College Physics 2e

# Chapter 3

### Problems & Exercises

1.

(a) $480 m480 m$

(b) $379 m379 m$, $18.4º18.4º$ east of north

3.

north component 3.21 km, east component 3.83 km

5.

$19.5 m19.5 m$, $4.65º4.65º$ south of west

7.

(a) $26.6 m26.6 m$, $65.1º65.1º$ north of east

(b) $26.6 m26.6 m$, $65.1º65.1º$ south of west

9.

$52.9 m52.9 m$, $90.1º90.1º$ 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) $30.8 m30.8 m$, $54.2º54.2º$ south of west

(b) $30.8 m30.8 m$, $54.2º54.2º$ north of east

21.

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

23.

$7.34 km7.34 km$, $63.5º63.5º$ 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, $50.2º50.2º$ below horizontal

29.

(a) $18.4º18.4º$

(b) The arrow will go over the branch.

31.

$R=v02 sin 2θ 0 g Forθ=45º, R= v02g R=v02 sin 2θ 0 g Forθ=45º, R= v02g$

$R=91.8 mR=91.8 m$ for $v0=30 m/sv0=30 m/s$; $R=163 mR=163 m$ for $v0=40 m/sv0=40 m/s$; $R=255mR=255m$ for $v0=50 m/sv0=50 m/s$.

33.

(a) 560 m/s

(b) $8.00 × 103 m8.00 × 103 m$

(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 $45º45º$

37.

$θ = 6.1º θ = 6.1º$

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.

$y−y0=0=v0yt−12gt2=( v0sinθ)t−12gt2y−y0=0=v0yt−12gt2=( v0sinθ)t−12gt2$,

so that $t=2(v0 sin θ)gt=2(v0 sin θ)g$

$x−x0=v0xt=(v0 cos θ)t=R,x−x0=v0xt=(v0 cos θ)t=R,$ and substituting for $tt$ gives:

$R = v 0 cos θ 2 v 0 sin θ g = 2 v 0 2 sin θ cos θ g R = v 0 cos θ 2 v 0 sin θ g = 2 v 0 2 sin θ cos θ g$

since $2sinθcosθ=sin2θ,2sinθcosθ=sin2θ,$ the range is:

$R= v 0 2 sin2θgR= v 0 2 sin2θg$.

52.

(a) $35.8 km35.8 km$, $45º45º$ south of east

(b) $5.53 m/s5.53 m/s$, $45º45º$ south of east

(c) $56.1 km56.1 km$, $45º45º$ south of east

54.

(a) 0.70 m/s faster

(b) Second runner wins

(c) 4.17 m

56.

$17.0 m/s17.0 m/s$, $22.1º22.1º$

58.

(a) $230 m/s230 m/s$, $8.0º8.0º$ 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.

$6.68 m/s6.68 m/s$, $53.3º53.3º$ south of west

64.

(a) $H average = 14 . 9 km/s Mly H average = 14 . 9 km/s Mly$

(b) 20.2 billion years

66.

$1.72 m/s1.72 m/s$, $42.3º42.3º$ north of east

71.

(a) Since $sin 2 50> sin 2 40 sin 2 50> sin 2 40$, then B reaches the greatest height.

(b) i. Yes, it is consistent because $sin50>sin40 sin50>sin40$. ii. No, it does not make sense because y is proportional to $sin 2 θ sin 2 θ$.

(c)

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