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we are adding vectors rather than numbers. Note that the resultant Fig.3.4 Addition of vectorsvector is specified by both a magnitude and a direction. When addingdistances travelled, the resultant vector is called the total displace- d៬2 v៬2ment. The resultant vectors in Fig. 3.4 are described properly in Fig.3.5. Notice how a magnitude and a direction are given. You should d៬1be able to specify direction in another way for each resultant vector. ៬dTotalFig.3.5 Resultant vectors v៬1 v៬Total 800 m 15 km/s ៬dTotal d៬3 d៬2 N 26˚ d៬1 WEd៬T ϭ 800 m [E26˚N] N S W E 53˚ v៬T ϭ 15 km/s [W53˚N] S3.1 km52˚d៬T ϭ 3.1 km [E52˚N] In summary, the “ϩ” sign in vector addition is an instruction to connect TRIGONOMETRY REVIEWthe vectors head to tail. When the vectors are collinear, you can simply addthem. When the vectors are not collinear, you must construct a vector dia- Fig.3.6A Bgram or use trigonometry. To calculate the resultant of two vectors, thetrigonometry for solving the triangle is straightforward. If your problem cacontains more than two vectors to add, you will normally find it easier touse components. Using this method, you resolve each vector into compo- ␪ b Cnents in the x and y directions. Then add the magnitudes of the components Ain each direction, and solve the final right-angle triangle. Its hypotenuse isthe resultant vector (see Fig. 3.6A). For a right-angle triangle with sides a and b forming the right angle, c is e x a m p l e 1 Finding the resultant vector the hypotenuse. Pythagoras’ theo- rem states A student runs to school, covering a distance of 0.50 km in a direction given by [N20°E] as measured from the starting point. She then trudges a2 ϩ b2 ϭ c2 over to a friend’s house 0.30 km away due west from the school. And finally the two friends go off to the mall 0.80 km away in a direction Using the basic trigonometric given by [W50°S] from the friend’s house. What is the total displacement ratios, ␪ can be found in one of of the student? three ways: sine: sin␪ ϭ ᎏaᎏ ϭ ᎏoppoᎏsite c hypotenuse cosine: cos␪ ϭ ᎏbᎏ ϭ ᎏadjacᎏent c hypotenuse tangent: tan␪ ϭ ᎏaᎏ ϭ ᎏoppoᎏsite b adjacent chapt e r 3: Motion in Two Dimensions 79

VECTOR COMPONENTS Solution and Connection to Theory A vector can be resolved into com- Given →d2 ϭ 0.30 km [W] →d3 ϭ 0.80 km [W50°S] ponents, which are perpendicular to →d1 ϭ 0.50 km [N20°E] each other. When they are added, their sum is equal to the vector. In The vector equation is Fig. 3.6B, vector →v is resolved into components vx and vy, where → ϩ → ϩ → ϭ → vx ϭ v cos␪ and vyϭ v sin␪ d1 d2 d3 d total You could also use the complemen- First set a scale: 1 cm represents 0.10 km. Then construct the vector dia- tary angle and get gram, as shown in Fig. 3.7, drawing the vectors to scale. Make sure the vectors are connected head to tail, with the correct angles. vx ϭ v sin␤ and vyϭ v cos␤ Fig.3.7 Choose whichever angle is most suitable in the given problem. 0.10 km d៬2 NFig. 3.6B 50˚ y WE ␤ ៬v vy S d៬1 x␪ d៬3 vx ៬dtotal 20˚ ␪ Find the resultant vector (total displacement) by drawing a line from the starting point to the end point. It should measure 6.6 cm. From the scale, this represents 0.66 km. Using a protractor, now measure the angle, ␪, of the total displacement at its tail from one of the cardinal directions (i.e., N, S, E, W). It is 78° westward from south. So, the net displacement is 0.66 km [S78°W] (or 0.66 km [W12°S]). For a trigonometric solution, you could break the figures into triangles and solve for the length and direction of the resultant. However, a more common way is to calculate the net displacement using a pair of compo- nents that are perpendicular to each other. Ordinarily, you will choose standard directions for the pair of components. As shown in Fig. 3.8, choose components along the N-S and E-W directions. Choose north and east as positive. Add the components first in the E-W direction (call the resultant dx), then those in the N-S direction (for dy ). Finally, add the two component sums using Pythagoras’ theorem and trigonometric ratios.80 u n i t a : M ot i o n a n d Forc es

