v F 3.50 m/s
xS 250  45 m

 70.24 s
vS
4.20 m/s
The second runner will win.
(c) xSF  x0,SF  vSF tS  45.0 m  0.70 m/s 70.24 s   4.17 m
55.
Verify that the coin dropped by the airline passenger in the Example 3.8 travels 144 m
horizontally while falling 1.50 m in the frame of reference of the Earth.
Solution
1
y  y 0   gt 2  t 
2
 2y
, so substituting for t gives:
g
x  x0  v x t  (260 m/s)
 2(1.50 m)
 144 m
9.80 m/s 2
56.
A football quarterback is moving straight backward at a speed of 2.00 m/s when he
throws a pass to a player 18.0 m straight downfield. The ball is thrown at an angle of
25.0 relative to the ground and is caught at the same height as it is released. What is
the initial velocity of the ball relative to the quarterback?
Solution
vA  vB  v0  vA
v Ax  v Bx  v 0 x  v 0 cos 
v Ay  v By  v0 y  v0 sin 
vB

vA = 2 m/s
v0 = 15.2 m/s
(relative to ground)
 

vB is relative to quarterback
Openstax College Physics
Instructor Solutions Manual
Chapter 3
 2 m/s  vBx  (15.2 m/s)cos25. 0  vBx  15.8 m/s
0  vBy  (15.2 m/s) sin 25.0  vBy  6.42 m/s
vB  vBx  vBy  17.0 m/s
2
2
 6.42 
  22.1
 15.8 
  tan 1 
57.
A ship sets sail from Rotterdam, The Netherlands, heading due north at 7.00 m/s
relative to the water. The local ocean current is 1.50 m/s in a direction 40.0 north of
east. What is the velocity of the ship relative to the Earth?
Solution B = boat, W = water, G = ground, so that v BG  v BW  v WG
1.50 m/s
N
40°
7.00 m/s

E
v BGx  v BWx  v WGx  0  1.50 m/s  cos 40  1.149 m/s
v BGy  v BWy  v WGy  7.00 m/s  1.50 m/s sin 40  7.964 m/s

v BG  v BGx  v BGy  (1.149 m/s) 2  (7.964 m/s) 2
2
 v BGy
α  tan 1 
 v BGx
2

1/2

7.964 m/s 
  tan 1 
  81.8 N of E
 1.149 m/s 

 8.05 m/s
Openstax College Physics
58.
Instructor Solutions Manual
Chapter 3
(a) A jet airplane flying from Darwin, Australia, has an air speed of 260 m/s in a
direction 5.0 south of west. It is in the jet stream, which is blowing at 35.0 m/s in a
direction 15 south of east. What is the velocity of the airplane relative to the Earth?
(b) Discuss whether your answers are consistent with your expectations for the effect
of the wind on the plane’s path.
Solution (a) P = plane; A = air; G = ground
v PG, x  v PA, x  v AG, x
v PG,x  (260 m/s)cos185   (35.0 m/s)cos(15 )  225.20 m/s
v PG, y  v PA, y  v AG, y
v PG,y  (260 m/s)sin185   (35.0 m/s)sin( 15)  31.72 m/s
260 m/s
5° 
20°
45 m/s
v PG  v PG, x  v PG, y  (225.2 m/s) 2  (31.72 m/s) 2  227 m/s  230 m/s
2
2
 v BG,y 
 31.72 m/s 
  tan 1 
  8.0 , S of W

  225.20 m/s 
 v BG,x 
  tan 1 
(b) The wind should make the plane travel slower and more to the south, which is
what was calculated.
59.
(a) In what direction would the ship in Exercise 3.57 have to travel in order to have a
velocity straight north relative to the Earth, assuming its speed relative to the water
remains 7.00 m/s ? (b) What would its speed be relative to the Earth?
Solution
v  vc  vs
Openstax College Physics
Instructor Solutions Manual
Chapter 3

v
vs = 7.00 m/s
40°
vc = 1.50 m/s
(a)
v x  0  vsx  v c cos 40  vs sin   vc cos 40
sin φ 
vc
1.50 m/s
cos40 
cos40  9.45 W of N
vs
7.00 m/s
(b) v y  vsy  vc sin 40
 vs cos   vc sin 40  (7.00 m/s) cos 9.45  (1.50 m/s) sin 40  7.87 m/s
60.
Solution
(a) Another airplane is flying in a jet stream that is blowing at 45.0 m/s in a direction
20.0 south of east (as in Exercise 3.58). Its direction of motion relative to the Earth is
45.0 south of west, while its direction of travel relative to the air is 5.00 south of
west. What is the airplane’s speed relative to the air mass? (b) What is the airplane’s
speed relative to the Earth?
(a)
20°
45 m/s

