Holt Physics Book PDF - Falling Objects MCQ

SECTION 3 Falling Objects SECTION OBJECTIVES Relate the motion of a freely falling body to motion with FREE FALL constant acceleration. On August 2, 1971, a demonstration was conducted on the moon by astronaut Calcu...

SECTION 3 Falling Objects SECTION OBJECTIVES Relate the motion of a freely falling body to motion with FREE FALL constant acceleration. On August 2, 1971, a demonstration was conducted on the moon by astronaut Calculate displacement, velocity, and time at various David Scott. He simultaneously released a hammer and a feather from the same points in the motion of a height above the moon’s surface. The hammer and the feather both fell straight freely falling object. down and landed on the lunar surface at exactly the same moment. Although Compare the motions of dif- the hammer is more massive than the feather, both objects fell at the same rate. ferent objects in free fall. That is, they traveled the same displacement in the same amount of time. Freely falling bodies undergo constant acceleration In Figure 14, a feather and an apple are released from rest in a vacuum cham- free fall ber. The two objects fell at exactly the same rate, as indicated by the horizontal the motion of a body when only alignment of the multiple images. the force due to gravity is acting The amount of time that passed between the first and second images is on the body equal to the amount of time that passed between the fifth and sixth images. The picture, however, shows that the displacement in each time interval did not remain constant. Therefore, the velocity was not constant. The apple and the feather were accelerating. Compare the displacement between the first and second images to the displacement between the second and third images. As you can see, within each time interval the displacement of the feather increased by the same amount as the displacement of the apple. Because the time intervals are the same, we know that the velocity of each object is increasing by the same amount in each time interval. In other words, the apple and the feather are falling with the same con- stant acceleration. If air resistance is disregarded, all objects dropped near the surface of a planet fall with the same constant acceleration. This acceleration is due to gravitational force, and the motion is referred to as free fall. The acceleration due to gravity is denoted with the symbols ag (gener- ally) or g (on Earth’s surface). The magnitude of g is about 9.81 m/s2, or 32 ft/s2. Unless stated otherwise, this book will use the value 9.81 m/s2 for calculations. This acceleration is directed downward, toward the center of the Earth. In our usual choice of coordinates, the Figure 14 downward direction is negative. Thus, the acceleration of objects in When there is no air resistance, all objects fall with the same accelera- free fall near the surface of the Earth is ag = −g = −9.81 m/s2. Because tion regardless of their masses. an object in free fall is acted on only by gravity, ag is also known as free-fall acceleration. 60 Chapter 2 Acceleration is constant during upward and downward motion Figure 15 is a strobe photograph of a ball thrown up into the air with an ini- tial upward velocity of +10.5 m/s. The photo on the left shows the ball mov- ing up from its release toward the top of its path, and the photo on the right shows the ball falling back down. Everyday experience shows that when we throw an object up in the air, it will continue to move upward for some time, stop momentarily at the peak, and then change direction and begin to fall. Because the object changes direction, it may seem that the velocity and accel- eration are both changing. Actually, objects thrown into the air have a down- ward acceleration as soon as they are released. In the photograph on the left, the upward displacement of the ball between each successive image is smaller and smaller until the ball stops and finally begins to move with an increasing downward velocity, as shown on the right. As soon as the ball is released with an initial upward velocity of +10.5 m/s, it has an acceleration of −9.81 m/s2. After 1.0 s (∆t = 1.0 s), the ball’s velocity will change by −9.81 m/s to 0.69 m/s upward. After 2.0 s (∆t = 2.0 s), the ball’s velocity will again change by −9.81 m/s, to −9.12 m/s. (a) (b) The graph in Figure 16 shows the velocity of the ball plotted against time. Figure 15 As you can see, there is an instant when the velocity of the ball is equal to At the very top of its path, the ball’s 0 m/s. This happens at the instant when the ball reaches the peak of its velocity is zero, but the ball’s accel- eration is –9.8 1 m/s2 at every upward motion and is about to begin moving downward. Although the veloc- point—both when it is moving up ity is zero at the instant the ball reaches the peak, the acceleration is equal to (a) and when it is moving down (b). −9.81 m/s2 at every instant regardless of the magnitude or direction of the velocity. It is important to note that the acceleration is −9.81 m/s2 even at the peak where the velocity is zero. The straight-line slope of the graph indicates that the acceleration is constant at every moment. 