Fig.3.8 Ϫ0.3 km 50˚Ϫ(0.8 km) sin 50˚ ϭ Ϫ0.61 kmϪ(0.8 km) cos 50˚ ϭ Ϫ0.15 km 0.8 km 0.5 km50˚ (0.5 km) sin 70˚ ϭ 0.47 km Ϫ0.61 km 0.47 km 70˚ (0.5 km) cos 70˚ ϭ 0.17 km dTx 70˚ ␪៬dTy dTotal dTy Ϫ0.51 km Ϫ0.3 km dTx 0.17 kmE-W: Adding all the horizontal components, dTx ϭ d1x ϩ d2x ϩ d3x ϭ (0.5 km)cos 70° Ϫ 0.3 km Ϫ (0.8 km)cos 50° ϭ 0.17 km Ϫ 0.3 km Ϫ 0.51 km ϭ Ϫ0.64 kmN-S: Adding all the vertical components, dTy ϭ d1y ϩ d2y ϩ d3y ϭ (0.5 km)sin 70° ϩ 0 Ϫ (0.8 km)sin 50° ϭ 0.47 km Ϫ 0.61 km ϭ Ϫ0.14 kmNow use Pythagoras’ theorem to calculate the sum of the two compo-nents, and the tangent ratio to find the angle of the resultant vector. → ϭ dTx ϩ dTy d2 ϭ d2Tx ϩ d2Ty total d total dtotal ϭ ͙dෆ2Tx ϩෆd2Ty ϭ ͙(ෆϪ0.64ෆkm)2ෆϩ (Ϫෆ0.14 kmෆ)2 ϭ ͙ෆ0.4292ෆkm2 ϭ 0.66 kmTo calculate the angle, substitute magnitudes of sides into the equation. ␪ ϭ tanϪ1 ᎏdTᎏy ϭ tanϪ1 ᎏ0.14ᎏkm ϭ tanϪ1 0.219 ϭ 12° dTx 0.64 kmYou can get the correct direction by checking the signs of dTx and dTy. Sincethey are both negative, the components of the resultant displacement are0.64 km [W] and 0.14 km [S]. Combine those directions with ␪ ϭ 12º tofind the student’s resultant displacement, which is 0.66 km [W12°S]. chapt e r 3: Motion in Two Dimensions 81

Fig.3.9 Method for Solving Vector Addition Problems utting Are it all gethTop Givens givens YES Set reference Solve by pplyiner addition/subtraction the collinear? direction ncep NO82 Use vector addition Connect vectors head to tail Resultant vector is directed from start of first to end of last component vector Find Find angle magnitude of direction g 1. Find the components in the x and y directions for the following dis-Co placements:a a) 20 km [S30°E] b) 40 km [W60°N] c) 10 m [N10°E] ts d) 5 km [S24°W] e) 12 m [N45°W] f) 10 km [N90°E] 2. A person walks 20 m [N20°E], then 120 m [N50°W], then 150 m [W], and finally 30 m [S75°E]. Find the person’s final displacement using a scaled diagram and the trigonometric method. 3. Using a real example, explain in your own words the trigonometric method of solving two-dimensional displacement problems. 3.2 Parabolic Motion When you throw a ball in outer space, the ball will leave your hands and travel in a straight line forever, unless it runs into an object or encounters a gravitational field. When you throw the same ball on Earth, it doesn’t travel in a straight line but in an arc, finally coming to rest on the ground. Of course, we know that Earth causes all objects to fall. Our ball is moving forward, as it would in outer space, and falling at the same time. The com- bination of these two actions produces the curved motion we see. The term projectile motion is used to describe events where the object moves under the influence of gravity and is not self powered. unit a: Motion and Forces

