Vectors and Motion in Two Dimensions

Transcription

1 KNIG549_0_ch03_pp067-0.qd 5//09 4:33 PM Page 67 3 Vectos and Motion in Two Dimensions Once the leopad jmps, its tajecto is fied b the initial speed and angle of the jmp. How can we wok ot whee the leopad will land? LOOKING AHEAD The goals of Chapte 3 ae to lean moe abot vectos and to se vectos as a tool to analze motion in two dimensions. Tools fo Descibing Motion In the last chapte, we discssed motion along a line. In this chapte, we ll look at motion in which the diection changes. We'll need new tools. Vectos We se vectos to descibe qantities, like velocit, fo which both the magnitde and diection ae impotant. Vectos specif a diection as well as a magnitde. Looking Back.5 Vectos Vecto Math O basic kinematic vaiables ae vectos. Woking with them means leaning how to wok with vectos. Net displacement d Stat net 0 d End 4 d 4 d d 3 Yo ll lean how to add, sbtact, and pefom othe mathematical opeations on vectos. 3 Vecto Components We make measements in a coodinate sstem. How do vectos fit in? A A A 5 A sin A 5 A cos Yo ll lean how to find the components of vectos and how to add and sbtact vectos sing components. Tpes of Motion Thee ae a few basic tpes of motion that we ll conside. In each case thee is an acceleation de to a change in speed o diection o both. Motion on a Ramp Gavit cases the motion, bt it s not staight down it s at an angle. Pojectile Motion Objects lanched thogh the ai follow a paabolic path. Wate going ove the falls, the leaping salmon, footballs, and jmping leopads all follow simila paths. Cicla Motion Motion in a cicle at a constant speed involves acceleation, bt not the constant acceleation o stdied in Chapte. Looking Back.. Basic motion concepts A speed skie is acceleating down a amp. How fast is he moving at the end of his n? Looking Back.5 Motion with constant acceleation The salmon is moving hoizontall and veticall as do all objects ndegoing pojectile motion. Looking Back.7 Fee fall The ides ae moving at a constant speed bt with an eve-changing diection. It is the changing diection that cases the acceleation and makes the ide fn. 4th Poof

2 KNIG549_0_ch03_pp067-0.qd 5//09 4:33 PM Page CHAPTER 3 Vectos and Motion in Two Dimensions FIGURE 3. The velocit vecto a magnitde and a diection. Magnitde of vecto Stat P v 5 5 m/s v Diection of vecto v B Name of vecto The vecto epesents the paticle s velocit at this one point. FIGURE 3. Displacement vectos. (a) Sam s actal path Sam s hose (b) Beck s hose Sam s hose FIGURE 3.3 The net displacement eslting fom two displacements and B. Net displacement S End C 4 mi d S A B Q 3 mi Individal displacements has both Sam s displacement Displacement is the staight-line connection fom the initial to the final position. d B d S C B A B N N d B and d S have the same magnitde and diection, so d B 5 d S. 3. Using Vectos In the pevios chapte, we solved man poblems in which an object moved in a staight-line path. In this chapte, we will look at paticles that take cving paths motion in two dimensions. Becase the diection of motion will be so impotant, we need to develop an appopiate mathematical langage to descibe it the langage of vectos. We intodced the concept of a vecto in Chapte, and in the net few sections we will develop that concept into a sefl and powefl tool. We will pactice sing vectos b analzing a poblem of motion in one dimension (that of motion on a amp) and b stding the inteesting notion of elative velocit. We will then be ead to analze the two-dimensional motion of pojectiles and of objects moving in a cicle. Recall fom Chapte that a vecto is a qantit with both a size (magnitde) and a diection. FIGURE 3. shows how to epesent a paticle s velocit as a vecto v B. The paticle s speed at this point is 5 m/s and it is moving in the diection indicated b the aow. The magnitde of a vecto is epesented b the lette withot an aow. In this case, the paticle s speed the magnitde of the velocit vecto v B is v = 5 m/s. The magnitde of a vecto, a scala qantit, cannot be a negative nmbe. NOTE Althogh the vecto aow is dawn acoss the page, fom its tail to its tip, this aow does not indicate that the vecto stetches acoss this distance. Instead, the aow tells s the vale of the vecto qantit onl at the one point whee the tail of the vecto is placed. We fond in Chapte that the displacement of an object is a vecto dawn fom its initial position to its position at some late time. Becase displacement is an eas concept to think abot, we can se it to intodce some of the popeties of vectos. Howeve, all the popeties we will discss in this chapte (addition, sbtaction, mltiplication, components) appl to all tpes of vectos, not jst to displacement. Sppose that Sam, o old fiend fom Chapte, stats fom his font doo, walks acoss the steet, and ends p 00 ft to the notheast of whee he stated. Sam s displacement, which we will label d B S, is shown in FIGURE 3.a. The displacement vecto is a staight-line connection fom his initial to his final position, not necessail his actal path. The dashed line indicates a possible ote Sam might have taken, bt his displacement is the vecto d B S. To descibe a vecto we mst specif both its magnitde and its diection. We can wite Sam s displacement as d B S = (00 ft, notheast) whee the fist nmbe specifies the magnitde and the second item gives the diection. The magnitde of Sam s displacement is d S = 00 ft, the distance between his initial and final points. Sam s net-doo neighbo Beck also walks 00 ft to the notheast, stating fom he own font doo. Beck s displacement d B B = (00 ft, notheast) has the same magnitde and diection as Sam s displacement d B S. Becase vectos ae defined b thei magnitde and diection, two vectos ae eqal if the have the same magnitde and diection. This is te egadless of the individal stating points of the vectos. Ths the two displacements in FIGURE 3.b ae eqal to each othe, and we can wite d B B = d B S. Vecto Addition As we saw in Chapte, we can combine sccessive displacements b vecto addition. Let s eview and etend this concept. FIGURE 3.3 shows the displacement of a hike who stats at point P and ends at point S. She fist hikes 4 miles to the east, then 3 miles to the noth. The fist leg of the hike is descibed b the displacement vecto A B = (4 mi, east). The second leg of the hike has displacement B = (3 mi, noth). B definition, a vecto fom he initial position P to he final position S is also a displacement. This is vecto C B on the fige. C B is the net displacement becase it descibes the net eslt of the hike s having fist displacement A B, then displacement B. 4th Poof