12 Developed and maintained by the National Science Teachers Association 10 For a variety of links related to this 8 chapter, go to www.scilinks.org 6 4 Topic: Free Fall 2 SciLinks Code: HF60620 0 Velocity (m/s) −2 0.50 1.00 1.50 2.00 2.50 3.00 −4 −6 −8 −10 −12 Figure 16 −14 On this velocity-time graph, the −16 slope of the line, which is equal to −18 the ball’s acceleration, is constant −20 from the moment the ball is Time (s) released (t = 0.00) and throughout its motion. Motion in One Dimension 61  Freely falling objects always have the same downward acceleration PHYSICS It may seem a little confusing to think of something that is moving upward, like the ball in the example, as having a downward acceleration. Thinking of this Module 1 motion as motion with a positive velocity and a negative acceleration may help. “One-Dimensional Motion” provides an interactive lesson The downward acceleration is the same when an object is moving up, when it is with guided problem-solving at rest at the top of its path, and when it is moving down. The only things practice to teach you about all kinds of one-dimensional changing are the position and the magnitude and direction of the velocity. motion, including free fall. When an object is thrown up in the air, it has a positive velocity and a nega- tive acceleration. From Table 3 in Section 2, we see that this means the object is slowing down as it rises in the air. From the example of the ball and from everyday experience, we know that this makes sense. The object continues to move upward but with a smaller and smaller speed. In the photograph of the ball, this decrease in speed is shown by the smaller and smaller displacements as the ball moves up to the top of its path. At the top of its path, the object’s velocity has decreased until it is zero. Although it is impossible to see this because it happens so quickly, the object is actually at rest at the instant it reaches its peak position. Even though the velocity is zero at this instant, the acceleration is still –9.81 m/s2. When the object begins moving down, it has a negative velocity and its acceleration is still negative. From Table 3, we see that a negative acceleration and a negative velocity indicate an object that is speeding up. In fact, this is what happens when objects undergo free-fall acceleration. Objects that are falling toward Earth move faster and faster as they fall. In the photograph of the ball in Figure 15 (on the previous page), this increase in speed is shown by the greater and greater displacements between the images as the ball falls. Knowing the free-fall acceleration makes it easy to calculate the velocity, time, and displacement of many different motions using the equations for constantly accelerated motion. Because the acceleration is the same through- out the entire motion, you can analyze the motion of a freely falling object during any time interval. SAFETY CAUTION your open hand. The meterstick should be Avoid eye injury; do not swing held so that the zero mark is between metersticks. your fingers with the 1 cm mark above it. Time Interval of Free Fall You should not be touching the meter- MATERIALS LIST Your reaction time affects your perfor- stick, and your catching hand must be rest- mance in all kinds of activities—from ing on a table.Without warning you, your 1 meterstick or ruler sports to driving to catching something friend should release the meterstick so that you drop. Your reaction time is the that it falls between your thumb and your time interval between an event and your finger. Catch the meterstick as quickly as response to it. you can. You can calculate your reaction Determine your reaction time by hav- time from the free-fall acceleration and ing a friend hold a meterstick vertically the distance the meterstick has fallen between the thumb and index finger of through your grasp. 62 Chapter 2  SECTION 1 Introduction to Vectors SECTION OBJECTIVES Distinguish between a scalar and a vector. SCALARS AND VECTORS Add and subtract vectors by In the chapter “Motion in One Dimension,” our discussion of motion was lim- using the graphical method. ited to two directions, forward and backward. Mathematically, we described Multiply and divide vectors by scalars. these directions of motion with a positive or negative sign. That method works only for motion in a straight line. This chapter explains a method of describing the motion of objects that do not travel along a straight line. Vectors indicate direction; scalars do not scalar Each of the physical quantities encountered in this book can be categorized as a physical quantity that has either a scalar quantity or a vector quantity. A scalar is a quantity that has magnitude but no direction magnitude but no direction. Examples of scalar quantities are speed, volume, and the number of pages in this textbook. A vector is a physical quantity that vector has both direction and magnitude. As we look back to the chapter “Motion in One Dimension,” we can see that a physical quantity that has both magnitude and direction displacement is an example of a vector quantity. An airline pilot planning a trip must know exactly how far and which way to fly. Velocity is also a vector quantity. If we wish to describe the velocity of a bird, we must specify both its speed (say, 3.5 m/s) and the direction in which the bird is flying (say, north- east). Another example of a vector