# Physics 2A HW #3 Solutions

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1 Chper 3 Focus on Conceps: 3, 4, 6, 9 Problems: 9, 9, 3, 41, 66, 7, 75, 77 Phsics A HW #3 Soluions Focus On Conceps 3-3 (c) The ccelerion due o grvi is he sme for boh blls, despie he fc h he hve differen velociies. Focus On Conceps 3-4 (d) The ccelerion of projecile is he sme ll poins on he rjecor. I poins downwrd, owrd he erh, nd hs mgniude of 9.8 m/s. Focus On Conceps 3-6 (c) The ime for projecile o rech he ground depends onl on he componen (or vericl componen) of is vribles, i.e.,, v, nd. These vribles re he sme for boh blls. The fc h Bll 1 is moving horizonll he op of is rjecor does no pl role in he ime i kes for i o rech he ground. Focus On Conceps 3-9 () The ime projecile is in he ir is equl o wice he ime i kes o fll from is mimum heigh. Projecile 1 reches he greer heigh, so i spends he greer moun of ime in he ir Problem 3-9 REASONING. We designe he direcion down nd prllel o he rmp s he + direcion, nd he ble shows he vribles h re known. Since hree of he five kinemic vribles hve vlues, one of he equions of kinemics cn be emploed o find he ccelerion. -Direcion D v v +1. m? +7.7 m/s m/s b. The ccelerion vecor poins down nd prllel o he rmp, nd he ngle of he rmp is 5. relive o he ground (see he drwing). Therefore, rigonomer cn be used o deermine he componen prllel of he ccelerion h is prllel o he ground. 5. prllel

2 SOLUTION. Equion 3.6 ( v v ) = + cn be used o find he ccelerion in erms of he hree known vribles. Solving his equion for gives ( + ) ( ) ( + ) v v 7.7 m/s m/s = = =.47 m/s 1. m b. The drwing shows h he ccelerion vecor is oriened 5. relive o he ground. The componen prllel of he ccelerion h is prllel o he ground is prllel ( ) = cos 5. =.47 m/s cos 5. =.4 m/s Problem 3-9 REASONING The vericl displcemen of he bll depends on he ime h i is in he ir before being cugh. These vribles depend on he -direcion d, s indiced in he ble, where he + direcion is "up." -Direcion D v v? 9.8 m/s m/s? Since onl wo vribles in he direcion re known, we cnno deermine his poin. Therefore, we emine he d in he direcion, where + is ken o be he direcion from he picher o he ccher. -Direcion D v v +17. m m/s +41. m/s? Since his ble conins hree known vribles, he ime cn be evlued b using n equion of kinemics. Once he ime is known, i cn hen be used wih he -direcion d, long wih he pproprie equion of kinemics, o find he vericl displcemen. SOLUTION Using he -direcion d, Equion 3.5 cn be emploed o find he ime h he bsebll is in he ir: Solving for gives ( ) = v + = v since = m/s m = = =.415 s v +41. m/s

3 The displcemen in he direcion cn now be evlued b using he -direcion d ble nd he vlue of =.415 s. Using Equion 3.5b, we hve ( )( ) ( )( ) = v + = + = 1 1 m/s.415 s 9.8 m/s.415 s.844 m The disnce h he bll drops is given b he mgniude of his resul, so Disnce =.844 m. Problem 3-3 REASONING Since we know he lunch ngle θ = 15., he lunch speed v cn be obined using rigonomer, which gives he componen of he lunch veloci s v = v sin θ. Solving his equion for v requires vlue for v, which we cn obin from he vericl heigh of = 13.5 m b using Equion 3.6b from he equions of kinemics. SOLUTION From Equion 3.6b we hve Using rigonomer, we find Problem 3-41 REASONING The speed v of he soccer bll jus before he golie cches i is given b v v v = +, where v nd v re he nd componens of he finl veloci of he bll. The d for his problem re (he + direcion is from he kicker o he golie, nd he + direcion is he up direcion): -Direcion D v v m m/s? +(16. m/s) cos 8. = m/s -Direcion D v v 9.8 m/s? +(16. m/s) sin 8. = m/s