3 KNIG549_0_ch03_pp067-0.qd 5//09 4:33 PM Page Using Vectos 69 is an initial displace- The wod net implies addition. The net displacement ment A B pls a second displacement B, o C B = A B + B (3.) The sm of two vectos is called the esltant vecto. Vecto addition is commtative: A B + B = B + A B. Yo can add vectos in an ode o wish. Look back at Tactics Bo.4 on page 9 to eview the thee-step pocede fo adding two vectos. This tip-to-tail method fo adding vectos, which is sed to find C B = A B + B in Fige 3.3, is called gaphical addition. An two vectos of the same tpe two velocit vectos o two foce vectos can be added in eactl the same wa. When two vectos ae to be added, it is often convenient to daw them with thei tails togethe, as shown in FIGURE 3.4a. To evalate D B + E B, o cold move vecto E B ove to whee its tail is on the tip of D B, then se the tip-to-tail le of gaphical addition. This gives vecto F B = D B + E B in FIGURE 3.4b. Altenativel, FIGURE 3.4c shows that the vecto sm D B + E B can be fond as the diagonal of the paallelogam defined b D B and E B. This method is called the paallelogam le of vecto addition. C B FIGURE 3.4 Two vectos can be added sing the tip-to-tail le o the paallelogam le. (a) E D What is D E? (b) F 5 D E D Tip-to-tail le: Slide the tail of E to the tip of D. E (c) E F 5 D E D Paallelogam le: Find the diagonal of the paallelogam fomed b D and E. Vecto addition is easil etended to moe than two vectos. FIGURE 3.5 shows the path of a hike moving fom initial position 0 to position, then position, then position 3, and finall aiving at position 4. These fo segments ae descibed b displacement vectos d B d B d B and d B,, 3, The hike s net displacement, an aow fom position 0 to position 4, is the vecto d B 4. net. In this case, d B net = d B + d B + d B 3 + d B 4 (3.) The vecto sm is fond b sing the tip-to-tail method thee times in sccession. Mltiplication b a Scala The hike in Fige 3.3 stated with displacement A B = (4 mi, east). Sppose a second hike walks twice as fa to the east. The second hike s displacement will then cetainl be A B = (8 mi, east). The wods twice as indicate a mltiplication, so we can sa A B = A B Mltipling a vecto b a positive scala gives anothe vecto of diffeent magnitde bt pointing in the same diection. Let the vecto A B be specified as a magnitde A and a diection A ; that is, A B = (A, Now let B = ca B A )., whee c is a positive scala constant. Then B = ca B means that (B, B ) = (ca, A ) (3.3) The vecto is stetched o compessed b the facto c (i.e., vecto has magnitde B = ca), bt B points in the same diection as A B. This is illstated in FIGURE 3.6. Sppose we mltipl A B b zeo. Using Eqation 3.3, we get B 0 # A = 0 B = (0 m, diection ndefined) (3.4) The podct is a vecto having zeo length o magnitde. This vecto is known as the zeo vecto, denoted 0 B. The diection of the zeo vecto is ielevant; o cannot descibe the diection of an aow of zeo length! B B FIGURE 3.5 The net displacement afte fo individal displacements. Net displacement Stat 0 d net d End 4 FIGURE 3.6 Mltiplication of a vecto b a positive scala. d 4 d The length of B is stetched b the facto c; that is, B 5 ca. A A 5 B A B points in the same diection as A. B 5 ca d 3 3 4th Poof

4 KNIG549_0_ch03_pp067-0.qd 5//09 4:34 PM Page CHAPTER 3 Vectos and Motion in Two Dimensions FIGURE 3.7 Vecto -A B. Tail of A at tip of A A Vecto A is eqal in magnitde bt opposite in diection to A. Ths A (A) 5 0. A Tip of A etns to the stating point. The esltant vecto is 0. FIGURE 3.8 Vectos A B, A B, and -3A B. A What happens if we mltipl a vecto b a negative nmbe? Eqation 3.3 does not appl if c < 0 becase vecto B cannot have a negative magnitde. Conside the vecto -A B, which is eqivalent to mltipling A B b -. Becase B A + (-A B ) = 0 B (3.5) The vecto -A B mst be sch that, when it is added to A B, the esltant is the zeo vecto 0 B. In othe wods, the tip of -A B mst etn to the tail of A B, as shown in FIGURE 3.7. This will be te onl if -A B is eqal in magnitde to A B bt opposite in diection. Ths we can conclde that -A B = (A, diection opposite A B ) (3.6) Mltipling a vecto b - eveses its diection withot changing its length. As an eample, FIGURE 3.8 shows vectos A B, A B, and -3A B. Mltiplication b dobles the length of the vecto bt does not change its diection. Mltiplication b -3 stetches the length b a facto of 3 and eveses the diection. 3A A Vecto Sbtaction How might we sbtact vecto fom vecto to fom the vecto A B - B? With nmbes, sbtaction is the same as the addition of a negative nmbe. That is, 5-3 is the same as 5 + (-3). Similal, A B - B = A B + (-B B ). We can se the les fo vecto addition and the fact that -B B is a vecto opposite in diection to B to fom les fo vecto sbtaction. B B A B TACTICS BOX 3. Sbtacting vectos To sbtact B fom A: A B Daw A. A B Place the tail of B at the tip of A. A 3 Daw an aow fom the tail of A to the tip of B. This is vecto A B. AB A B Eecises 5 8 STOP TO THINK 3. Which fige shows P B - Q B? P Q A. B. C. D. E. 4th Poof

5 KNIG549_0_ch03_pp067-0.qd 5//09 4:34 PM Page 7 3. Using Vectos on Motion Diagams In Chapte, we defined velocit fo one-dimensional motion as an object s displacement the change in position divided b the time inteval in which the change occs: v = t = f - i t In two dimensions, an object s displacement is a vecto. Sppose an object ndegoes displacement d B ding the time inteval t. Let s define an object s velocit vecto to be 3. Using Vectos on Motion Diagams 7 v B = db t = a d t, same diection as db b (3.7) Definition of velocit in two o moe dimensions Notice that we ve mltiplied a vecto b a scala: The velocit vecto is simpl the displacement vecto mltiplied b the scala / t. Conseqentl, as we fond in Chapte, the velocit vecto points in the diection of the displacement. As a eslt, we can se the dot-to-dot vectos on a motion diagam to visalize the velocit. NOTE Stictl speaking, the velocit defined in Eqation 3.7 is the aveage velocit fo the time inteval t. This is adeqate fo sing motion diagams to visalize motion. As we did in Chapte, when we make t ve small, we get an instantaneos velocit we can se in pefoming some calclations. EXAMPLE 3. Finding the velocit of an aiplane A small plane is 00 km de east of Denve. Afte ho of fling at a constant speed in the same diection, it is 00 km de noth of Denve. What is the plane s velocit? PREPARE The initial and final positions of the plane ae shown in FIGURE 3.9; the displacement d B is the vecto that points fom the initial to the final position. FIGURE 3.9 Displacement vecto fo an aiplane. End 00 km d SOLVE The length of the displacement vecto is the hpotense of a ight tiangle: d = (00 km) + (00 km) = 4 km The diection of the displacement vecto is descibed b the angle in Fige 3.9. Fom tigonomet, this angle is = tan - 00 km a 00 km b = tan- (.00) = 63.4 Ths the plane s displacement vecto is d B = (4 km, 63.4 noth of west) Becase the plane ndegoes this displacement ding ho, its velocit is v B = a d t, same diection as 4 km db b = a, 63.4 noth of westb h = (4 km/h, 63.4 noth of west) Denve 00 km Stat ASSESS The plane s speed is the magnitde of the velocit, v = 4 km/h. This is appoimatel 40 mph, which is a easonable speed fo a small plane. 4th Poof