quantity is acceleration. Vectors are represented by boldface symbols In physics, quantities are often represented by symbols, such as t for time. To help you keep track of which symbols represent vector quan- tities and which are used to indicate scalar quantities, this book will use boldface type to indicate vector quantities. Scalar quantities will be in italics. For example, the speed of a bird is written as v = 3.5 m/s. But a velocity, which includes a direction, is written as v = 3.5 m/s to the northeast. When writing a vector on your paper, you can distin- guish it from a scalar by drawing an arrow above the abbreviation for a quantity, such as v៮៬ = 3.5 m/s to the northeast. One way to keep track of vectors and their directions is to use diagrams. In diagrams, vectors are shown as arrows that point in the direction of the vector. The length of a vector arrow in a diagram is proportional to the vector’s magnitude. For example, in Figure 1 Figure 1 the arrows represent the velocities of the two soccer players running The lengths of the vector arrows represent the magnitudes of these toward the soccer ball. two soccer players’ velocities. 82 Chapter 3 A resultant vector represents the sum of two or more vectors Did you know? When adding vectors, you must make certain that they have the same units The word vector is also used by air- and describe similar quantities. For example, it would be meaningless to add a line pilots and navigators. In this velocity vector to a displacement vector because they describe different physi- context, a vector is the particular cal quantities. Similarly, it would be meaningless, as well as incorrect, to add path followed or to be followed, two displacement vectors that are not expressed in the same units. For exam- given as a compass heading. ple, you cannot add meters and feet together. Section 1 of the chapter “Motion in One Dimension” covered vector addi- tion and subtraction in one dimension. Think back to the example of the gecko that ran up a tree from a 20 cm marker to an 80 cm marker. Then the gecko reversed direction and ran back to the 50 cm marker. Because the two parts of this displacement are opposite, they can be added together to give a total displacement of 30 cm. The answer found by adding two vectors in this resultant way is called the resultant. a vector that represents the sum of two or more vectors Vectors can be added graphically Consider a student walking 1600 m to a friend’s house and then 1600 m to school, as shown in Figure 2. The student’s total displacement during his walk to school is in a direction from his house to the school, as shown by the dotted line. This direct path is the vector sum of the student’s displacement from his house to his friend’s house and his displacement from the friend’s house to school. How can this resultant displacement be found? One way to find the magnitude and direction of the student’s total dis- placement is to draw the situation to scale on paper. Use a reasonable scale, such as 50 m on land equals 1 cm on paper. First draw the vector representing the student’s displacement from his house to his friend’s house, giving the proper direction and scaled magnitude. Then draw the vector representing his walk to the school, starting with the (b) tail at the head of the first vector. Again give its scaled magnitude and the right direction. The magnitude of (c) the resultant vector can then be determined by using a ruler. Measure the length of the vector pointing from (a) the tail of the first vector to the head of the second vec- tor. The length of that vector can then be multiplied by 50 (or whatever scale you have chosen) to get the actu- al magnitude of the student’s total displacement in Figure 2 meters. A student walks from his house to his friend’s house (a), then The direction of the resultant vector may be deter- from his friend’s house to the school (b). The student’s resultant mined by using a protractor to measure the angle displacement (c) can be found by using a ruler and a protractor. between the resultant and the first vector or between the resultant and any chosen reference line. Two-Dimensional Motion and Vectors 83  vwalkway = 1.5 m/s PROPERTIES OF VECTORS (c) Now consider a case in which two or more vectors act at the same point. vcar = 0.80 m/s When this occurs, it is possible to find a resultant vector that has the same net (a) effect as the combination of the individual vectors. Imagine looking down (b ) from the second level of an airport at a toy car moving at 0.80 m/s across a vresultant walkway that moves at 1.5 m/s. How can you determine what the car’s resul- tant velocity will look like from your view point? Car Figure 3 Vectors can be moved parallel to themselves in a diagram The resultant velocity (a) of a toy car moving at a velocity of 0.80 m/s Note that the car’s resultant velocity while moving from one side of the walk- (b) across a moving walkway with a way to the other will be the combination of two independent motions. Thus, velocity of 1.5 m/s (c) can be found the moving car can be thought of as traveling first at 0.80 m/s across the walk- using a ruler and a protractor. way and then at 1.5 m/s down the walkway. In this way, we can draw a given vector anywhere in the diagram as long as the vector is parallel