4 Since here is no ccelerion in he direcion ( = m/s ), v remins he sme s v, so v = v = m/s. The ime h he soccer bll is in he ir cn be found from he -direcion d, since hree of he vribles re known. Wih his vlue for he ime nd he -direcion d, he componen of he finl veloci cn be deermined. SOLUTION Since = m/s, he ime cn be clculed from Equion 3.5 s m = = = 1.19 s. The vlue for v v m/s cn now be found b using Equion 3.3b wih his vlue of he ime nd he -direcion d: ( )( ) v = v + = m/s m/s 1.19 s = 4.15 m/s The speed of he bll jus s i reches he golie is ( ) ( ) v = v + v = m/s m/s = 14.7 m/s Problem 3-66 REASONING The mgniude nd direcion of he iniil veloci v cn be obined using he Phgoren heorem nd rigonomer, once he nd componens of he iniil veloci v nd v re known. These componens cn be clculed using Equions 3.3 nd 3.3b. SOLUTION Using Equions 3.3 nd 3.3b, we obin he following resuls for he veloci componens: Using he Phgoren heorem nd rigonomer, we find

5 Problem 3-7 REASONING When he bll is hrown srigh up wih n iniil speed v, he mimum heigh h i reches cn be found b using wih he relion v = v + (Equion 3.6b). Since he bll is hrown srigh up, v = v, where v is he iniil speed of he bll. Also, he speed of he bll is momenril zero is mimum heigh, so v = m/s h poin. The ccelerion is h due o grvi, so he onl unknown besides is he iniil speed v of he bll. To deermine v we will emplo Equion 3.6b second ime, bu now i will be pplied o he cse where he bll is hrown upwrd n ngle of 5º bove he horizonl. In his cse he mimum heigh reched b he bll is 1 = 7.5 m, he iniil speed in he direcion is v = v sin 5º, nd he -componen of he speed he mimum heigh is v = m/s. SOLUTION We will sr wih he relion v = v + (Equion 3.6b) o find he mimum heigh h he bll ins when i is hrown srigh up. Solving his equion for, nd subsiuing in v = v nd v = m/s gives v = (1) To deermine v, we now ppl he equion v = v + o he siuion where he bll is hrown upwrd n ngle of 5º relive o he horizonl. In his cse we noe h v = v sin 5º, v = m/s, nd = 1 (he mimum heigh of 7.5 m reched b he bll). Solving for v, we find ( ) 1 sin 5 1 or v = v + v = sin 5 Subsiuing his epression for v ino Equion (1) gives 1 v sin m = = = = = sin 5 sin 5 1 m Problem 3-75 SSM REASONING The horizonl disnce covered b sone 1 is equl o he disnce covered b sone fer i psses poin P in he following digrm. Thus, he disnce beween he poins where he sones srike he ground is equl o, he horizonl disnce covered b sone when i reches P. In he digrm, we ssume up nd o he righ re posiive.

6 θ P θ θ 1 SOLUTION If P is he ime required for sone o rech P, hen = v = ( v cos θ ) P P For he vericl moion of sone, v = v sin θ +. Solving for gives v = v sin θ When sone reches P, Then, v = v sinθ, so he ime required o rech P is P v sinθ = v sinθ vp ( vcos θ ) = = Problem 3-77 REASONING Using he d given in he problem, we cn find he mimum fligh ime 1 of he bll using Equion 3.5b ( = v + ). Once he fligh ime is known, we cn use he definiion of verge veloci o find he minimum speed required o cover he disnce in h ime. SOLUTION Equion 3.5b is qudric in nd cn be solved for using he qudric formul. According o Equion 3.5b, he mimum fligh ime is (wih upwrd ken s he posiive direcion)

7 1 ( ) ( ) 1 ( ) ± ± + v v 4 v v = = ( ) ( ) 15. m/s sin 5. ± 15. m/s sin 5. +( 9.8 m/s ) (.1 m) = 9.8 m/s =. s nd.145 s where he firs roo corresponds o he ime required for he bll o rech vericl displcemen of = +.1 m s i rvels upwrd, nd he second roo corresponds o he ime required for he bll o hve vericl displcemen of = +.1 m s he bll rvels upwrd nd hen downwrd. The desired fligh ime is.145 s. During he.145 s, he horizonl disnce rveled b he bll is [ ] = v= ( v cos θ ) = (15. m/s) cos 5. (.145 s) =.68 m Thus, he opponen mus move.68 m 1. m = 1.68 m in.145 s.3 s = s. The opponen mus, herefore, move wih minimum verge speed of v min 1.68 m = = s 5.79 m / s

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