6 KNIG549_0_ch03_pp067-0.qd 5//09 4:34 PM Page 7 7 CHAPTER 3 Vectos and Motion in Two Dimensions We defined an object s acceleation in one dimension as a = v / t. In two dimensions, we need to se a vecto to descibe acceleation. The vecto definition of acceleation is a staightfowad etension of the one-dimensional vesion: a B = vb f - v B i = vb t f - t i t (3.8) Definition of acceleation in two o moe dimensions Lnging vess veeing The top photo shows a baacda, a tpe of fish that catches pe with a apid linea acceleation, a qick change in speed. The baacda s bod shape is optimized fo sch a staight-line stike. The bttefl fish in the bottom photo has a ve diffeent appeaance. It can t apidl change its speed, bt its bod shape lets it qickl change its diection. Once the baacda gets p to speed, it can t change its diection ve easil, so the bttefl fish can, b emploing this othe tpe of acceleation, avoid capte. Thee is an acceleation wheneve thee is a change in velocit. Becase velocit is a vecto, it can change in eithe o both of two possible was:. The magnitde can change, indicating a change in speed.. The diection of motion can change. In Chapte we saw how to compte an acceleation vecto fo the fist case, in which an object speeds p o slows down while moving in a staight line. In this chapte we will eamine the second case, in which an object changes its diection of motion. Sppose an object has an initial velocit v B i at time t i and late, at time t f, has velocit v B f. The fact that the velocit changes tells s the object ndegoes an acceleation ding the time inteval t = t f - t i. We see fom Eqation 3.8 that the acceleation points in the same diection as the vecto v B. This vecto is the change in the velocit v B = v B f - v B i, so to know which wa the acceleation vecto points, we have to pefom the vecto sbtaction v B f - v B i. Tactics Bo 3. showed how to pefom vecto sbtaction. Tactics Bo 3. shows how to se vecto sbtaction to find the acceleation vecto. TACTICS BOX 3. Finding the acceleation vecto To find the acceleation between velocit v and velocit v : i f v i v f Daw the velocit vecto v f. v f v i Daw v i at the tip of v f. v f 3 Daw Dv 5 v f v i 5 v f (v i ) This is the diection of a. Dv v f v i 4 Retn to the oiginal motion diagam. Daw a vecto at the middle point in the diection of Dv; label it a. This is the aveage acceleation at the midpoint between v i and v f. v i a v f Eecises, 4th Poof

7 KNIG549_0_ch03_pp067-0.qd 5//09 4:34 PM Page Using Vectos on Motion Diagams 73 Now that we know how to detemine acceleation vectos, we can make a complete motion diagam with dots showing the position of the object, aveage velocit vectos fond b connecting the dots with aows, and acceleation vectos fond sing Tactics Bo 3.. Note that thee is one acceleation vecto linking each two velocit vectos, and a B is dawn at the dot between the two velocit vectos it links. Let s look at two eamples, one with changing speed and one with changing diection. EXAMPLE 3. Dawing the acceleation fo a Mas descent A spacecaft slows as it safel descends to the sface of Mas. Daw a complete motion diagam fo the last few seconds of the descent. PREPARE FIGURE 3.0 shows two vesions of a motion diagam: a pofessionall dawn vesion like o geneall find in this tet and a simple vesion simila to what o might daw fo a homewok assignment. As the spacecaft slows in its descent, the dots get close togethe and the velocit vectos get shote. FIGURE 3.0 Motion diagam fo a descending spacecaft. (a) Atist vesion We daw the dots epesenting the sccessive positions connected b velocit vectos then se sccessive velocit vectos accoding to Tactics Bo 3. to find the acceleation. 4 v a (b) Stdent sketch SOLVE The inset in Fige 3.0 shows how Tactics Bo 3. is sed to detemine the acceleation at one point. All the othe acceleation vectos will be simila, becase fo each pai of velocit vectos the ealie one is longe than the late one. 3 Δv v v ASSESS As the spacecaft slows, the acceleation vectos and velocit vectos point in opposite diections, consistent with what we leaned abot the sign of the acceleation in Chapte. The acceleation vecto is the same Stops diection as Δv. EXAMPLE 3.3 Dawing the acceleation fo a Feis wheel ide Anne ides a Feis wheel at an amsement pak. Daw a complete motion diagam fo Anne s ide. PREPARE FIGURE 3. shows 0 points of the motion ding one complete evoltion of the Feis wheel. A peson iding a Feis wheel moves in a cicle at a constant speed, FIGURE 3. Motion diagam fo Anne on a Feis wheel. Velocit vectos Acceleation vectos v a v Sccessive velocit vectos 4 point in diffeent diections, so thee is an acceleation. a a Dv v v v 3 Contined 4th Poof