to its previous alignment (so that it still points in the same direction). Thus, you can draw one vector with its tail starting at the tip of the other as long as the size and direction of each vector do not change. This process is illustrated in Figure 3. Although both vectors act on the car at the same point, the horizontal vector has been moved up so that its tail begins at the tip of the vertical vector. The resultant vector can then be drawn from the tail of the first vector to the tip of the last vector. This method is known as the triangle (or polygon) method of addition. Again, the magnitude of the resultant vector can be measured using a ruler, and the angle can be measured with a protractor. In the next section, we will develop a technique for adding vectors that is less time-consuming because it d involves a calculator instead of a ruler and protractor. Vectors can be added in any order When two or more vectors are added, the sum is independent of the order of the addition. This idea is demonstrated by a runner practicing for a marathon (a) along city streets, as represented in Figure 4. The runner executes the same four displacements in each case, but the order is different. Regardless of which d path the runner takes, the runner will have the same total displacement, expressed as d. Similarly, the vector sum of two or more vectors is the same regardless of the order in which the vectors are added, provided that the mag- nitude and direction of each vector remain the same. (b) To subtract a vector, add its opposite Figure 4 Vector subtraction makes use of the definition of the negative of a vector. The A marathon runner’s displacement, negative of a vector is defined as a vector with the same magnitude as the origi- d, will be the same regardless nal vector but opposite in direction. For instance, the negative of the velocity of whether the runner takes path (a) or (b) because the vectors can of a car traveling 30 m/s to the west is −30 m/s to the west, or +30 m/s to the be added in any order. east. Thus, adding a vector to its negative vector gives zero. 84 Chapter 3  When subtracting vectors in two dimensions, first draw the negative of the vector to be subtracted. Then add that negative vector to the other vector by Developed and maintained by the National Science Teachers Association using the triangle method of addition. For a variety of links related to this chapter, go to www.scilinks.org Multiplying or dividing vectors by scalars results in vectors Topic: Vectors There are mathematical operations in which vectors can multiply other vec- SciLinks Code: HF61597 tors, but they are not needed in this book. This book does, however, make use of vectors multiplied by scalars, with a vector as the result. For example, if a cab driver obeys a customer who tells him to go twice as fast, that cab’s origi- nal velocity vector, vcab, is multiplied by the scalar number 2. The result, writ- ten 2vcab, is a vector with a magnitude twice that of the original vector and pointing in the same direction. On the other hand, if another cab driver is told to go twice as fast in the opposite direction, this is the same as multiplying by the scalar number −2. The result is a vector with a magnitude two times the initial velocity but pointing in the opposite direction, written as −2vcab. SECTION REVIEW 1. Which of the following quantities are scalars, and which are vectors? a. the acceleration of a plane as it takes off b. the number of passengers on the plane c. the duration of the flight d. the displacement of the flight e. the amount of fuel required for the flight 2. A roller coaster moves 85 m horizontally, then travels 45 m at an angle of 30.0° above the horizontal. What is its displacement from its starting point? Use graphical techniques. 3. A novice pilot sets a plane’s controls, thinking the plane will fly at 2.50 × 102 km/h to the north. If the wind blows at 75 km/h toward the southeast, what is the plane’s resultant velocity? Use graphical techniques. 4. While flying over the Grand Canyon, the pilot slows the plane down to one-half the velocity in item 3. If the wind’s velocity is still 75 km/h toward the southeast, what will the plane’s new resultant velocity be? Use graphical techniques. 5. Critical Thinking The water used in many fountains is recycled. For instance, a single water particle in a fountain travels through an 85 m system and then returns to the same point. What is the displacement of this water particle during one cycle? Two-Dimensional Motion and Vectors 85  RESOLVING VECTORS INTO COMPONENTS In the pyramid example, the horizontal and vertical parts that add up to give components of a vector the tourist’s actual displacement are called components. The x component is the projections of a vector along parallel to the x-axis. The y component is parallel to the y-axis. Any vector can the axes of a coordinate system be completely described by a set of perpendicular components. In this textbook, components of vectors are shown as outlined, open arrows. Components have arrowheads to indicate their direction. Compo- nents are scalars (numbers), but they are signed numbers, and the direction is important to determine their sign in a particular coordinate system. vplane You can often describe an object’s motion more