va
5°
5°
40°
45°

vg
115°
a = airplane, w = wind, g = ground
va, x  va cos 5  (45.0 m/s) cos 20  vg cos 45
va,y  va sin 5  (45.0 m/s) sin 20  vg sin 45
va cos 5  sin 5  (45.0 m/s)( cos 20  sin 20)
va  (45.0 m/s)
(since cos 45  sin 45)
cos 20  sin 20
 63.5 m/s
cos 5  sin 5
Openstax College Physics
Instructor Solutions Manual
Chapter 3
(b) vg  va  v w  2va v w cos 
2
2
2
 (63.45 m/s) 2  (45.0 m/s) 2


vg  

  2(63.45 m/s)(45.0 m/s)cos25  
61.
1/ 2
 29.6 m/s
A sandal is dropped from the top of a 15.0-m-high mast on a ship moving at 1.75 m/s
due south. Calculate the velocity of the sandal when it hits the deck of the ship: (a)
relative to the ship and (b) relative to a stationary observer on shore. (c) Discuss how
the answers give a consistent result for the position at which the sandal hits the deck.
Solution (a) v 2  2gy  y   v   2 gy 1/ 2
y
0
y

  2(9.80 m/s 2 )( 15.0 m)

1/2
 17.15 m/s
v y  17.1 m straight down

(b) v  v x  v y
2
  tan 1
vy
vx
  (1.75 m/s)
2 1/2
2
 (17.15 m/s) 2

1/2
 17.2 m/s
 84.2 below horizontal and to the south.
(c) The sandal hits the ship going straight down (according to the ship), but the ship is
moving south, so the observer on the shore sees the sandal moving mainly down,
but also a bit to the south.
62.
The velocity of the wind relative to the water is crucial to sailboats. Suppose a sailboat
is in an ocean current that has a velocity of 2.20 m/s in a direction 30.0° east of north
relative to the Earth. It encounters a wind that has a velocity of 4.50 m/s in a direction
of 50.0° south of west relative to the Earth. What is the velocity of the wind relative
to the water?
Solution
v WO  v WE  v EO  v WE  v OE
Openstax College Physics
Instructor Solutions Manual
Chapter 3
v WOx  v WEx  v0Ex   4.50 m/s  cos 50  2.20 m/s  cos 60  3.993 m/s
v WOy  v WEy  vOEy   4.50 m/s sin 50  2.20 m/s sin 60  5.352 m/s
v  v x  v y  (3.993 m/s) 2  (5.352 m/s ) 2  6.68 m/s
2
α  tan 1
2
vy
vx
 tan 1
-5.352 m/s
 53.3 S of W
-3.993 m/s
63.
The great astronomer Edwin Hubble discovered that all distant galaxies are receding
from our Milky Way Galaxy with velocities proportional to their distances. It appears
to an observer on the Earth that we are at the center of an expanding universe. Figure
3.64 illustrates this for five galaxies lying along a straight line, with the Milky Way
Galaxy at the center. Using the data from the figure, calculate the velocities: (a)
relative to galaxy 2 and (b) relative to galaxy 5. The results mean that observers on all
galaxies will see themselves at the center of the expanding universe, and they would
likely be aware of relative velocities, concluding that it is not possible to locate the
center of expansion with the given information.
Solution

(a) v1

v2

v3

v4

v5
 4500 km/s   2200 km/s    2300 km/s
0
 0   2200 km/s   2200 km/s
 2830 km/s  2200 km/s  5030 km/s
 6700 km/s  2200 km/s  8900 km/s
Openstax College Physics
Instructor Solutions Manual
Chapter 3
(b) v1  4500 km/s  6700 km/s   11,200 km/s

v 2  2200 km/s  6700 km/s   8900 km/s

v3  0  6700 km/s    6700 km/s

v 4  2830 km/s  6700 km/s   3870 km/s

v5  0
64.
Solution
(a) Use the distance and velocity data in Figure 3.64 to find the rate of expansion as a
function of distance. (b) If you extrapolate back in time, how long ago would all of the
galaxies have been at approximately the same position? The two parts of this problem
give you some idea of how the Hubble constant for universal expansion and the time
back to the Big Bang are determined, respectively.
(a)
v
,
d
 4500 km/s
1 ly
H1 