8 KNIG549_0_ch03_pp067-0.qd 5//09 4:34 PM Page CHAPTER 3 Vectos and Motion in Two Dimensions so we ve shown eqal distances between sccessive dots. As befoe, the velocit vectos ae fond b connecting each dot to the net. Note that the velocit vectos ae staight lines, not cves. We see that all the velocit vectos have the same length, bt each has a diffeent diection, and that means Anne is acceleating. This is not a speeding p o slowing down acceleation, bt is, instead, a change of diection acceleation. SOLVE The inset to Fige 3. shows how to se the steps of Tactics Bo 3. to find the acceleation at one paticla position, at the bottom of the cicle. Vecto v B is the velocit vecto that leads into this dot, while v B moves awa fom it. Fom the cicla geomet of the main fige, the two angles maked a ae eqal. Ths we see that v B and -v B fom an isosceles tiangle and vecto v B = v B - v B is eactl vetical, towad the cente of the cicle. If we did a simila calclation fo each point of the motion, we d find a simila eslt: In each case, the acceleation points towad the cente of the cicle. ASSESS The speed is constant bt the diection is changing, so thee is an acceleation, as we epect. No matte which dot o select on the motion diagam of Fige 3., the velocities change in sch a wa that the acceleation vecto a B points diectl towad the cente of the cicle. An acceleation vecto that alwas points towad the cente of a cicle is called a centipetal acceleation. We will have mch moe to sa abot centipetal acceleation late in this chapte. 3.3 Coodinate Sstems and Vecto Components In the past two sections, we have seen how to add and sbtact vectos gaphicall, sing these opeations to dedce impotant details of motion. Bt the gaphical combination of vectos is not an especiall good wa to find qantitative eslts. In this section we will intodce a coodinate desciption of vectos that will be the basis fo doing vecto calclations. Achaeologists establish a coodinate sstem so that the can pecisel detemine the positions of objects the ecavate. Coodinate Sstems As we saw in Chapte, the wold does not come with a coodinate sstem attached to it. A coodinate sstem is an atificiall imposed gid that o place on a poblem in ode to make qantitative measements. The ight choice of coodinate sstem will make a poblem easie to solve. We will geneall se Catesian coodinates, the familia ectangla gid with pependicla aes, as illstated in FIGURE 3.. FIGURE 3. A Catesian coodinate sstem. 90 Coodinate aes have a positive end and a negative end, sepaated b zeo at the oigin whee the two aes coss. When o daw a coodinate sstem, it is impotant to label the aes. This is done b placing and labels at the positive ends of the aes, as in Fige 3.. The ppose of the labels is twofold: To identif which ais is which. To identif the positive ends of the aes. This will be impotant when o need to detemine whethe the qantities in a poblem shold be assigned positive o negative vales. 4th Poof

9 KNIG549_0_ch03_pp067-0.qd 5//09 4:34 PM Page Coodinate Sstems and Vecto Components 75 Component Vectos FIGURE 3.3 shows a vecto and an -coodinate sstem that we ve chosen. Once the diections of the aes ae known, we can define two new vectos paallel to the aes that we call the component vectos of A B. Vecto A B called the -component vecto, is the pojection of A B along the -ais. Vecto A B,, the -component vecto, is the pojection of A B along the -ais. Notice that the component vectos ae pependicla to each othe. Yo can see, sing the paallelogam le, that A B is the vecto sm of the two component vectos: B A = A B + A B (3.9) In essence, we have boken vecto A B into two pependicla vectos that ae paallel to the coodinate aes. We sa that we have decomposed o esolved vecto A B into its component vectos. NOTE It is not necessa fo the tail of to be at the oigin. All we need to know is the oientation of the coodinate sstem so that we can daw A B and A B paallel to the aes. Components Yo leaned in Chapte to give the one-dimensional kinematic vaiable v a positive sign if the velocit vecto v B points towad the positive end of the -ais and a negative sign if v B points in the negative -diection. The basis of this le is that v is the -component of v B. We need to etend this idea to vectos in geneal. Sppose we have a vecto A B that has been decomposed into component vectos A B and paallel to the coodinate aes. We can descibe each component vecto with a A B single nmbe (a scala) called the component. The -component and -component of vecto A B, denoted and A, ae detemined as follows: TACTICS BOX 3.3 A A B Detemining the components of a vecto The absolte vale ƒ A ƒ of the -component A is the magnitde of the component vecto A B. The sign of A is positive if A B points in the positive -diection, negative if A B points in the negative -diection. 3 The -component A is detemined similal. Eecises 6 8 In othe wods, the component A tells s two things: how big A B is and which end of the ais A B points towad. FIGURE 3.4 shows thee eamples of detemining the components of a vecto. NOTE A B and A B ae component vectos; the have a magnitde and a diection. A and A ae simpl components. The components A and A ae scalas jst nmbes (with nits) that can be positive o negative. Mch of phsics is epessed in the langage of vectos. We will feqentl need to decompose a vecto into its components o to eassemble a vecto fom its components, moving back and foth between the gaphical and the component epesentations of a vecto. Let s stat with the poblem of decomposing a vecto into its - and -components. FIGURE 3.5a on the net page shows a vecto A B at an angle above hoizontal. It is essential to se a picte o diagam sch as this to define the angle o ae sing to descibe a vecto s diection. A B points to the ight and p, so Tactics Bo 3.3 tells s that the components and ae both positive. A A A B FIGURE 3.3 Component vectos A B and A B ae dawn paallel to the coodinate aes sch that A B = A B + A B. A The -component vecto is paallel to the -ais. (m) A points in the positive -diection, 3 so A 5 m. B Magnitde 5 m B (m) 3 (m) 3 A 5 A A FIGURE 3.4 Detemining the components of a vecto. A B A A The -component vecto is paallel to the -ais. Magnitde 5 m (m) 3 4 A points in the positive -diection, so A 5 3 m. B points in the positive -diection, so B 5 m. Magnitde 5 m (m) 3 4 B points in the negative -diection, so B 5 m. The -component of C is C 5 4 m. Magnitde Magnitde 5 4 m 5 3 m (m) 3 4 C C The -component of C is C 5 3 m. C A Magnitde 5 3 m A 4th Poof