conveniently by breaking a single vector into two components, or resolving the vector. Resolving a vector allows you to analyze the motion in each direction. This point may be illustrated by examining a scene on the set of a new action 20˚ movie. For this scene, a biplane travels at 95 km/h at an angle of 20° relative to v t ruck the ground. Attempting to film the plane from below, a camera team travels in a Figure 10 truck that is directly beneath the plane at all times, as shown in Figure 10. A truck carrying a film crew must To find the velocity that the truck must maintain to stay beneath the plane, be driven at the correct velocity to enable the crew to film the under- we must know the horizontal component of the plane’s velocity. Once more, side of a biplane. The plane flies at the key to solving the problem is to recognize that a right triangle can be 95 km/h at an angle of 20° relative to the ground. drawn using the plane’s velocity and its x and y components. The situation can then be analyzed using trigonometry. The sine and cosine functions are defined in terms of the lengths of the sides of such right triangles. The sine of an angle is the ratio of the leg oppo- site that angle to the hypotenuse. DEFINITION OF THE SINE FUNCTION FOR RIGHT TRIANGLES opp opposite leg sin q = ⎯⎯ sine of an angle = ⎯⎯ hyp hypotenuse In Figure 11, the leg opposite the 20° angle represents the y component, vy , vplane = 95 km/h which describes the vertical speed of the airplane. The hypotenuse, vplane, is vy 20° the resultant vector that describes the airplane’s total velocity. vx The cosine of an angle is the ratio between the leg adjacent to that angle and the hypotenuse. Figure 11 To stay beneath the biplane, the truck must be driven with a velocity DEFINITION OF THE COSINE FUNCTION FOR RIGHT TRIANGLES equal to the x component (vx) of the biplane’s velocity. adj adjacent leg cos q = ⎯⎯ cosine of an angle = ⎯⎯ hyp hypotenuse In Figure 11, the adjacent leg represents the x component, vx , which describes the airplane’s horizontal speed. This x component equals the speed that the truck must maintain to stay beneath the plane. Thus, the truck must maintain a speed of vx = (cos 20°)(95 km/h) = 90 km/h. 90 Chapter 3  PRACTICE B Resolving Vectors 1. How fast must a truck travel to stay beneath an airplane that is moving 105 km/h at an angle of 25° to the ground? 2. What is the magnitude of the vertical component of the velocity of the plane in item 1? 3. A truck drives up a hill with a 15° incline. If the truck has a constant speed of 22 m/s, what are the horizontal and vertical components of the truck’s velocity? 4. What are the horizontal and vertical components of a cat’s displacement when the cat has climbed 5 m directly up a tree? ADDING VECTORS THAT ARE NOT PERPENDICULAR Until this point, the vector-addition problems concerned vectors that are per- pendicular to one another. However, many objects move in one direction and then turn at an angle before continuing their motion. Suppose that a plane initially travels 5 km at an angle of 35° to the ground, then climbs at only 10° relative to the ground for 22 km. How can you deter- mine the magnitude and direction for the vector denoting the total displace- ment of the plane? Because the original displacement vectors do not form a right triangle, you can not apply the tangent function or the Pythagorean theorem when adding the original two vectors. Determining the magnitude and the direction of the resultant can be achieved by resolving each of the plane’s displacement vectors into its x and y components. Then the components along each axis can be added together. As shown in Figure 12, these sums will be the two perpendicular components of the resultant, d. The resultant’s magnitude can then be found by using the PHYSICS Pythagorean theorem, and its direction can be found by using the inverse tan- gent function. Module 2 “Vectors” provides an interactive lesson with guided d2 problem-solving practice to ∆y2 Figure 12 teach you how to add different vectors, especially those that d1 d Add the components of the original dis- are not at right angles. ∆y1 placement vectors to find two components that form a right triangle with the resul- ∆x1 ∆x2 tant vector. 92 Chapter 3 Projectile Motion SECTION 3 SECTION OBJECTIVES Recognize examples of pro- TWO-DIMENSIONAL MOTION jectile motion. In the last section, quantities such as displacement and velocity were shown to Describe the path of a pro- jectile as a parabola. be vectors that can be resolved into components. In this section, these compo- Resolve vectors into their nents will be used to understand and predict the motion of objects thrown components and apply the into the air. kinematic equations to solve problems involving projectile motion. Use of components avoids vector multiplication How can you know the displacement, velocity, and acceleration of a ball at any point in time during its flight? All of the kinematic equations could be rewrit- ten in terms of vector quantities. However, when an object is propelled into the air in a direction other than straight up or down, the velocity, acceleration, and displacement of the object do not all point in the same direction. This makes the vector forms of the equations difficult to solve. One way to deal with