# Giancoli Physics (5th ed) Chapter 3

Main Page > Giancoli Physics (5th ed) Solutions > Giancoli Physics (5th ed) Chapter 3

## Contents

- 1 About this page
- 2 Problems
- 2.1 1. Question
- 2.2 2. Question
- 2.3 3. Question
- 2.4 4. Question
- 2.5 5. Question
- 2.6 6. Question
- 2.7 7. Question
- 2.8 8. Question
- 2.9 9. Question
- 2.10 10. Question
- 2.11 11. Question
- 2.12 12. Question
- 2.13 13. Question
- 2.14 14. Question
- 2.15 15. Question
- 2.16 16. Question
- 2.17 17. Question
- 2.18 18. Question
- 2.19 19. A tiger leaps horizontally form a 7.5-m-high rock with a speed of 4.5 m/s. How far from the base of the rock will she land?
- 2.20 20. Question
- 2.21 21. Question
- 2.22 22. Romeo is chucking pebbles gently up to Juliet's window, and he wants the pebbles to hit the window with only a horizontal component of velocity. He is standing at the edge of a rose garden 8.0 m below her window and 9.0 m from the base of the wall (Fig. 3-38). How fast are the pebbles going when they hit her window?
- 2.23 23. Question
- 2.24 24. A ball is thrown horizontally from the roof of a building 56 m tall lands 45 m from the base . What was the ball's initial speed?
- 2.25 25. Question
- 2.26 26. Question
- 2.27 27. A ball thrown horizontally at 22.2 m/s from the roof of a building lands 36.0 m from the base of the building. How high is the building?
- 2.28 28. Question
- 2.29 29. Question
- 2.30 30. Question
- 2.31 31. The pilot of an airplane traveling160 km/h wants to drop supplies to flood victims isolated on a patch of land 160 m below. The supplies should be dropped how many how many seconds before the plane is directly overhead?
- 2.32 32. A hunter aims directly at a target (on the same level) 120 m away. (a) If the bullet leaves the gun at a speed of 250 m/s, by how much will it miss the target? (b) At what angle should the gun be aimed so the target will be hit?
- 2.33 33. Question
- 2.34 34. Question
- 2.35 35. Question
- 2.36 36. Question
- 2.37 37. Question
- 2.38 38. Question
- 2.39 39. Question
- 2.40 40. Question
- 2.41 41. Question
- 2.42 42. Question
- 2.43 43. Question
- 2.44 44. Question
- 2.45 45. Question
- 2.46 46. Question
- 2.47 47. Question
- 2.48 48. Question
- 2.49 49. Question
- 2.50 50. Question
- 2.51 51. Question
- 2.52 52. Question
- 2.53 53. Question
- 2.54 54. Question
- 2.55 55. Question
- 2.56 56. Question
- 2.57 57. Question
- 2.58 58. Question
- 2.59 59. Question
- 2.60 60. Question
- 2.61 61. Question
- 2.62 62. Question
- 2.63 63. Question
- 2.64 64. Question
- 2.65 65. Question
- 2.66 66. Question
- 2.67 67. Question
- 2.68 68. Question
- 2.69 69. Question
- 2.70 70. Question
- 2.71 71. Question
- 2.72 72. Question
- 2.73 73. Question

## About this page

All images uploaded for this page must start with the string "Gp5_3_" so the image p19.jpg associated with chapter 3 should be uploaded as Gp5_3_p19.jpg. This way we can avoid conflicts in the image directory, and we can find images easily.

