hw2s - P3.5 5 = 3.00} m/s2 ; 0,. = 5003 m/s ; j. = oi + 03...

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Unformatted text preview: P3.5 5 = 3.00} m/s2 ; 0,. = 5003 m/s ; j. = oi + 03 A 1 A. (a) =i‘i+V,-t+%§t2 = [5.00ti+-2—3.00t2]]m \1 vf =v,. +§t = (5,003+30013) m/s (b) t = 2.00 s, :3 = 5.00(2.00)i + l(3.00)(2.00)23 = (10.03 + 6.003) m 2 6f = 5.00i + 3.00(2.00)i = (500$ + 6003) m/s vf =|vf|=,/vf,f +v§f =‘/(5.00)2 +(6.00)2 = P3.10 From Equation 3.16 with R = 15.0 m, vi = 3.00 m/s, Bmax = 45.0° P3.11 Take the origin at the mouth of the cannon. xf =vxit 2000 m=(1000 m/s)cos€,~t Therefore, t= 2'00 5 cos 0i 1 . 1 y}: =vyit+—2—ayt2: 800 m=(1000 m/s)sm0it+E(—9.80 m/s2)t2 800 m=(1000 m/s)sin0,~[2'00 SJ—aaso m/sz)[ 2'00 5] cos 0i cos 6,- 800 m(cos2 19,-) = 2 000 m(sin 6i cos (9,) — 19.6 m 19.6 m+ 800 m(cos2 19,-) = 2 000 m‘ll — cos2 0i (cos 0,) 334 + (31 360) cos.2 a,- + (640 000) cos4 0 = (4000 000) cos2 0,. — (4 000 000) cos4 0,. 4640 000 cos4 0,. —3968640cosz 01+ 384: 0 2 3968 640 i: "(3 968 640)2 — 4(4 640 000)(384) cos (9,. = 9280000 cos 0,- = 0.925 or 0.00984 9,- = 22.4° or 894° (Both solutions are valid.) P3.16 P327 P3.45 The horizontal component of displacement is x f = vxit = (vi cos 6,-)t. Therefore, the time required to (1 vi COS 0i . At this time, the altitude of the water is reach the building a distance d away is t = 2 _ 1 2_ . d _g d yf—int+Euyt —vism0,-[vicosei] 2[vicosai . Therefore the water strikes the building at a height h above ground level of 2 gd h: = dt 6i—————. yf an ngzcoszfli The satellite is in free fall. Its acceleration is due to gravity and is by effect a centripetal acceleration. “C = 8 Dz 3 _ = - - _ _ I 2 _ 3 so r g. Solv1ng for the velocrty, v — fig — (6,400+ 600)(10 m)(8.21 m/s ) — 7.58 X 10 m/s . v = 2—7” and T 2,” 27r(7,000 x 103 m) T=—=—= 5.80x103 0 7.58x103 m/s T = 5.80 x103 5(1 mm) = 96.7 min. 60 5 Refer to the sketch: (b) A x = vxl-t ; substitution yields 130 = (0,- cos 35.0°)t. A y = vyit + —12—at2; substitution yields 20.0 = (2;,- sin 35.0°)t + %(—9.80)t2. . . 2 _ 130tan35°—20 _ Solv1ng gives t — ————4.9 :> t — . (a) v,- = ITI/S (c) vyf = v,- sint9l- —gt, vx = vi c050,- At t=3.81 s, vyf = 41.7 sin35.0°—(9.80)(3.81)= —13.4 m/s vx = (41.7 cos 350°) = FIG. P3.45 P3.47 x f = vixt = vit cos 40.0° Thus, when xf =10.0 m, t=&. vi cos 40.0° At this time, yf should be 3.05 m— 2.00 m = 1.05 m. ' 2 ~ 400° 10.0 ThuS,1.05 m=w+1(_930 m/Sz) _m t 0,- cos 40.0° 2 01. cos 4000 From this, vi = . ...
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This note was uploaded on 04/02/2008 for the course PHYS 6A taught by Professor Mahaashour-abdalla during the Winter '07 term at UCLA.

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hw2s - P3.5 5 = 3.00} m/s2 ; 0,. = 5003 m/s ; j. = oi + 03...

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