Giancoli Physics (5th ed) Chapter 3

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Problems

1. Question


177 Km, 15o S of W

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2. Question

20 blocks and 37o S of E

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3. Question

Solution goes here

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4. Question

24.95 and 41.10o below the x-axis

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5. Question

31 M, 44o N of E

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6. Question

B) -14 and 19.9
C) 24.3 and 54.8o above -x-axis

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7. Question

A) 2.76 in the +x-direction
B)13.86 in the +x-direction
C) 13.86 in the -x-direction

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8. Question

A) V1x=-8.08, V1y=0
V2x=V2cos 45o=4.51 cos 45o=3.19
V2y=V2 sin 45o=3.19
B) 5.84 and 33.1o above -x-axis

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9. Question

A) 614 Km/Hr and 489 Km/Hr
B) 1.84E3 Km and 1.47E3 Km

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10. Question

(5.9, -1.4, -1.4) and 6.2

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11. Question

Gp5 3 3-35.JPG
A) 35.9 and 17.3
B)39.9 and 25.8o above +x-axis

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12. Question

97.2 and 53.1o above the +x-axis

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13. Question

Gp5 3 3-35.JPG
A) 80.7 and 1.56o above -x-axis
B) 80.7 and 1.56o below +x-axis

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14. Question

A) 94.4 and 31.3o below +x-axis
B) 117 and 72.1o above +x-axis
C) 142 and 11.7o below -x-axis

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15. Question

Gp5 3 3-35.JPG
A) 117 and 72.1o below -x-axis
B) 226 and 35.5o below +x-axis

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16. Question

A) 1.90 m/s2 down
B) t=18.8 s

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17. Question

-2454 m
+3867 m
+2085 m
5032 m

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18. Question

A) Ax=-/+ 71.2
B) Bx=-151.2
By=+55.0
161
20o above -x-axis

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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?

Horizontal Vertical
s =
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 1/2at2
-7.5 = 1/2(-9.8)t2
t =1.2372 s

Use:

vav = \Deltas/\Deltat, as we now know t

4.5 m/s = s/(1.2372s)
s = 5.567 m = 5.6m


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20. Question

44 m and 4.8 m

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21. Question

14o and 76o

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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?

Horizontal Vertical
s =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

v2 = u2 +2as
0 = u2 +2(-9.8)(8.0)
u = 12.522m/s

The 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 = \sqrt{V_x^2 + V_y^2} = \sqrt{7^2 + 0^2} = 7m/s </span>


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23. Question

Unsuccessful and 34.7m

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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?

Horizontal Vertical
s = 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 + 1/2at2
t = \sqrt{\frac{2s}{g}} = \sqrt{\frac{2 * 56}{9.8}} = 3.38s


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25. Question

Solution goes here

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26. Question

2.46 s

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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?

Horizontal Vertical
s =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 1/2at2

y0 = 0

s = 1/2gt2 = 1/2 * 9.8 * 1.62162 = 12.9m


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28. Question

22m

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29. Question

6 times as far

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30. Question

Solution goes here

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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?

Horizontal Vertical
s = 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 + 1/2at2
-160 = 0 + 1/2(-9.8)t2

So t = 5.714 s = 5.7 s


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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.

Horizontal Vertical
s = 120 m
u = 250 m/s
v = "
a = 0
t = ?
s = ????
u = 0 (purely horizontal)
v = ?
a = -9.8 m/s/s
t = ?

Use:
vav = Ds/Dt

250 m/s = 120 m/t,
t = .48 s

Use

s = ut + 1/2at2

s = 0 + 1/2(-9.8)(.48) = 1.129 = 1.1 m

b. use the range equation
Range = \frac{v^2}{g}sin(2\Theta)
so
\Theta = \frac{sin^{-1}(\frac{g*Range}{v^2})}{2} = \frac{sin^{-1}(\frac{9.8*120}{250^2})}{2} = .539o
of course, 90o - .539o would hit too.


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33. Question

Solution goes here

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34. Question

Solution goes here

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35. Question

a. h=92.6 m

b. 8.69 s


c.539 m
d. 68.0 m/s
24.2° above the horizontal.

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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

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37. Question

Solution goes here

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38. Question

a. 180 m
b. -8.41 m/s (down)
97.5 m/s

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39. Question

Solution goes here

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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

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41. Question

2.9 m/s
20° from the river bank

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42. Question

VSC = 29 m/s
VSG = 14 m/s (down)

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43. Question

a. 2.59 m/s
63.4° from the shore
b. 7.77 m at 62.4° to the shore

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44. Question

0.00600 h = 21.6 s

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45. Question

a. 435 km/h
9.36° east of south
b. 12 km

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46. Question

8.13° west of south

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47. Question

Solution goes here

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48. Question

VBS = 1.41 m/s

11° above the water

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49. Question

a. 1.66 m/s

b. 3.19 m/s


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50. Question

0.90 m/s


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51. Question

a. 120 m

b. 150 s (2.5 min)


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52. Question

53°


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53. Question

42.9° N of E


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54. Question

a = 0.366 m/s2

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55. Question

58 km/h

31°

58 km/h opposite to V12


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56. Question

15.8 s

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57. Question

Vsg = 109 km/h


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58. Question

6.2°

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59. Question

a. parallel

b. perpendicular

c. v2 = 0

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60. Question

Dx = 50 m, Dy = -25m, Dz = -10,

57 m

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

10° below the horizontal

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61. Question

-2.4 m/s2 (opposite to the trucks motion)

-1.4 m/s2 (down)


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62. Question

+/- 31.6°

+/- 46.3

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63. Question

Vr = VT/tanθ

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64. Question

111 km/h

θ = 37.0°N of E

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65. Question

t1 = 1.8x102 s

4.8 km

t2 = 21 s

0.56 km

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66. Question

0.88 s

hmax = 0.95 m

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67. Question

1.6 m/s2

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68. Question

33 m/s


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69. Question

t = 2.7 s

v0 = 1.9 m/s

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70. Question

v0 = 26.3 m/s

t2 = 0.714 s

3.8 m beyond the net

good

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71. Question

θ = 61.8° below the horizontal

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72. Question

Dv/(v2-u2)


Dv/(v2-u2)1/2

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73. Question

63 m/s, 66° above the horizontal

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