# ch3 - :2 The truck has a displacement of 18(—16 = 2...

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Unformatted text preview: :2. The truck has a displacement of 18 + (—16) = 2 blocks north and 10 blocks east. The resultant has a magnitude of 422 + 102 = 10 blocks and a direction of tan" 2/10= 11° north of east . 5. The vectors for the problem are drawn approximately to scale. The / resultant has a length of and a direction I Enorth of east. If calculations are done, the actual resultant should be 57.4 m at 47. 5° north of east. , ‘ t} “7% 7. (a) See the accompanying_ diagram (b) V=-l4..°3cos348 =_ll.7units V=l4.3sin34.8°= (c) V=./V,,2 +V2= ./( 117) +(816 =m _,8.16 0: tan ——= 34. 8°above the —x axis 11.7 A, = 44.0COS 28.0° = 38.85 A = 44.0 sin 28.0° = 20.66 B. = —26.5 cos 56.0° = —l4.82 B, = 26.5 sin 56.0° = 21.97 C, = 31.0cos270° = 0.0 Cy = 31.0sin 270° -31.0 (a) (K+ﬁ+é)x =38.85+(-l4.82)+0.0= 24.03:- (K+f3+é)y =20.66+21.97+(—31.0)=11.63:- (b) [A + 1‘1 + 6| = «24.03): +(11.63)’ = a: [M41193 = 24.03 Peas—"1‘ 11. A, = 44.0 cos 28.0": 38.85 A, = 44.0 sin 28.0° = 20.66 Cx=31.0c05270°=0.0 Cy=31.0sin 270°=-31.0' - (27,"; ¢3\$=f§~0 =32'35‘ 14. A; = 44.0cos 28.0° = 38.85 A, = 44.05in 28.0° = 20.66 Bx = -26.5 cos 56.0° = —14.82 By = 26.5 sin 56.0” = 21.97 Cx = 31.0cos 270° = 0.0 C, = 31.0sin 270° = -3”) (a) (1‘; — 23) = -—14.82 — 2(38.85) = —92.52 (1”; — 2;), = 21.97 — 2(20.66) = _19.35 Z Note that since both components are negative, the vector is in the 3“i quadrant. Iﬁ —2Kl= J(-92-52)’ +1935)” = a = "‘9'” =— -92.52 (b) (28 — sﬁ + 26); = 2(38.85)—3(—l4.82) + 2(o.o) = 122.16 (23 — 31‘; + 26)’ = 2(20.66)-— 3(21.97) + 2(-31.o) = —86.59 ' Note that since the x component is positive and the y component is negative, the vector is in the 4"I quadrant. - —86.5 0‘ " , I23. — 3]? + 2Cl = 122.16)2 + (—86.59 2 = t9 = tam’l 9 =, 35.3 below + 1: ans Since the ball’s height is less than 3.00 m, he football does not clear the b It is 0.76 m too low when it reaches the horizontal location of the goalposts. 30. Choose the origin to be where the projectile is launched, and upwards to be the pcisitive y direction. The initial velocity of the projectile is v0 , the launching angle is 00 , a y = -g , and vyo = v0 sin 00 . (a) The maximum height is found from Eq. 2-]1c, v: = v30 + 2a, (y — yo) , with vy = 0 at the maximum height. 2 2 2 - 2 2 - 2 2 - 2 o v —v —v 8111 0 v sm 0 (65.2 m/s) sm 34.5 = 0+ y yo = 0 o = -—2——1 = ——-—————_—"- = m . y“ 2ay —2g 2g 2(9.80m/s’) - (b) The total time in the air is found from Eq. 2-11b, with a total vertical displacement of 0 for the ball to reach the ground. y = y0 +v’ot+-;-a’tz —) 0=vo sinﬁot—g-gtz —) I: 2V0 Sin 00 = 2(65'2 m/S)Sin 34-5 = and t = 0 g (9.80m/s2 ) The time of 0 represents the launching of the ball. (c) The total horizontal distance covered is found from the horizontal motion at constant velocity. Ax = vxt = (v0 cos 60)t = (65.2 m/s)(cos 34.5°)(7.54 s) = (d) The velocity of the projectile 1.50 s after ﬁring is found as the vector sum of the horizontal and vertical velocities at that time. The horizontal velocity is a constant vo cos 00 = (65.2 m/s) (cos 34.5° ) = 53 .7 m/s . The vertical velocity is found from Eq. 2-11a. . vy = vyo + at = v(, sin 00 - gt = (65.2 m/s)sin 345° - (9.80 m/s‘)(1.so s) = 22.2 m/s Thus the speed of the projectile is v = . Iv: + v; = J53]: + 22.22 = . 4 v _ The direction above the horizontal is given by 0 = tan'l —’— = tan‘l -:—32—:-= . vx . M9 N at . 34. Choose the origin to be the location from which the balloon is ﬁred, and choose upward as the positive y direction. Assume the boy in the tree is a distance H up from the point at which the balloon is ﬁred, and that the tree is a distance D horizontally from the point at which the balloon is ﬁred. The equations of motion for the balloon and boy are as follows, using constant acceleration relationships. xBIlhon =vo cc’sgnt ymm =0+VOSin00t—-;-gt2 yaoy =H_.;. 2 Use the horizontal motion at constant velocity to ﬁnd the elapsed time after the balloon has traveled D to the right. D D = v0 cos 00:0 —> ID = vo cos0o Where is the balloon vertically at that time? 2 2 . . D D D =v smﬂt —-'-gt2 =v srn€ -——-—-Lg —— =Dtan0—J- ° "D 2 D ° °v0cost9o 2 vocostS’o ° 2g vocosﬂo y Balloon Where is the boy vertically at that time? Note that H = D tan (9,. 2 2 D D =H-J-gt2 =H—J- =Dtant9 —1 —— ya” 2 D 2g[vocos00) ° 2g(vo c0360) The boy and the balloon are at the same height and the same horizontal location at the same time. Thus they collide! 71. (a) Choose the origin to be the location where the car leaves the ramp, and choose upward to be the positive y direction. At the end of its ﬂight over the 8 cars, the car must be at y = -1.5 m . Also for the car, vyo = 0 , a’ = —g , v,‘ = vo , and Ax = 20 m. The time of ﬂight is found from the horizontal motion at constant velocity: Ax = vxt —) t = Ax/vo. That expression for the time is used in Eq. 2-1 lb for the vertical motion. 2. Ax y-_-_-yo+vyot+-;-ayt2 —+ y=0+0+-§-(—g)(v—] ——) 0 v0: -283) = -(9.80m/s )(20m) :- 2(—1.5 m) (b) Again choose the origin to be the location where the car leaves the ramp, and choose upward to be the positive y direction. The y displacement of the car at the end of its ﬂight. over the 8 cars must again be y = —l.5 m. Forthe car, v,o =vosin00, a, =-g , V; =v°c059° , and~ Ax = 20 m. The launch angle is 00 = 10°. The time of ﬂight is found from the horizontal motion at constant veloci . _,_ k W _><.—)lo+\/w‘t +23fo ’ ‘ :3" '2 a / . a. 5 ¢qu {9695/0 ‘?W’%@ ...
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