10 KNIG549_0_ch03_pp067-0.qd 5//09 4:34 PM Page CHAPTER 3 Vectos and Motion in Two Dimensions FIGURE 3.5 Beaking a vecto into components. (a) (b) Magnitde: A The vecto is specified b its magnitde and diection. Angle: The components ae sides of a ight tiangle with hpotense A and angle. The -component is the opposite side of the tiangle, so we se sin. A A 5 A sin A 5 A cos The -component is the adjacent side of the tiangle, so we se cos. FIGURE 3.6 Specifing a vecto fom its components. If we know the components, we can detemine the magnitde and the angle. A 5 "A A A A A 5 tan (A /A ) FIGURE 3.7 Relationships fo a vecto with a negative component. Diection of C f5 tan C /0C 0 C 5 C cos f f Magnitde C 5 "C C C 5 C sin f C We can find the components sing tigonomet, as illstated in FIGURE 3.5b. Fo this case, we find that A = A cos A = A sin (3.0) whee A is the magnitde, o length, of A B. These eqations convet the length and angle desciption of vecto A B into the vecto s components, bt the ae coect onl if is meased fom hoizontal. Altenativel, if we ae given the components of a vecto, we can detemine the length and angle of the vecto fom the - and -components, as shown in FIGURE 3.6. Becase A in Fige 3.6 is the hpotense of a ight tiangle, its length is given b the Pthagoean theoem: A = A + A (3.) Similal, the tangent of angle is the atio of the opposite side to the adjacent side, so = tan - a A A b (3.) Eqations 3. and 3. can be thoght of as the invese of Eqations 3.0. How do things change if the vecto isn t pointing to the ight and p that is, if one of the components is negative? FIGURE 3.7 shows vecto C B pointing to the ight and down. In this case, the component vecto C B is pointing down, in the negative -diection, so the -component C is a negative nmbe. The angle f is dawn meased fom the -ais, so the components of ae C B C = C sin f C =-C cos f (3.3) The oles of sine and cosine ae evesed fom those in Eqations 3.0 becase the angle f is meased with espect to vetical, not hoizontal. NOTE Whethe the - and -components se the sine o cosine depends on how o define the vecto s angle. As noted above, o mst daw a diagam to define the angle that o se, and o mst be se to efe to the diagam when compting components. Don t se Eqations 3.0 o 3.3 as geneal les the aen t! The appea as the do becase of how we defined the angles. Net, let s look at the invese poblem fo this case: detemining the length and diection of the vecto given the components. The signs of the components don t matte fo detemining the length; the Pthagoean theoem alwas woks to find the length o magnitde of a vecto becase the sqaes eliminate an concens ove the signs. The length of the vecto in Fige 3.7 is simpl C = C + C (3.4) When we detemine the diection of the vecto fom its components, we mst conside the signs of the components. Finding the angle of vecto C B in Fige 3.7 eqies the length of withot the mins sign, so vecto C B has diection C f = tan - a C C b (3.5) Notice that the oles of and diffe fom those in Eqation 3.. 4th Poof

11 KNIG549_0_ch03_pp067-0.qd 5//09 4:34 PM Page Coodinate Sstems and Vecto Components 77 EXAMPLE 3.4 Find the - and -components of the acceleation vecto in FIGURE 3.8. Finding the components of an acceleation vecto FIGURE 3.8 Acceleation vecto a B of Eample a 6.0 m/s a B shown FIGURE 3.9 The components of the acceleation vecto. a is negative. a is negative. PREPARE It s impotant to daw the vectos. Making a sketch is ccial to setting p this poblem. FIGURE 3.9 shows the oiginal vecto a B decomposed into component vectos paallel to the aes. SOLVE The acceleation vecto a B = (6.0 m/s, 30 below the negative -ais) points to the left (negative -diection) and down (negative -diection), so the components a and a ae both negative: a =-a cos 30 =-(6.0 m/s ) cos 30 =-5. m/s a =-a sin 30 =-(6.0 m/s ) sin 30 =-3.0 m/s ASSESS The magnitde of the -component is less than that of the -component, as seems to be the case in Fige 3.9, a good check on o wok. The nits of a and a ae the same as the nits of vecto a B. Notice that we had to inset the mins signs manall b obseving that the vecto points down and to the left. STOP TO THINK 3. What ae the - and -components C and C of vecto C B? C (cm) 4 3 (cm) Woking with Components We ve seen how to add vectos gaphicall, bt thee s an easie wa: sing components. To illstate, let s look at the vecto sm C B = A B + B fo the vectos shown in FIGURE 3.0. Yo can see that the component vectos of C B ae the sms of the component vectos of A B and B. The same is te of the components: C = A + B and C = A + B. In geneal, if D B = A B + B + C B + Á, then the - and -components of the esltant vecto D B ae D = A + B + C + Á (3.6) D = A + B + C + Á This method of vecto addition is called algebaic addition. FIGURE 3.0 Using components to add vectos. B A A A C 5 A B B C 5 A B B C 5 A B EXAMPLE 3.5 Using algebaic addition to find a bid s displacement A bid flies 00 m de east fom a tee, then 00 m nothwest (that is, 45 noth of west). What is the bid s net displacement? PREPARE FIGURE 3.a on the net page shows the displacement vectos A B = (00 m, east) and B = (00 m, nothwest) and also the net displacement C B. We daw vectos tip-to-tail if we ae going to add them gaphicall, bt it s sall easie to daw them all fom the oigin if we ae going to se algebaic addition. FIGURE 3.b edaws the vectos with thei tails togethe. Contined 4th Poof

12 KNIG549_0_ch03_pp067-0.qd 5//09 4:34 PM Page CHAPTER 3 Vectos and Motion in Two Dimensions FIGURE 3. Finding the net displacement. (a) The bid s net displacement is C 5 A B. (b) Angle descibes the diection of vecto C. B 00 m Net displacement C 5 A B End C N 00 m 45 Stat 00 m A N C B 00 m SOLVE To add the vectos algebaicall we mst know thei components. Fom the fige these ae seen to be A = 00 m A = 0 m A We leaned fom the fige that has a negative -component. Adding and b components gives A B B B B =-(00 m) cos 45 =-4 m B = (00 m) sin 45 = 4 m C = A + B = 00 m - 4 m =-4 m C = A + B = 0 m + 4 m = 4 m The magnitde of the net displacement C = C + C = (-4 m) + (4 m) = 47 m The angle, as defined in Fige 3., is = tan - a C b = tan - a 4 m ƒ C ƒ 4 m b = 74 Ths the bid s net displacement is C B = (47 m, 74 noth of west). ASSESS The final vales of C and C match what we wold epect fom the sketch in Fige 3.. The geometic addition was a valable check on the answe we fond b algebaic addition. B B C B is Vecto sbtaction and the mltiplication of a vecto b a scala ae also easil pefomed sing components. To find D B = P B - Q B we wold compte Similal, T B = cs B is D = P - Q D = P - Q (3.7) When o look at a tail map fo a hike in a montainos egion, it will give the length of a tail and the elevation gain an impotant vaiable! The elevation gain is simpl d, the vetical component of the displacement fo the hike. FIGURE 3. A coodinate sstem with tilted aes. C C C The component vectos of C ae fond with espect to the tilted aes. (3.8) The net few chaptes will make feqent se of vecto eqations. Fo eample, o will lean that the eqation to calclate the net foce on a ca skidding to a stop is (3.9) Eqation 3.9 is eall jst a shothand wa of witing the two simltaneos eqations: (3.0) In othe wods, a vecto eqation is intepeted as meaning: Eqate the -components on both sides of the eqals sign, then eqate the -components. Vecto notation allows s to wite these two eqations in a moe compact fom. Tilted Aes T = cs T = cs F B = n B + w B + f B F = n + w + f F = n + w + f Althogh we ae sed to having the -ais hoizontal, thee is no eqiement that it has to be that wa. In Chapte, we saw that fo motion on a slope, it is often most convenient to pt the -ais along the slope. When we add the -ais, this gives s a tilted coodinate sstem sch as that shown in FIGURE 3.. 4th Poof