Fig.3.10 Projectile components Further analyzing the motion of the throw, we will Fig.3.11 Launch angle affects rangeassume that the air resistance is negligible. Therefore, the balldoes not slow down, i.e., the velocity in the horizontal direction (x) isconstant. But the velocity in the downward direction (y) is increasing inmagnitude because of Earth’s gravitational pull. Its acceleration due to grav-ity is 9.8 m/s2. This value will be derived in Chapter 5. The trajectory of a ball in Fig. 3.10 shows the two components of projectilemotion. Since the horizontal velocity is constant, the x velocity vector is thesame length throughout the motion. Since gravity causes a downward acceler-ation, the y velocity vector gets progressively longer. When you add these two velocity vectors, you obtain the resultant veloc-ity at any given moment. This is indicated by a hollow vector on the diagram(Fig. 3.10). In summary, as the motion proceeds in time, the velocity compo-nent in the y direction increases in length. The velocity vector in the x direc-tion stays the same length. When you combine a horizontal constant motionwith an accelerating vertical motion, you produce parabolic (curved) motion. In Fig. 3.11, you can see how the angle of launch of a volleyball or a foot-ball can affect the range of a parabolic trajectory. The range is the displace-ment in the horizontal direction, x.␪ Ͼ 45˚ ␪ ϭ 65˚ ␪ ϭ 45˚ ␪ ϭ 45˚ ␪ ϭ 25˚ ␪ Ͻ 45˚ Range Range(a) (b)Timing a Parabolic Trajectory Fig.3.12 Vertical and horizontalHow long will it take the baseball thrown in Fig. 3.10 to hit the ground? The components are independentanswer may surprise you. If the ball is thrown exactly horizontally, it willstay in the air for exactly the same length of time as a ball dropped from the For imagesame height, no matter how fast you throw it! see student Figure 3.12 is a strobe photo of two balls released at exactly the same text.moment. Ball A drops straight down, while ball B is projected horizontally.Horizontal lines are superimposed on the photograph. You can see that chapt e r 3: Motion in Two Dimensions 83

Fig.3.13 Changing the initial v៬1 v៬1 v៬1vertical velocity v1x v1x v1x v1y 0 v1y v1y (a) (b) (c) both balls cross the lines together. The objects are falling at the same rate and will land at the same time. The initial push only caused ball B to travel farther in the horizontal (x) direction. It still fell the same vertical distance, at the same rate as the dropped ball (A). If you wish to have a clear winner, Fig. 3.13 shows two other possible cases. If you wish ball B to land first, then you could throw it downward, thereby giving it an initial velocity in the y direction (as in (b), in the direc- tion of acceleration). If you throw the object upward, as in (c), again it will have an initial velocity in the y direction. However, this time the ball will cover a greater vertical distance (go up, stop, then come down) and land last. Fig.3.14 2D Motion Summary guttinit allTogethp 2D vector Is there Add horizontal Parabolic motion er problem, and vertical horizontal acceleration YES Vertical velocity vector velocity constant is increasing components due to gravity? NO Vertical velocity is constant Add horizontal and vertical vector components Linear motion pplying 1. A cannonball is shot from a cannon simultaneously as another can- theCo nonball is dropped from the same height. Why do both cannonballs ncepa land at the same time? Assume the cannon is pointed horizontally. ts84 2. What would happen to your answer for Question 1 if the cannon was not pointed horizontally? 3. How would air and spin affect the answer to Question 1? unit a: Motion and Forces

