College Physics for AP® Courses

# Chapter 3

### Problems & Exercises

1.

(a) $480 m480 m size 12{"480 m"} {}$

(b) $379 m379 m size 12{"379 m"} {}$, $18.4°18.4° size 12{"18" "." "4° east of north"} {}$ east of north

3.

north component 3.21 km, east component 3.83 km

5.

$19.5 m19.5 m size 12{"19" "." "5 m"} {}$, $4.65°4.65° size 12{4 "." "65°"} {}$ south of west

7.

(a) $26.6 m26.6 m size 12{"26" "." "6 m"} {}$, $65.1°65.1° size 12{"65" "." "1°"} {}$ north of east

(b) $26.6 m26.6 m size 12{"26" "." "6 m"} {}$, $65.1°65.1° size 12{"65" "." "1°"} {}$ south of west

9.

$52.9 m52.9 m size 12{"52" "." "9 m"} {}$, $90.1°90.1° size 12{"90" "." "1°"} {}$ with respect to the x-axis.

11.

x-component 4.41 m/s

y-component 5.07 m/s

13.

(a) 1.56 km

(b) 120 m east

15.

North-component 87.0 km, east-component 87.0 km

17.

30.8 m, 35.8 west of north

19.

(a) $30.8 m30.8 m size 12{"30" "." "8 m"} {}$, $54.2º54.2º size 12{"54" "." 2°} {}$ south of west

(b) $30.8 m30.8 m size 12{"30" "." "8 m"} {}$, $54.2º54.2º size 12{"54" "." 2°} {}$ north of east

21.

18.4 km south, then 26.2 km west(b) 31.5 km at $45.0º45.0º size 12{"45.0º"} {}$ south of west, then 5.56 km at $45.0º45.0º size 12{"45.0º"} {}$ west of north

23.

$7.34 km7.34 km size 12{2 "." "97 km"} {}$, $63.5º63.5º size 12{"22" "." 2°} {}$ south of east

25.

$x = 1.30 m×102 y = 30.9 m. x = 1.30 m×102 y = 30.9 m.$

27.

(a) 3.50 s

(b) 28.6 m/s (c) 34.3 m/s

(d) 44.7 m/s, $50.2°50.2°$ below horizontal

29.

(a) $18.4°18.4° size 12{"18" "." "4°"} {}$

(b) The arrow will go over the branch.

31.

$R = v02 sin2θ0g For θ = 45° , R=v02g R = v02 sin2θ0g For θ = 45° , R=v02g$

$R=91.8 mR=91.8 m size 12{R=91.8} {}$ for $v0=30 m/sv0=30 m/s size 12{"30 m/s"} {}$; $R=163 mR=163 m size 12{R=91.8} {}$ for $v0=40 m/sv0=40 m/s size 12{"40 m/s"} {}$; $R=255mR=255m size 12{R} {}$ for $v0=50 m/sv0=50 m/s size 12{"50 m/s"} {}$.

33.

(a) 560 m/s

(b) $8.00 × 103 m8.00 × 103 m size 12{8 "." "00 " times " 10" rSup { size 8{3} } " m"} {}$

(c) 80.0 m. This error is not significant because it is only 1% of the answer in part (b).

35.

1.50 m, assuming launch angle of $45°45°$

37.

$θ = 6.1° θ = 6.1° size 12{θ} {}$

yes, the ball lands at 5.3 m from the net

39.

(a) −0.486 m

(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.

41.

4.23 m. No, the owl is not lucky; he misses the nest.

43.

No, the maximum range (neglecting air resistance) is about 92 m.

45.

15.0 m/s

47.

(a) 24.2 m/s

(b) The ball travels a total of 57.4 m with the brief gust of wind.

49.