13 KNIG549_0_ch03_pp067-0.qd 5//09 4:34 PM Page Motion on a Ramp 79 Finding components with tilted aes is no hade than what we have done so fa. Vecto C B in Fige 3. can be decomposed into component vectos C B and C B, with C = C cos and C = C sin. STOP TO THINK 3.3 Angle f that specifies the diection of is compted as A. tan - (C B. tan - /C ). (C /ƒ C ƒ). C. tan - (ƒ C D. tan - ƒ/ƒ C ƒ). (C /C ). E. tan - (C F. tan - /ƒ C ƒ). (ƒ C ƒ/ƒ C ƒ). C B C f 3.4 Motion on a Ramp In this section, we will eamine the poblem of motion on a amp o incline. Thee ae thee easons to look at this poblem. Fist, it will povide good pactice at sing vectos to analze motion. Second, it is a simple poblem fo which we can find an eact soltion. Thid, this seemingl abstact poblem has eal and impotant applications. We begin with a constant-velocit eample to give s some pactice with vectos and components befoe moving on to the moe geneal case of acceleated motion. EXAMPLE 3.6 Finding the height gained on a slope A ca dives p a steep 0 slope at a constant speed of 5 m/s. Afte 0 s, how mch height has the ca gained? PREPARE FIGURE 3.3 is a visal oveview, with - and -aes defined. The velocit vecto v B points p the slope. We ae inteested in the vetical motion of the ca, so we decompose v B into component vectos and as shown. v B v B FIGURE 3.3 Visal oveview of a ca moving p a slope. i, i, t i v v f, f, t f v Known i 5 i 5 0 m t i 5 0 s, t f 5 0 s v 5 5 m/s 5 0 Find D SOLVE The velocit component we need is v ; this descibes the vetical motion of the ca. Using the les fo finding components otlined above, we find v = v sin = (5 m/s) sin(0 ) =.6 m/s Becase the velocit is constant, the ca s vetical displacement (i.e., the height gained) ding 0 s is = v t = (.6 m/s)(0 s) = 6 m ASSESS The ca is taveling at a pett good clip 5 m/s is a bit faste than 30 mph p a steep slope, so it shold climb a espectable height in 0 s. 6 m, o abot 80 ft, seems easonable. Acceleated Motion on a Ramp FIGURE 3.4a on the net page shows a cate sliding down a fictionless (i.e., smooth) amp tilted at angle. The cate acceleates de to the action of gavit, bt it is constained to acceleate paallel to the sface. What is the acceleation? A motion diagam fo the cate is dawn in FIGURE 3.4b. Thee is an acceleation becase the velocit is changing, with both the acceleation and velocit vectos paallel to the amp. We can take advantage of the popeties of vectos to find the cate s acceleation. To do so, FIGURE 3.4C sets p a coodinate sstem with the -ais along the amp and the -ais pependicla. All motion will be along the -ais. 4th Poof

14 KNIG549_0_ch03_pp067-0.qd 5//09 4:34 PM Page CHAPTER 3 Vectos and Motion in Two Dimensions FIGURE 3.4 Acceleation on an inclined plane. (a) (b) v a Angle of incline If the incline sddenl vanished, the object wold have a fee-fall acceleation a B fee fall staight down. As Fige 3.4c shows, this acceleation vecto can be decomposed into two component vectos: a vecto a B that is paallel to the incline and a vecto a B that is pependicla to the incline. The vecto addition les stdied ealie in this chapte tell s that a B fee fall = a B + a B. The motion diagam shows that the object s actal acceleation a B is paallel to the incline. The sface of the incline somehow blocks the othe component of the acceleation a B, thogh a pocess we will eamine in Chapte 5, bt a B is nhindeed. It is this component of a B fee fall, paallel to the incline, that acceleates the object. We can se tigonomet to wok ot the magnitde of this acceleation. Fige 3.4c shows that the thee vectos a B a B and a B fee fall,, fom a ight tiangle with angle as shown; this angle is the same as the angle of the incline. B definition, the magnitde of a B fee fall is g. This vecto is the hpotense of the ight tiangle. The vecto we ae inteested in, a B, is opposite angle. Ths the vale of the acceleation along a fictionless slope is a =;g sin (3.) (c) This component of afee fall acceleates the cate down the incline. a fee fall a a This ight tiangle elates the fee-fall acceleation and its components. Same angle NOTE The coect sign depends on the diection in which the amp is tilted. The acceleation in Fige 3.4 is +g sin, bt pcoming eamples will show sitations in which the acceleation is -g sin. Let s look at Eqation 3. to veif that it makes sense. A good wa to do this is to conside some limiting cases in which the angle is at one end of its ange. In these cases, the phsics is clea and we can check o eslt. Let s look at two sch possibilities:. Sppose the plane is pefectl hoizontal, with = 0. If o place an object on a hoizontal sface, o epect it to sta at est with no acceleation. Eqation 3. gives a = 0 when = 0, in ageement with o epectations.. Now sppose o tilt the plane ntil it becomes vetical, with = 90. Yo know what happens the object will be in fee fall, paallel to the vetical sface. Eqation 3. gives a = g when = 90, again in ageement with o epectations. NOTE Checking o answe b looking at sch limiting cases is a ve good wa to see if o answe makes sense. We will often do this in the Assess step of a soltion. Eteme phsics A speed skie, on wide skis with little fiction, weaing an aeodnamic helmet and coched low to minimize ai esistance, moves in a staight line down a steep slope pett mch like an object sliding down a fictionless amp. Thee is a maimm speed that a skie cold possibl achieve at the end of the slope. Cose designes set the stating point to keep this maimm speed within easonable (fo this spot!) limits. EXAMPLE 3.7 Maimm possible speed fo a skie The Willamette Pass ski aea in Oegon was the site of the 993 U.S. National Speed Skiing Competition. The skies stated fom est and then acceleated down a stetch of the montain with a easonabl constant slope, aiming fo the highest possible speed at the end of this n. Ding this acceleation phase, the skies taveled 360 m while dopping a vetical distance of 70 m. What is the fastest speed a skie cold achieve at the end of this n? How mch time wold this fastest n take? 4th Poof