these situations is to avoid using the complicated vec- tor forms of the equations altogether. Instead, apply the technique of resolving vectors into components. Then you can apply the simpler one-dimensional forms of the equations for each component. Finally, you can recombine the components to determine the resultant. Components simplify projectile motion When a long jumper approaches his jump, he runs along a straight line, which can be called the x-axis. When he jumps, as shown in Figure 13, his velocity has both horizontal and vertical components. Movement in this plane can be depicted by using both the x- and y-axes. Figure 13 Note that in Figure 14(b), a jumper’s velocity vector is resolved into its two When the long jumper is in the air, vector components. This way, the jumper’s motion can be analyzed using the his velocity has both a horizontal kinematic equations applied to one direction at a time. and a vertical component. vy Figure 14 vx (a) A long jumper’s velocity while sprinting along the runway can be v represented by a horizontal vector. (b) Once the jumper is airborne, the jumper’s velocity at any instant (a) (b) can be described by the compo- nents of the velocity. Two-Dimensional Motion and Vectors 95  In this section, we will focus on the form of two-dimensional motion called projectile motion projectile motion. Objects that are thrown or launched into the air and are the curved path that an object subject to gravity are called projectiles. Some examples of projectiles are soft- follows when thrown, launched, or otherwise projected near the balls, footballs, and arrows when they are projected through the air. Even a long surface of Earth jumper can be considered a projectile. Path without air resistance Path with air resistance Figure 15 (a) Without air resistance, the (a) soccer ball would travel along a (b) parabola. (b) With air resistance, the soccer ball would travel along a shorter path. Projectiles follow parabolic trajectories The path of a projectile is a curve called a parabola, as shown in Figure 15(a). Many people mistakenly believe that projectiles eventually fall straight down in much the same way that a cartoon character does after running off a cliff. But if an object has an initial horizontal velocity in any given time interval, there will be horizontal motion throughout the flight of the projectile. Note that for the purposes of samples and exercises in this book, the horizontal velocity of the projec- tile will be considered constant. This velocity would not be constant if we accounted for air resistance. With air resistance, a projectile slows down as it col- lides with air particles, as shown in Figure 15(b). Projectile motion is free fall with an initial horizontal velocity To understand the motion a projectile undergoes, first examine Figure 16. The red ball was dropped at the same instant the yellow ball was launched horizon- tally. If air resistance is disregarded, both balls hit the ground at the same time. By examining each ball’s position in relation to the horizontal lines and to one another, we see that the two balls fall at the same rate. This may seem impossible because one is given an initial velocity and the other begins from rest. But if the motion is analyzed one component at a time, it makes sense. First, consider the red ball that falls straight down. It has no motion in the horizontal direction. In the vertical direction, it starts from rest (vy,i = 0 m/s) Figure 16 This is a strobe photograph of two and proceeds in free fall. Thus, the kinematic equations from the chapter table-tennis balls released at the “Motion in One Dimension” can be applied to analyze the vertical motion of same time. Even though the yellow the falling ball, as shown on the next page. Note that on Earth’s surface the ball is given an initial horizontal veloc- ity and the red ball is simply dropped, acceleration (ay ) will equal −g (−9.81 m/s2) because the only vertical compo- both balls fall at the same rate. nent of acceleration is free-fall acceleration. Note also that ∆y is negative. 96 Chapter 3  VERTICAL MOTION OF A PROJECTILE THAT FALLS FROM REST Did you know? vy,f = ay ∆t The greatest distance a regulation- size baseball has ever been thrown vy,f 2 = 2ay ∆y is 1 35.9 m, by Glen Gorbous in 1 957. 1 ∆y = 2ay (∆t)2 Now consider the components of motion of the yellow ball that is launched in Figure 16. This ball undergoes the same horizontal displacement during each time interval. This means that the ball’s horizontal velocity remains constant (if air resistance is assumed to be negligible). Thus, when the kinematic equations are used to analyze the horizontal motion of a pro- jectile, the initial horizontal velocity is equal to the horizontal velocity throughout the projectile’s flight. A projectile’s horizontal motion is described by the following equation. HORIZONTAL MOTION OF A PROJECTILE vx = vx,i = constant ∆x = vx ∆t Developed and maintained by the Next consider the initial motion of the launched yellow ball in Figure 16. National Science Teachers Association Despite having an initial horizontal velocity, the launched ball has no initial For a variety of links related to this chapter, go to www.scilinks.org velocity in the vertical direction. Just like the red ball that falls straight