Hw2 - _— PROBLEM 11.1fl5 I ‘ firm A Emmemmer uses a 51m“ I‘JEL‘HHQJ In dual-11H drixcua Rmm i113[hul pm 3:53[he:me E5 diachurgcd 111

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Unformatted text preview: [____.____—___.__ _ PROBLEM 11.1fl5 I ‘ firm A Emmemmer uses a 51m“ I‘JEL‘HHQJ' In dual-11H drixcua}. Rmm i113 [hul pm 3:53. * [he :me E5 diachurgcd 111 an .11. urugt Hug]: L11" 4W" with tlu: luriznnlul. ' Llctunninu {Etc :ni1ia] speed In, ut'The snow. TSOLUTION liming r: 4H" 1|1I: Eaurimnm] and ~. urlitul mulimm arr: . l a .I; — [1',CL‘I5rr]L and _'.' — _I.',I -.- [1}..wlr1ufl — 1-53! . I -.- _ | . I “r r = '. arm _'I' = 'I', ' I151“ IQ" _L{F rurngrr _ I I .. v a 1[I.,,—.1'tm1rx- l] l'1'nll1ufltch :- : -——_ :3" Using. - [Mn 11]. Th = 4.2 In. "ml _1'_..). = H m -_. EHJfi—ijtan-J-lW—ll] HUN)“. 9'” [lg] - — .:a r x . I“... ' .r_._, -U."-H51 S From the Imrimnlnl moliun. | .1: 31.3 {n.TH‘ilms-LLFJ r. - (L‘Jt'i r11; '1 I I' CUE if PROBLEM 11.105 .3 _ A. basketball player shoots when ethe i5. 5 m flotn the backboard. Knott'htg I I that the: ball has an initial 1u'e1oojt1v' T” at ttl'l tangle ol' 3U" with the ' lf__ horizontal. t1t¢1erntino the value of en when n" is equal to [:1] 22?; mm. ' - ' to} 431'} mm. SOLUTION . . _".' . I-lomoutol mutton: :L' = [rumour]! or t = .z ' 1}: cm: a ‘Ie'ettieol motion: _t' = y” + t'.;_.sitmt -- ,3 it!" .Il- ' l 2 _t' 2 _L-'.. + J-‘tttnt: — :yo‘ 4'. . Ely.) + .ttttncr — j H to} When n“ = 213 mm = 0.228 m. .t = 5 41.223 = 4.??? m 2 2 + 4.???[31‘130‘3 -- some} _ 9.31 _ r' : 0.3.430 :5 r = {L590 5 . 4.7?2 1.. = —.. ” [Ligflcogifln v” = 9.34 Itt-‘Lx' *1 {h} when .5! = 430 mm = H.431} m. .r = 5 — {mm = 4.5? m . 2E1 + 4.5Ttflnfifl“ — M148] a ._ - r' = — = U314: 5' 9.81 t = 0.569 5 _ 4.5? tifiogeofifl“ 1'.) | — — -—. | PROBLEM 11.11: |: 1 f1 3 The initial velocity v” iif‘a huekey pit-ck is ITIEI lam-"h. Determine lti'll the l“! 39* - . e _ _ I 1; - litrgexl value {less than 45°} 01 the :mgle or lor which the pmk wall enter ' the net, Hr} the etirreepflndittg time required For the puck to reach the net. | SOLUTION l . . i- | Ilei'imntal nmtmn: .r = [1}, Cesar}: m' i = I. _ t'._. ensa '- 1'ltlat'llt:ttl nmtmn: y = [t=”$t|3(;r}.t — fi—gi‘ gr: : x tan a — 5 | 2v; we" ex .t'lEil'Ifif — ,ll |- I-Ettl' a} Data: 1'“ = lTll‘I lam-“Ii — 47.2321113'5. .t' = 4.8 113 ill point t". l‘“ ,i- = 1.21m ut pm'nt t". E __ {21:41:22} 2 94 m m _ {9.3mm} ' ; — — 24.07-‘3 gr 4.8 gt-n'..- _ [fl-'IEIZHIJE} in} :ou r: — lat-L'i'lla * 25.0?3 = e I tuttrx : 112mm? and 94.45 a = 14.869” or 89.4“ re = 14.9“ *1 “I” F: I _. 4-3 _ F .- tiittj: x 4 enema {41222}e0514.369° PROBLEM 11.11?r J? 1"“ I AL an intE-1'5ecrim1 cur .-t is tux-ding 5;]th with a velocity 01‘ 4U lam-“n when it is struck by cm H Iran-fling 3U“ north uJ' e391 with a $610?in 01‘ SE} kms'h. Determinu ELL- l'L'IEJEi‘L’r.‘ velocity “fear B with IEHPEH to Car .-f. SOL UTION v Sketch wcmr :deitmn on a diugl'um 2:5 xhi‘mn. A Law nt‘ctminw: 1. -. I .I I 1'” - 5-}; + rt, -- 1-.-I.41-_,cns|1{1” I = 50‘ + 403 -[2][SEHIII_4EI']I;:35]2[}” 4"- I - . .2 = (alt'IME-ims'hj . . 1'31. _| : THJ huh I 1‘ . SUN: sinllfl“ .3ch smus: I : _. -— ~- .- Iw | - . 2| I . 405m 11”“ a- M“ (X = T = | .IL . rx — 25.3“. c: 'I' 30‘ = 56.3“ 1."..__! — '.-':-‘s..! kmsh iififi” '1 PROBLEM 11.125 n" . . . -. As the truck shtrwrt his-guts to back up mth a constant EtCE‘E1EI'i'IElOl'1 of i.“ a -. 4 fish". the nutter section B of its hottht Starts lU I'ulmct with i1 L'fll'lh'lilm 1 1 ': ___“_ acceleration at H: I‘E-"s' t'clzttit-fl m the Lruck. lJutL‘J‘m'lIt: [as] the O ttcrelet'atmh flt'auut'tmt H. [M the VCIUEiI}' uf hL‘L‘IttHI B when t = 2 s. _ — . _—_ SOLUTION Furthutt'tick. 3i : 4 fig: .+ Fur the hm‘lm. all“ ._ H] {pg-,3 :; 5”."- [;r} :3_r5 - :1_| -+- an . J' I Skutuh 1111-: ruclttl' attld'ttitm. B} lam nt'cnsittcs‘. a}; = (F. + HIT? i — 3-5! In” .CI'JS 5”” = 4-: +1.wa — 1E:4']|::| .ft_]m.~' 5U“ “It. = 3.314 ttssfi 5“ _ _ I”. Kilt 5f)" H .A' EU3 Law whines: sum: = ii '-' = —j m = [LIN-HM “I... 3.214 a = 22.42 3,... = 3.21 11:33 ‘7; 12w» 4 1h] 1-“ —- {t-I.,}” + NIH-i" = {J -I [REHHEJ =tt.431t-'E 22.4” «1 ...
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This note was uploaded on 01/09/2009 for the course ENGINEERIN 440:222 taught by Professor - during the Summer '06 term at Rutgers.

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Hw2 - _— PROBLEM 11.1fl5 I ‘ firm A Emmemmer uses a 51m“ I‘JEL‘HHQJ In dual-11H drixcua Rmm i113[hul pm 3:53[he:me E5 diachurgcd 111

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