15 KNIG549_0_ch03_pp067-0.qd 5//09 4:34 PM Page Motion on a Ramp 8 PREPARE We begin with the visal oveview in FIGURE 3.5. The motion diagam shows the acceleation of the skie and the pictoial epesentation gives an oveview of the poblem inclding the dimensions of the slope. As befoe, we pt the -ais along the slope. FIGURE 3.5 Visal oveview of a skie acceleating down a slope. v 0 a i, (v ) i, t i Slope f, (v ) f, t f length Height 360 m 70 m Known t 5 0 s i 5 0 m i (v ) 5 0 m/s i m f Find (v ) f, t f SOLVE The fastest possible n wold be one withot an fiction o ai esistance, meaning the acceleation down the slope is given b Eqation 3.. The acceleation is in the positive -diection, so we se the positive sign. What is the angle in Eqation 3.? Fige 3.5 shows that the 360-m-long slope is the hpotense of a tiangle of height 70 m, so we se tigonomet to find sin = 70 m 360 m which gives = sin - (70/360) = 8. Eqation 3. then gives a =+g sin = (9.8 m/s )(sin 8 ) = 4.6 m/s Fo linea motion with constant acceleation, we can se the thid of the kinematic eqations in Table.4: (v (v ) ) f = i + a. The initial velocit (v ) i is zeo; ths: This is the distance along the slope, the length of the n. v f 5!a D 5!4.6 m/s 360 m 5 58 m/s This is the fastest that an skie cold hope to be moving at the end of the n. An fiction o ai esistance wold decease this speed. Becase the acceleation is constant and the initial velocit (v ) i is zeo, the time of the fastest-possible n is t = (v ) f a A speed skiing event is a qick affai! = 58 m/s 4.6 m/s = 3 s ASSESS The final speed we calclated is 58 m/s, which is abot 30 mph, easonable becase we epect a high speed fo this spot. In the competition noted, the actal winning speed was mph, not mch slowe than the eslt we calclated. Obviosl, the effots to minimize fiction and ai esistance ae woking! Skis on snow have ve little fiction, bt thee ae othe was to edce the fiction between sfaces. Fo instance, a olle coaste ca olls along a tack on lowfiction wheels. No dive foce is applied to the cas afte the ae eleased at the top of the fist hill: the speed changes de to gavit alone. The cas speed p as the go down hills and slow down as the climb. EXAMPLE 3.8 Speed of a olle coaste A classic wooden coaste has cas that go down a big fist hill, gaining speed. The cas then ascend a second hill with a slope of 30. If the cas ae going 5 m/s at the bottom and it takes them.0 s to climb this hill, how fast ae the going at the top? PREPARE We stat with the visal oveview in FIGURE 3.6, which incldes a motion diagam, a pictoial epesentation, and a list of vales. We ve done this with a sketch sch as o might daw fo o homewok. Notice how the motion diagam of Fige 3.6 diffes fom that of the pevios eample: The velocit deceases as the ca moves p the hill, so the acceleation vecto is opposite the diection of the velocit vecto. The motion is along the -ais, as befoe, bt the acceleation vecto points in the negative -diection, so the component a is negative. In the motion diagam, notice that we dew onl a single acceleation vecto a easonable shotct becase we know that the acceleation is constant. One vecto can epesent the acceleation fo the entie motion. FIGURE 3.6 The coaste s speed deceases as it goes p the hill. Contined 4th Poof

16 KNIG549_0_ch03_pp067-0.qd 5//09 4:34 PM Page 8 8 CHAPTER 3 Vectos and Motion in Two Dimensions SOLVE To detemine the final speed, we need to know the acceleation. We will assme that thee is no fiction o ai esistance, so the magnitde of the olle coaste s acceleation is given b Eqation 3. sing the mins sign, as noted: a =-g sin =-(9.8 m/s ) sin 30 =-4.9 m/s The speed at the top of the hill can then be compted sing o kinematic eqation fo velocit: (v ) f = (v ) i + a t = 5 m/s + (-4.9 m/s )(.0 s) = 5 m/s ASSESS The speed is less at the top of the hill than at the bottom, as it shold be, bt the coaste is still moving at a pett good clip at the top almost 35 mph. This seems easonable; a fast ide is a fn ide. STOP TO THINK 3.4 A block of ice slides down a amp. Fo which height and base length is the acceleation the geatest? Height A. Height 4 m, base m B. Height 3 m, base 6 m C. Height m, base 5 m D. Height m, base 3 m Base FIGURE 3.7 Am, Bill, and Calos each mease the velocit of the nne. The velocities ae shown elative to Am. Am Bill Rnne 5 m/s 5 m/s Calos 5 m/s Thowing fo the gold An athlete thowing the javelin does so while nning. It s hade to thow the javelin on the n, bt thee s a ve good eason to do so. The distance of the thow will be detemined b the velocit of the javelin with espect to the gond which is the sm of the velocit of the thow pls the velocit of the athlete. A faste n means a longe thow. 3.5 Relative Motion Yo ve now dealt man times with poblems that sa something like A ca tavels at 30 m/s o A plane tavels at 300 m/s. Bt, as we will see, we ma need to be a bit moe specific. In FIGURE 3.7, Am, Bill, and Calos ae watching a nne. Accoding to Am, the nne s velocit is v = 5 m/s. Bt to Bill, who s iding alongside, the nne is lifting his legs p and down bt going neithe fowad no backwad elative to Bill. As fa as Bill is concened, the nne s velocit is v = 0 m/s. Calos sees the nne eceding in his eaview mio, in the negative -diection, getting 0 m fathe awa fom him eve second. Accoding to Calos, the nne s velocit is v =-0 m/s. Which is the nne s te velocit? Velocit is not a concept that can be te o false. The nne s velocit elative to Am is 5 m/s; that is, his velocit is 5 m/s in a coodinate sstem attached to Am and in which Am is at est. The nne s velocit elative to Bill is 0 m/s, and the velocit elative to Calos is -0 m/s. These ae all valid desciptions of the nne s motion. Relative Velocit Sppose we know that the nne s velocit elative to Am is 5 m/s; we will call this velocit (v ) RA. The second sbscipt RA means Rnne elative to Am. We also know that the velocit of Calos elative to Am is 5 m/s; we wite this as (v ) CA = 5 m/s. It is eqall valid to compte Am s velocit elative to Calos. Fom Calos s point of view, Am is moving to the left at 5 m/s; we wite Am s velocit elative to Calos as (v ) AC =-5 m/s; note that (v ) AC =-(v ) CA. Given the nne s velocit elative to Am and Am s velocit elative to Calos, we can compte the nne s velocit elative to Calos b combining the two velocities we know. The sbscipts as we have defined them ae o gide fo this combination: v RC 5 v RA v AC The A appeas on the ight of the fist epession and on the left of the second; when we combine these velocities, we cancel the A to get (v ) RC. (3.) 4th Poof