3.3 Projectile Motion CalculationsIn Chapter 2, we used the kinematics equations for one-dimensionalmotions only. The key to solving two-dimensional problems is to breakthem up into two one-dimensional parts, then recombine them to producea final answer. The problem set-up will now have a set of givens in the xdirection and another set in the y direction. We will develop the process for solving such two-dimensional projectileproblems in a series of steps. First, consider a horizontally fired cannon. Wewill set up the situation in Example 2, and then perform the calculations inExample 3. Next, consider a cannon that points upward at an angle in orderto gain a greater range. Example 4 contains calculations to find the range.Example 5 calculates the maximum height achieved in the cannon shot.e x a m p l e 2 Setting up a projectile problemThe Great Projecto, a circus duck, is shot at 40 m/s out of a horizontalcannon at the top of a cliff 100 m high. Derive the equations that repre-sent the horizontal and vertical components of the motion.Solution and Connection to TheoryWe will assume a standard sign reference system. The first step is to setup givens for components in the x and y directions.x} v1x ϭ v2x ϭ 40 m/s (horizontal velocity is constant)ax ϭ 0 ⌬dx ϭ ? ⌬t ϭ ?y} v1y ϭ 0 ay ϭ Ϫ9.8 m/s2 ⌬dy ϭ Ϫ100 m. ⌬t ϭ ?⌬dy is negative because Projecto travels downward, the negative direction.The problem is now ready to be solved. The unknowns are time, and dis-placement in the horizontal direction, which is referred to as the range.Figure 3.15 shows Projecto shooting out with an initial horizontal veloc-ity only; as he sails through the air, he develops an ever-increasing verti-cal velocity downward. ϩFig.3.15 40 m/s ϩ 9.8 m/s2 100 m chapt e r 3: Motion in Two Dimensions 85

m etho From the givens, you can see that there is a ⌬t in each dimension. Since s of this is the time from firing to landing, ⌬t is the same in both directions. do ces So, if you can find a value for time in one of the directions, you can usepr that value in equations for the other direction. Choice of Equation. The variables used in the x direction are ⌬t, ⌬d→, a→, and v→1. The equation containing all four of these variables is ⌬d→ ϭ v→1⌬t ϩ ᎏ21ᎏa→⌬t2 This equation simplifies to ⌬dx ϭ vx⌬t because the acceleration in the x direction is zero and the velocity is constant. The variables in the y direction are the same as those in the x direction, hence we can use the same equation: ⌬dy ϭ v1y⌬t ϩ ᎏ21ᎏay⌬t2 Fig.3.16 Solving Projectile Problems ax ϭ 0 ⌬dx ϭ vx⌬t vx ϭ constantGiven Split velocity ⌬dx ϭ range Is YES Transfer to Find heightvalues into horizontal ⌬t y direction ⌬t found and vertical components ay ϭ Ϫ9.8 m/s2 in x ? vy is changing ⌬dy ϭ height ⌬dy ϭ vy • ⌬t ϩ 1/2ay⌬t2 NO Calculate ⌬t using equation in y Transfer to x direction Find range Now let’s finish the cannon problem. e x a m p l e 3 Finding the range The Great Projecto is shot out of a horizontal cannon at 40 m/s. If the can- non is sitting at the top of a cliff 100 m high, how far will the duck travel?86 u n i t a : M ot i o n a n d Forc es

Solution and Connection to TheoryGiven ax ϭ 0 ⌬dx ϭ ? ⌬t ϭ ?x} v1x ϭ v2x ϭ 40 m/sy} v1y ϭ 0 ay ϭ Ϫ9.8 m/s2 ⌬dy ϭ Ϫ100 m ⌬t ϭ ?In the y direction, there is only one unknown. Solve in this direction first.⌬dy ϭ v1y⌬t ϩ ᎏ1ᎏ ay⌬t2 2v1y ϭ 0, therefore ⌬t2 ϭ ᎏ2⌬ᎏdy ay ᎏ2⌬ᎏdy ϭ Ϯ ᎏ2Ϫ(Ϫ9.180ᎏm0 /ms2) ϭ Ϯ4.5 s ayΊ๶ Ί๶๶⌬t ϭ ϮSince time must be positive, ⌬t ϭ ϩ4.5 s. Now we use this value of timeto calculate the range.⌬dx ϭ vx⌬t ax ϭ 0; therefore, ⌬dx ϭ 40 m/s (4.5 s) ϭ 180 mDuring the 4.5 s that Projecto took to fall 100 m in the y direction, hetravelled 180 m in the x direction.e x a m p l e 4 Projection angled upwardA cannonball is shot out of a cannon with a horizontal velocity Fig.3.17Avy ϭ 20 m/s ϩcomponent of 40 m/s and a vertical velocity component of 20 m/s v៬0[up]. If the cannon is sitting at the top of a cliff 100 m high, how farwill the cannonball travel?Solution and Connection to Theory ϩThis problem is very much like Example 3, except it has a vertical 100 m vx ϭ 40 m/svelocity component. This means that the cannonball is being fired Rangeat an angle upwards from the cliff top.Given ax ϭ 0 ⌬dx = ? ⌬t ϭ ?x} v1x ϭ v2x ϭ 40 m/sy} v1y ϭ 20 m/s ay ϭ Ϫ9.8 m/s2 ⌬dy ϭ Ϫ100 m ⌬t ϭ ?Notice in Fig. 3.17A that the height of the cliff is still 100 m, as it was inExamples 2 and 3. In this example, the object travelled higher than thecliff top, but the definition of displacement is final position minus initialposition, which is just the height of the cliff. If we were to stop the prob-lem at a different moment, then the displacement would be different. Asbefore, solve first in the y direction. chapt e r 3: Motion in Two Dimensions 87