down, the launched yellow ball is in free fall. The vertical motion of the launched yel- Topic: Projectile Motion SciLinks Code: HF61223 low ball is described by the same free-fall equations. In any time interval, the launched ball undergoes the same vertical displacement as the ball that falls straight down. For this reason, both balls reach the ground at the same time. To find the velocity of a projectile at any point during its flight, find the vector that has the known components. Specifically, use the Pythagorean the- orem to find the magnitude of the velocity, and use the tangent function to find the direction of the velocity. SAFETY CAUTION roll the first ball off the table. Does vary- Perform this experiment away from walls ing the speed affect whether the two balls and furniture that can be damaged. strike the ground at the same time? Next Projectile Motion roll one of the balls down a slope. Drop MATERIALS LIST Roll a ball off a table. At the instant the the other ball from the base of the slope rolling ball leaves the table, drop a second at the instant the first ball leaves the slope. 2 identical balls ball from the same height above the floor. Which of the balls hits the ground first in slope or ramp Do the two balls hit the floor at the same this situation? time? Try varying the speed at which you Two-Dimensional Motion and Vectors 97  SAMPLE PROBLEM D Projectiles Launched Horizontally PROBLEM The Royal Gorge Bridge in Colorado rises 321 m above the Arkansas River. Suppose you kick a rock horizontally off the bridge. The magnitude of the rock’s horizontal displacement is 45.0 m. Find the speed at which the rock was kicked. SOLUTION 1. DEFINE Given: ∆y = −321 m ∆x = 45.0 m ay = −g = −9.81 m/s2 Unknown: vi = vx = ? y vx x Diagram: The initial velocity vector of the ay rock has only a horizontal compo- nent. Choose the coordinate system –321 m oriented so that the positive y direc- tion points upward and the positive x direction points to the right. 2. PLAN Choose an equation or situation: 45.0 m Because air resistance can be neglected, the rock’s horizontal velocity remains constant. ∆x = vx ∆t Because there is no initial vertical velocity, the following equation applies. 1 ∆y = 2ay(∆t)2 Rearrange the equations to isolate the unknowns: Note that the time interval is the same for the vertical and horizontal dis- placements, so the second equation can be rearranged to solve for ∆t. ∆t = 冪莦莦 2∆y  ay Next rearrange the first equation for vx , and substitute the above value of ∆t into the new equation. 冢冪莦2∆莦y 冣∆x ∆x ay vx =  = The value for vx can be either ∆t positive or negative because of 3. CALCULATE Substitute the values into the equation and solve: the square root. Because the direction was not asked for, use vx = 冪莦 −9.81 m/s2  (45.0 m) = 5.56 m/s (2)(−321 m) the positive root. 4. EVALUATE To check your work, estimate the value of the time interval for ∆x and solve for ∆y. If vx is about 5.5 m/s and ∆x = 45 m, ∆t ≈ 8 s. If you use an approxi- mate value of 10 m/s2 for g, ∆y ≈ −320 m, almost identical to the given value. 98 Chapter 3  PRACTICE D Projectiles Launched Horizontally 1. A baseball rolls off a 0.70 m high desk and strikes the floor 0.25 m away from the base of the desk. How fast was the ball rolling? 2. A cat chases a mouse across a 1.0 m high table. The mouse steps out of the way, and the cat slides off the table and strikes the floor 2.2 m from the edge of the table. When the cat slid off the table, what was its speed? 3. A pelican flying along a horizontal path drops a fish from a height of 5.4 m. The fish travels 8.0 m horizontally before it hits the water below. What is the pelican’s speed? 4. If the pelican in item 3 was traveling at the same speed but was only 2.7 m above the water, how far would the fish travel horizontally before hitting the water below? Use components to analyze objects launched at an angle Let us examine a case in which a projectile is launched at an angle to the hori- vi zontal, as shown in Figure 17. The projectile has an initial vertical component vy,i of velocity as well as a horizontal component of velocity. θ Suppose the initial velocity vector makes an angle q with the horizontal. vx,i Again, to analyze the motion of such a projectile, you must resolve the initial Figure 17 velocity vector into its components. The sine and cosine functions can be used An object is projected with an initial to find the horizontal and vertical components of the initial velocity. velocity, vi, at an angle of q. Resolve the initial velocity into its x and y vx,i = vi cos q and vy,i = vi sin q components. Then, the kinematic equations can be applied to We can substitute these values for vx,i and vy,i into the kinematic equations describe the motion of the projec- to obtain a set of equations that can be used to analyze the motion of a pro- tile throughout its flight. jectile launched at an angle. PROJECTILES LAUNCHED AT AN ANGLE vx = vx,i = vi cos q = constant ∆x = (vi cos q )∆t vy,f = vi sin q + ay ∆t vy,f = vi2 (sin q )2 + 2ay ∆y 2 PHYSICS 1 ∆y = (vi sin q )∆t + a (∆t)2 2 y Module 3 “Two-Dimensional Motion” provides an interactive lesson As we have seen, the velocity of a projectile launched at an angle to the with guided problem-solving ground has both horizontal and vertical components. The vertical motion is practice to teach you about analyzing projectile motion. similar to that of an object that is thrown straight up with an initial velocity. Two-Dimensional Motion and Vectors 99  SAMPLE PROBLEM E STRATEGY Projectiles Launched at an Angle PROBLEM A zookeeper finds an escaped monkey hanging from a light pole. Aiming her tranquilizer gun at the monkey, she kneels 10.0 m from the light pole, which is 5.00 m high. The tip of her gun is 1.00 m above the ground. At the same moment that the monkey drops a banana, the zookeeper shoots. If the dart travels at 50.0 m/s, will the dart hit the monkey, the banana, or neither one? SOLUTION y 1. Select a coordinate system. 