17 KNIG549_0_ch03_pp067-0.qd 5//09 4:34 PM Page Relative Motion 83 Geneall, o can add two elative velocities in this manne, b canceling sbscipts as in Eqation 3.. In Chapte 7, when we lean abot elativit, we will have a moe igoos scheme fo compting elative velocities, bt this techniqe will seve o pposes at pesent. EXAMPLE 3.9 Speed of a seabid Reseaches doing satellite tacking of albatosses in the Sothen Ocean obseved a bid maintaining sstained flight speeds of 35 m/s neal 80 mph! This seems spisingl fast ntil o ealize that this paticla bid was fling with the wind, which was moving at 3 m/s. What was the bid s aispeed its speed elative to the ai? This is a te mease of its flight speed. PREPARE FIGURE 3.8 shows the wind and the albatoss moving to the ight, so all velocities will be positive. We ve shown the FIGURE 3.8 Relative velocities fo the albatoss and the wind fo Eample 3.9. v aw v bw Known (v ) bw 5 35 m/s (v ) aw 5 3 m/s Find (v ) ba velocit (v ) bw of the bid with espect to the wate, which is the meased flight speed, and the velocit (v ) aw of the ai with espect to the wate, which is the known wind speed. We want to find the bid s aispeed the speed of the bid with espect to the ai. SOLVE We need the sbscipt fo the wate to cancel, so, accoding to Eqation 3., we wite (v ) ba = (v ) bw + (v ) wa The tem (v ) wa, is the opposite of the second of o known vales, so we se (v ) wa =-(v ) aw =-3 m/s to find (v ) ba = (35 m/s) + (-3 m/s) = m/s ASSESS m/s abot 5 mph is a easonable aispeed fo a bid. And it s slowe than the obseved flight speed, which makes sense becase the bid is fling with the wind. This techniqe fo finding elative velocities also woks fo two-dimensional sitations, as we see in the net eample. Relative motion in two dimensions is anothe good eecise in woking with vectos. EXAMPLE 3.0 Finding the gond speed of an aiplane Cleveland is appoimatel 300 miles east of Chicago. A plane speed of the ai elative to the gond (v B ag); the speed of the plane leaves Chicago fling de east at 500 mph. The pilot fogot to elative to the gond will be the vecto sm of these velocities: check the weathe and doesn t know that the wind is blowing to v B pg = v B pa + v B ag the soth at 00 mph. What is the plane s velocit elative to the gond? This vecto sm is shown in Fige 3.9. PREPARE FIGURE 3.9 is a visal oveview of the sitation. We ae given the speed of the plane elative to the ai and the (v B pa) FIGURE 3.9 The wind cases a plane fling de east in the ai to move to the sotheast elative to the gond. SOLVE The plane s speed elative to the gond is the hpotense of the ight tiangle in Fige 3.9; ths: v pg = v pa + v ag = (500 mph) + (00 mph) = 50 mph The plane s diection can be specified b the angle meased fom de east: 00 mph = tan - a 500 mph b = tan- (0.0) = The velocit of the plane elative to the gond is ths v B pg = (50 mph, soth of east) ASSESS The good news is that the wind is making the plane move a bit faste elative to the gond; the bad news is that the wind is making the plane move in the wong diection! 4th Poof

18 KNIG549_0_ch03_pp067-0.qd 5//09 4:34 PM Page CHAPTER 3 Vectos and Motion in Two Dimensions FIGURE 3.30 The motion of a tossed ball. The inset shows how to find the diection of v B, the change in velocit. This is the diection in which the acceleation a B points. (a) (b) Release v a D a v v a v a a a a Gond The acceleation is the same at all points. FIGURE 3.3 The lanch and motion of a pojectile. (v ) i 5 v i sin Initial velocit v i v i is the initial velocit. Lanch angle (v ) i 5 v i cos We compte components of the initial velocit as shown. 3.6 Motion in Two Dimensions: Pojectile Motion Balls fling thogh the ai, long jmpes, and cas doing stnt jmps ae all eamples of the two-dimensional motion that we call pojectile motion. Pojectile motion is an etension to two dimensions of the fee-fall motion we stdied in Chapte. A pojectile is an object that moves in two dimensions nde the inflence of gavit and nothing else. Althogh eal objects ae also inflenced b ai esistance, the effect of ai esistance is small fo easonabl dense objects moving at modest speeds, so we can ignoe it fo the cases we conside in this chapte. As long as we can neglect ai esistance, an pojectile will follow the same tpe of path: a tajecto with the mathematical fom of a paabola. Becase the fom of the motion will alwas be the same, the stategies we develop to solve one pojectile poblem can be applied to othes as well. FIGURE 3.30a shows the paabolic ac of a ball tossed into the ai; the camea has capted its position at eqal intevals of time. In FIGURE 3.30b we show the motion diagam fo this toss, with velocit vectos connecting the points. The acceleation vecto points in the same diection as the change in velocit v B, which we can compte sing the techniqes of Tactics Bo 3.. Yo can see that the acceleation vecto points staight down; a caefl analsis wold show that it has magnitde 9.80 m/s. Conseqentl, the acceleation of a pojectile is the same as the acceleation of an object falling staight down namel, the fee-fall acceleation: Becase the fee-fall acceleation is the same fo all objects, it is no wonde that the shape of the tajecto a paabola is the same as well. As the pojectile moves, the fee-fall acceleation will change the vetical component of the velocit, bt thee will be no change to the hoizontal component of the velocit. Theefoe, the vetical and hoizontal components of the acceleation ae (3.3) The vetical component of acceleation a fo all pojectile motion is jst the familia -g of fee fall, while the hoizontal component is zeo. Analzing Pojectile Motion Sppose o toss a basketball down the cot, as shown in FIGURE 3.3. To std this pojectile motion, we ve established a coodinate sstem with the -ais hoizontal and the -ais vetical. The stat of a pojectile s motion is called the lanch, and the angle of the initial velocit v B i above the hoizontal (i.e., above the -ais) is the lanch angle. As o leaned in Section 3.3, the initial velocit vecto v B i can be epessed in tems of the - and -components (v ) i and (v ) i. Yo can see fom the fige that whee v i is the initial speed. a B fee fall = (9.80 m/s, staight down) a = 0 m/s a =-g =-9.80 m/s (v ) i = v i cos (v ) i = v i sin (3.4) NOTE The components (v ) i and (v ) i ae not alwas positive. A pojectile lanched at an angle below the hoizontal (sch as a ball thown downwad fom the oof of a bilding) has negative vales fo and (v ) i. Howeve, the speed v i is alwas positive. To see how the acceleation detemines the sbseqent motion, FIGURE 3.3 shows a pojectile lanched at a speed of.0 m/s at an angle of 63 fom the hoizontal. a 4th Poof