QUADRATIC FORMULA y} ⌬dy ϭ v1y⌬t ϩ ᎏ21ᎏay⌬t2 Ϫ100 m ϭ (20 m/s) ⌬t ϩ ᎏ21ᎏ(Ϫ9.8 m/s2) ⌬t2Given the quadratic equationax2 ϩ bx ϩ c ϭ 0,the solution is This is a quadratic equation in ⌬t.xϭ ᎏϪb Ϯ ͙ෆbᎏ2 Ϫ 4aෆc (Ϫ4.9 m/s2) ⌬t2 ϩ (20 m/s) ⌬t ϩ 100 m ϭ 0 2a Using the quadratic formula,ALTERNATIVE APPROACH ⌬t ϭ ᎏϪ20 m/ᎏs Ϯ ͙(ෆ20ᎏm/ෆs)2 Ϫᎏෆ4(Ϫ4ෆ.ᎏ9 m/s2ෆ)(ᎏ100ෆm) Use v22y ϭ v12y ϩ 2ay⌬dy to find v2y 2(Ϫ4.9 m/s2) Then use ϭ ᎏϪ20 m/ᎏs Ϯ ͙ෆ40ᎏ0 mෆ2/s2ᎏϩෆ1960ᎏෆm2/s2 v2y ϭ v1y ϩ ay ⌬t to find ⌬t. Ϫ9.8 m/s2 ϭ ᎏϪ20 mϪᎏ/9s.8Ϯm4/8sᎏ.26 m/s ϭ ᎏϪ68.6ᎏm/s or ᎏ28.6ᎏm/s Ϫ9.8 m/s2 Ϫ9.8 m/s2 This gives ⌬t ϭ 7.0 s and Ϫ2.9 s. Note that the units cancel to produce the correct unit for ⌬t. A negative time does not apply in real life, so the correct value for ⌬t is 7.0 s. Now substitute 7.0 s for ⌬t in the equation for the x direction. ax ϭ 0; therefore, ⌬dx ϭ v1x⌬t ⌬dx ϭ 40 m/s(7.0 s) ϭ 280 m Because this projectile spent more time in the air than the one in Example 3, it managed to travel a greater distance, 280 m instead of 180 m.NEGATIVE TIME Fig.3.17B We can give a meaning to the nega- 150 m ⌬d៬ tive time by illustrating the problem using a parabola. The quadratic 100 m equation for ⌬t describes a motion for a vertical displacement of 100 m 50 m for the initial velocity components given in the problem. Those same Ϫ4 s Ϫ2 s 0 ⌬t values could also be achieved by 2s 4s 6s 8s starting at the ground 2.9 s earlier, and firing a projectile upward. After To find the maximum height of the shot, modify the solution as shown 2.9 s, the upward velocity of the in Example 5. projectile would be 20 m/s. Mathematically, that earlier portion of the motion is the part of the parabola to the left of the vertical axis, as shown in Fig. 3.17B.88 u n i t a : M ot i o n a n d Forc es