 1.59  10 18 s 1
6
12
 300  10 ly 9.46  10 km
 2200 km/s
1 ly
H2 

 1.55  10 18 s 1
6
12
 150  10 ly 9.46  10 km
2830 km/s
1 ly
H4 

 1.57  10 18 s 1
6
12
190  10 ly 9.46  10 km
6700 km/s
1 ly
H5 

 1.57  10 18 s 1
6
450  10 ly 9.46  1012 km
km/s
H average  1.57  10 18 s 1  14.9
Mly
H
d
1
1
(b) t  v  H  1.57  10 18 s 1
1y
 6.37  1017 s 
 2.02  1010 y  20.2 billion years
7
3.16  10 s
Openstax College Physics
65.
Solution
Instructor Solutions Manual
Chapter 3
An athlete crosses a 25-m-wide river by swimming perpendicular to the water current
at a speed of 0.5 m/s relative to the water. He reaches the opposite side at a distance
40 m downstream from his starting point. How fast is the water in the river flowing
with respect to the ground? What is the speed of the swimmer with respect to a friend
at rest on the ground?
tan  
d w 40

ds
25
  58
tan  
vw
vs
vw
0.5 m/s
v w  0.80 m/s
tan 58 
2
2
vSG  vSW
 v WG
 0.5 2  0.80 2  0.94 m/s
66.
Solution
A ship sailing in the Gulf Stream is heading 25.0 west of north at a speed of 4.00 m/s
relative to the water. Its velocity relative to the Earth is 4.80 m/s 5.00 west of north.
What is the velocity of the Gulf Stream? (The velocity obtained is typical for the Gulf
Stream a few hundred kilometers off the east coast of the United States.)
vg
vgy N

vgx
4.8 m/s
4 m/s 20°
5°
v WE  v WS  v SE   v SW  v SE
v WEx  vSWx  vSEx  4.00 m/s  cos115  4.80 m/s  cos 95  1.272 m/s
v WEy  vSWy  vSEy  4.00 m/s sin 115  4.80 m/s sin 95  1.157 m/s
v WE  v 2 WE,x v 2 WE,y  (1.272 m/s) 2  (1.157 m/s) 2  1.72 m/s
 v WE,y
α  tan 1 
 v WE,x

  tan 1 1.157 m/s  42.3 N of E

1.272 m/s

Openstax College Physics
Instructor Solutions Manual
Chapter 3
67.
An ice hockey player is moving at 8.00 m/s when he hits the puck toward the goal. The
speed of the puck relative to the player is 29.0 m/s. The line between the center of the
goal and the player makes a 90.0 angle relative to his path as shown in Figure 3.65.
What angle must the puck’s velocity make relative to the player (in his frame of
reference) to hit the center of the goal?
Solution
v puck sin   v player
 v player 
  sin 1  8.00 m/s   16.01
θ  sin 1 
v

 29.0 m/s 
 puck 
  90  16.01  73.99  74.0
The puck will make an angle of 74.0 relative to the player’s motion.
68.
Solution
69.
Unreasonable Results Suppose you wish to shoot supplies straight up to astronauts in
an orbit 36,000 km above the surface of the Earth. (a) At what velocity must the supplies
be launched? (b) What is unreasonable about this velocity? (c) Is there a problem with
the relative velocity between the supplies and the astronauts when the supplies reach
their maximum height? (d) Is the premise unreasonable or is the available equation
inapplicable? Explain your answer.
(a) v0  2hg  23.6  10 7 m 9.80 m/s 2   26,563 m/s  27 km/s
(b) This velocity far exceeds present capabilities; the value is close to the actual
orbital speed of the earth about the sun.
(c) Yes, the supplies will be only momentarily at rest, so the timing of the arrival of
the ship must be just right (very small tolerance of error).
(d) This approach is unreasonable. You want to set up a scenario where there will be
a small relative velocity between the ship and the supplies for some reasonable
length of time. It cannot be done this way! Also, the assumption that
g  9.80 m/s 2 =constant is unreasonable for distances as large as 36,000 km
above the surface of the earth.
Unreasonable Results A commercial airplane has an air speed of 280 m/s due east
and flies with a strong tailwind. It travels 3000 km in a direction 5 south of east in
1.50 h. (a) What was the velocity of the plane relative to the ground? (b) Calculate the
magnitude and direction of the tailwind’s velocity. (c) What is unreasonable about
both of these velocities? (d) Which premise is unreasonable?
Openstax College Physics
Solution
Instructor Solutions Manual
Chapter 3
vp
(a)
5°

v
vtw
v
(b)
3000 km
 2000 km h  556 m/s
1.50 h
v x  v px  v twx ; v y  v py  v twy
v cos 5  v p  v twx  v twx  v cos 5  280 m/s  273.9 m s
 v sin 5  v twy   48.4 m s
 vtw 
273.9 m/s 2   48.4 m/s 2
 278 m/s
 48.4 m/s 
and θ  tan 1 
  10.0 S of E
 273.9 m/s 
(c) The results for (a) and (b) are unreasonably high. They are greater than the speed
of sound.
(d) The initial premise that the plane can travel 3000 km in 1.50 h is unreasonable.
Either the distance is too large, or the time is too short. (An air speed of 280 m/s
for a commercial plane is quite reasonable).
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