Table of Contents

## Problems

### 1. Question

177 Km, 15^{o}S of W

### 2. Question

20 blocks and 37

^{o}S of E

### 3. Question

Solution goes here

### 4. Question

24.95 and 41.10

^{o}below the x-axis

### 5. Question

31 M, 44

^{o}N of E

### 6. Question

B) -14 and 19.9

C) 24.3 and 54.8^{o}above -x-axis

### 7. Question

A) 2.76 in the +x-direction

B)13.86 in the +x-direction

C) 13.86 in the -x-direction

### 8. Question

A) V

_{1x}=-8.08, V_{1y}=0

V_{2x}=V_{2}cos 45^{o}=4.51 cos 45^{o}=3.19

V_{2y}=V_{2}sin 45^{o}=3.19

B) 5.84 and 33.1^{o}above -x-axis

### 9. Question

A) 614 Km/Hr and 489 Km/Hr

B) 1.84E3 Km and 1.47E3 Km

### 10. Question

(5.9, -1.4, -1.4) and 6.2

### 11. Question

### 12. Question

97.2 and 53.1

^{o}above the +x-axis

### 13. Question

A) 80.7 and 1.56^{o}above -x-axis

B) 80.7 and 1.56^{o}below +x-axis

### 14. Question

A) 94.4 and 31.3

^{o}below +x-axis

B) 117 and 72.1^{o}above +x-axis

C) 142 and 11.7^{o}below -x-axis

### 15. Question

A) 117 and 72.1^{o}below -x-axis

B) 226 and 35.5^{o}below +x-axis

### 16. Question

A) 1.90 m/s

^{2}down

B) t=18.8 s

### 17. Question

-2454 m

+3867 m

+2085 m

5032 m

### 18. Question

A) A

_{x}=-/+ 71.2

B) B_{x}=-151.2

B_{y}=+55.0

161

20^{o}above -x-axis

### 19. A tiger leaps horizontally form a 7.5-m-high rock with a speed of 4.5 m/s. How far from the base of the rock will she land?

HorizontalVerticals =

u = 4.5 m/s

v = "

a = 0

t =s = -7.5 m

u = 0 (Purely horizontal)

v = ? (don't care!)

a = -9.8 m/s/s

t = ?Use

s = ut

^{1}/_{2}at^{2}-7.5 =^{1}/_{2}(-9.8)t^{2 }

t =1.2372 sUse:

v

_{av = }s/t, as we now know t4.5 m/s = s/(1.2372s)

s = 5.567 m =5.6m

### 20. Question

44 m and 4.8 m

### 21. Question

14

^{o}and 76^{o}

### 22. Romeo is chucking pebbles gently up to Juliet's window, and he wants the pebbles to hit the window with only a horizontal component of velocity. He is standing at the edge of a rose garden 8.0 m below her window and 9.0 m from the base of the wall (Fig. 3-38). How fast are the pebbles going when they hit her window?

HorizontalVerticals =9.0 m

u = ???

v = "

a = 0

t =s = 8.0 m

u = ?

v = 0 (final is purely horizontal)

a = -9.8 m/s/s

t = ?From the vertical we've got: t = 1.278s Use

v

^{2}= u^{2 }+2as^{ }0 = u^{2 }+2(-9.8)(8.0)

u = 12.522m/sThe horizontal velocity is: x = ut = vt

v = x/t = 9/1.278 = 7m/s

v = u at

t = u/g = 12.522/9.8 = 1.278s

As the vertical final velocity is zero, the velocity of the pebble when it hits the windows is:

V = = =

7m/s</span>

### 23. Question

Unsuccessful and 34.7m

### 24. A ball is thrown horizontally from the roof of a building 56 m tall lands 45 m from the base . What was the ball's initial speed?

HorizontalVerticals = 45 m

u = ???

v = "

a = 0

t = ?s = -56 m

u = 0 (purely horizontal)

v = ?

a = -9.8 m/s/s

t = ?The ball's initial speed is horizontal. So we have:

s = vt = ut

==> u = s/t = 45/3.38 = '''13.31m/s'''

Use

s = ut +

^{1}/_{2}at^{2}

t = = = 3.38s

### 25. Question

Solution goes here

### 26. Question

2.46 s

### 27. A ball thrown horizontally at 22.2 m/s from the roof of a building lands 36.0 m from the base of the building. How high is the building?

HorizontalVerticals =36.0 m

u = 22.2 m/s

v = "

a = 0

t = ?s = ????

u = 0 (purely horizontal)

v = ?

a = g = 9.8 m/s/s

t = ?x = vt = ut

t = x/u = 36/22.2 = 1.6216s

The height of the building is:

s = ut 1/2at

^{2}y

_{0}= 0s = 1/2gt

^{2}= 1/2 * 9.8 * 1.6216^{2}= 12.9m

### 28. Question

22m

### 29. Question

6 times as far

### 30. Question

Solution goes here

### 31. The pilot of an airplane traveling160 km/h wants to drop supplies to flood victims isolated on a patch of land 160 m below. The supplies should be dropped how many how many seconds before the plane is directly overhead?

