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Problems & Exercises


(a) 480 m480 m

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


north component 3.21 km, east component 3.83 km


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


(a) 26.6 m26.6 m, 65.65. north of east

(b) 26.6 m26.6 m, 65.65. south of west


52.9 m52.9 m, 90.90. with respect to the x-axis.


x-component 4.41 m/s

y-component 5.07 m/s


(a) 1.56 km

(b) 120 m east


North-component 87.0 km, east-component 87.0 km


30.8 m, 35.8 west of north


(a) 30.8 m30.8 m, 54.54. south of west

(b) 30.8 m30.8 m, 54.54. north of east


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


7.34 km7.34 km, 63.63. south of east


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


(a) 3.50 s

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

(d) 44.7 m/s, 50.2º50.2º below horizontal


(a) 18.18.

(b) The arrow will go over the branch.


R=v02 sin 0 g Forθ=45º, R= v02g R=v02 sin 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.


(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).


1.50 m, assuming launch angle of 45º45º


θ = 6.1º θ = 6.1º

yes, the ball lands at 5.3 m from the net


(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.


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


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


15.0 m/s


(a) 24.2 m/s

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


yy0=0=v0yt12gt2=( v0sinθ)t12gt2yy0=0=v0yt12gt2=( v0sinθ)t12gt2,

so that t=2(v0 sin θ)gt=2(v0 sin θ)g

xx0=v0xt=(v0 cos θ)t=R,xx0=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=sin,2sinθcosθ=sin, the range is:

R= v 0 2 singR= v 0 2 sing.


(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


(a) 0.70 m/s faster

(b) Second runner wins

(c) 4.17 m


17.0 m/s17.0 m/s, 22.22.


(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.


(a) 63.5 m/s

(b) 29.6 m/s


6.68 m/s6.68 m/s, 53.53. south of west


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

(b) 20.2 billion years


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


(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.


The figure shows speed (absolute value of v) graphed as a function of position y. The plot of A is linear, with a negative slope, starting at v sub zero A and ending at speed zero. The plot of B is linear, with the same negative slope as that of A, starting at v sub zero B, which is greater than v sub zero A, and ending at speed zero.
Figure 3.61

Test Prep for AP® Courses




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.


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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