$y−y0=0=v0yt−12gt2=( v0sinθ)t−12gt2y−y0=0=v0yt−12gt2=( v0sinθ)t−12gt2 size 12{y - y rSub { size 8{0} } =0=v rSub { size 8{0y} } t - { {1} over {2} } ital "gt" rSup { size 8{2} } = $$v rSub { size 8{0} } "sin"θ$$ t - { {1} over {2} } ital "gt" rSup { size 8{2} } } {}$,

so that $t=2(v0 sin θ)gt=2(v0 sin θ)g size 12{t= { {2 $$v rSub { size 8{0} } "sin"θ$$ } over {g} } } {}$

$x−x0=v0xt=(v0 cos θ)t=R,x−x0=v0xt=(v0 cos θ)t=R, size 12{x - x rSub { size 8{0} } =v rSub { size 8{0x} } t= $$v rSub { size 8{0} } "cos"θ$$ t=R,} {}$ and substituting for $tt size 12{t} {}$ gives:

$R = v 0 cos θ 2 v 0 sin θ g = 2 v 0 2 sin θ cos θ g R = v 0 cos θ 2 v 0 sin θ g = 2 v 0 2 sin θ cos θ g size 12{R=v rSub { size 8{0} } "cos"θ left ( { {2v rSub { size 8{0} } "sin"θ} over {g} } right )= { {2v rSub { size 8{0} rSup { size 8{2} } } "sin"θ"cos"θ} over {g} } } {}$

since $2sinθcosθ=sin2θ,2sinθcosθ=sin2θ, size 12{2"sin"θ"cos"θ="sin"2θ,} {}$ the range is:

$R= v 0 2 sin2θgR= v 0 2 sin2θg size 12{ {underline {R= { {v rSub { size 8{0} rSup { size 8{2} } } "sin"2θ} over {g} } }} } {}$.

52.

(a) $35.8 km35.8 km size 12{"35" "." 8" km"} {}$, $45º45º size 12{"45º"} {}$ south of east

(b) $5.53 m/s5.53 m/s size 12{5 "." "53 m/s"} {}$, $45º45º size 12{"45º"} {}$ south of east

(c) $56.1 km56.1 km size 12{"56" "." 1" km"} {}$, $45º45º size 12{"45"º} {}$ south of east

54.

(a) 0.70 m/s faster

(b) Second runner wins

(c) 4.17 m

56.

$17.0 m/s17.0 m/s size 12{17 "." "0 m/s"} {}$, $22.1º22.1º size 12{"22" "." 1°} {}$

58.

(a) $230 m/s230 m/s size 12{2"20 m/s"} {}$, $8.0º8.0º size 12{9 "." "96"°} {}$ south of west

(b) The wind should make the plane travel slower and more to the south, which is what was calculated.

60.

(a) 63.5 m/s

(b) 29.6 m/s

62.

$6.68 m/s6.68 m/s size 12{6 "." "68"" m/s"} {}$, $53.3º53.3º size 12{"53" "." 3º} {}$ south of west

64.

(a) H average = 14 . 9 km/s Mly H average = 14 . 9 km/s Mly alignl { stack { size 12{H rSub { size 8{"average"} } ={}} {} # "14" "." "9 " { {"km/s"} over {"Mly"} } {} } } {}

(b) 20.2 billion years

66.

$1.72 m/s1.72 m/s size 12{1 "." "72"" m/s"} {}$, $42.3º42.3º size 12{"42" "." 3°} {}$ north of east

### Test Prep for AP® Courses

1.

(d)

3.

We would need to know the horizontal and vertical positions of each ball at several times. From that data, we could deduce the velocities over several time intervals and also the accelerations (both horizontal and vertical) for each ball over several time intervals.

5.

The graph of the ball's vertical velocity over time should begin at 4.90 m/s during the time interval 0 - 0.1 sec (there should be a data point at t = 0.05 sec, v = 4.90 m/s). It should then have a slope of -9.8 m/s2, crossing through v = 0 at t = 0.55 sec and ending at v = -0.98 m/s at t = 0.65 sec.

The graph of the ball's horizontal velocity would be a constant positive value, a flat horizontal line at some positive velocity from t = 0 until t = 0.7 sec.

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