HorizontalVerticals = 36.0 m

u = 160/3.6 = 44.4444 m/s (not used)

v = "

a = 0

t = ?s = -160 m

u = 0 (purely horizontal)

v = ?

a = -9.8 m/s/s

t = ??????UseÂ

s = ut +

^{1}/_{2}at^{2}

-160 = 0 +^{1}/_{2}(-9.8)t^{2}So t = 5.714 s =

5.7 s

### 32. A hunter aims directly at a target (on the same level) 120 m away. (a) If the bullet leaves the gun at a speed of 250 m/s, by how much will it miss the target? (b) At what angle should the gun be aimed so the target will be hit?

a.

HorizontalVerticals = 120 m

u = 250 m/s

v = "

a = 0

t = ?s = ????

u = 0 (purely horizontal)

v = ?

a = -9.8 m/s/s

t = ?Use:

v_{av = }Ds/Dt250 m/s = 120 m/t,

t = .48 sUse

s = ut +

^{1}/_{2}at^{2}s = 0 +

^{1}/_{2}(-9.8)(.48)^{2Â }= 1.129 =1.1 mb. use the range equation

Range =

so

= .539^{o}

of course, 90^{o}- .539^{o}would hit too.

### 33. Question

Solution goes here

### 34. Question

Solution goes here

### 35. Question

a. h=92.6 m

b. 8.69 s

c.539 m

d. 68.0 m/s

24.2° above the horizontal.

### 36. Question

a. t = 14.6 s

b. 1.22 km

c. 83.9 m/s -79.9 m/s

d. 116 m/s

e. 43.6° below the horizontal

### 37. Question

Solution goes here

### 38. Question

a. 180 m

b. -8.41 m/s (down)

97.5 m/s

### 39. Question

Solution goes here

### 40. Question

10.5 m/s in the direction of the ships motion

6.5 m/s in the direction of the ship's motion

### 41. Question

2.9 m/s

20° from the river bank

### 42. Question

V_{SC}= 29 m/s

V_{SG}= 14 m/s (down)

### 43. Question

a. 2.59 m/s

63.4° from the shore

b. 7.77 m at 62.4° to the shore

### 44. Question

0.00600 h = 21.6 s

### 45. Question

a. 435 km/h

9.36° east of south

b. 12 km

### 46. Question

**8.13° west of south**

### 47. Question

Solution goes here

### 48. Question

**V _{BS} = 1.41 m/s**

**11° above the water**

### 49. Question

**a. 1.66 m/s**

**b. 3.19 m/s**

### 50. Question

**0.90 m/s**

### 51. Question

**a. 120 m**

**b. 150 s (2.5 min)**

### 52. Question

**53°**

### 53. Question

**42.9° N of E**

### 54. Question

**a = 0.366 m/s ^{2}**

### 55. Question

**58 km/h**

**31°**

**58 km/h opposite to V _{12}**

### 56. Question

**15.8 s**

### 57. Question

**V _{sg} = 109 km/h**

### 58. Question

**6.2°**

### 59. Question

**a. parallel**

**b. perpendicular**

**c. v _{2} = 0**

### 60. Question

**D _{x} = 50 m, D_{y} = -25m, D_{z} = -10,**

**57 m**

**27° from the x-axis toward the -y-axis**

**10° below the horizontal**

### 61. Question

**-2.4 m/s ^{2} (opposite to the trucks motion)**

**-1.4 m/s ^{2} (down)**

### 62. Question

+/- 31.6°

+/- 46.3

### 63. Question

V_{r} = V_{T}/tanθ

### 64. Question

111 km/h

θ = 37.0°N of E

### 65. Question

t_{1} = 1.8x10^{2} s

4.8 km

t_{2} = 21 s

0.56 km

### 66. Question

0.88 s

h_{max} = 0.95 m

### 67. Question

1.6 m/s^{2}

### 68. Question

33 m/s

### 69. Question

t = 2.7 s

v_{0} = 1.9 m/s

### 70. Question

v_{0} = 26.3 m/s

t_{2} = 0.714 s

3.8 m beyond the net

good

### 71. Question

θ = 61.8° below the horizontal

### 72. Question

Dv/(v^{2}-u^{2})

Dv/(v^{2}-u^{2})^{1/2}

### 73. Question

63 m/s, 66° above the horizontal