10.0 m The positive y-axis points up, and the positive x-axis points along the ground vi 4.00 m toward the pole. Because the dart leaves θ the gun at a height of 1.00 m, the vertical 1.00 m distance is 4.00 m. x 2. Use the inverse tangent function to find the angle that the initial velocity makes with the x-axis. 冢 冣 冢 冣 ∆y 4.00 m q = tan−1  = tan−1  = 21.8° ∆x 10.0 m 3. Choose a kinematic equation to solve for time. Rearrange the equation for motion along the x-axis to isolate the unknown, ∆t, which is the time the dart takes to travel the horizontal distance. ∆x = (vi cos q )∆t ∆x 10.0 m ∆t =  =  = 0.215 s vi cosq (50.0 m/s)(cos 21.8°) 4. Find out how far each object will fall during this time. Use the free-fall kinematic equation in both cases. For the banana, vi = 0. Thus: 1 1 ∆yb = 2ay(∆t)2 = 2(−9.81 m/s2)(0.215 s)2 = −0.227 m The dart has an initial vertical component of velocity equal to vi sinq, so: 1 ∆yd = (vi sin q)∆t + 2ay(∆t)2 1 ∆yd = (50.0 m/s)(sin 21.8°)(0.215 s) + 2(−9.81 m/s2)(0.215 s)2 ∆yd = 3.99 m − 0.227 m = 3.76 m 5. Analyze the results. Find the final height of both the banana and the dart. ybanana, f = yb,i + ∆yb = 5.00 m + (−0.227 m) = 4.77 m above the ground ydart, f = yd,i + ∆yd = 1.00 m + 3.76 m = 4.76 m above the ground The dart hits the banana. The slight difference is due to rounding. 100 Chapter 3  SECTION 4 Relative Motion SECTION OBJECTIVES Describe situations in terms of frame of reference. FRAMES OF REFERENCE Solve problems involving If you are moving at 80 km/h north and a car passes you going 90 km/h, to relative velocity. you the faster car seems to be moving north at 10 km/h. Someone standing on the side of the road would measure the velocity of the faster car as 90 km/h toward the north. This simple example demonstrates that velocity measure- ments depend on the frame of reference of the observer. Velocity measurements differ in different frames of reference Observers using different frames of reference may measure different displace- ments or velocities for an object in motion. That is, two observers moving with respect to each other would generally not agree on some features of the motion. Consider a stunt dummy that is dropped from an airplane flying horizontal- ly over Earth with a constant velocity. As shown in Figure 19(a), a passenger on the airplane would describe the motion of the dummy as a straight line toward Earth. An observer on the ground would view the trajectory of the dummy as that of a projectile, as shown in Figure 19(b). Relative to the ground, the dummy would have a vertical component of velocity (resulting from free-fall acceleration and equal to the velocity measured by the observer in the airplane) and a horizontal component of velocity given to it by the airplane’s motion. If the airplane continued to move horizontally with the same velocity, the dummy would enter the swimming pool directly beneath the airplane (assuming negli- gible air resistance). (a) (b) Figure 19 When viewed from the plane (a), the stunt dummy (represented by the maroon dot) falls straight down. When viewed from a sta- tionary position on the ground (b), the stunt dummy follows a parabolic projectile path. 102 Chapter 3 PRACTICE F Relative Velocity 1. A passenger at the rear of a train traveling at 15 m/s relative to Earth throws a baseball with a speed of 15 m/s in the direction opposite the motion of the train. What is the velocity of the baseball relative to Earth as it leaves the thrower’s hand? 2. A spy runs from the front to the back of an aircraft carrier at a velocity of 3.5 m/s. If the aircraft carrier is moving forward at 18.0 m/s, how fast does the spy appear to be running when viewed by an observer on a nearby stationary submarine? 3. A ferry is crossing a river. If the ferry is headed due north with a speed of 2.5 m/s relative to the water and the river’s velocity is 3.0 m/s to the east, what will the boat’s velocity relative to Earth be? (Hint: Remember to include the direction in describing the velocity.) 4. A pet-store supply truck moves at 25.0 m/s north along a highway. Inside, a dog moves at 1.75 m/s at an angle of 35.0° east of north. What is the velocity of the dog relative to the road? SECTION REVIEW 1. A woman on a 10-speed bicycle travels at 9 m/s relative to the ground as she passes a little boy on a tricycle going in the opposite direction. If the boy is traveling at 1 m/s relative to the ground, how fast does the boy appear to be moving relative to the woman? 2. A girl at an airport rolls a ball north on a moving walkway that moves east. If the ball’s speed with respect to the walkway is 0.15 m/s and the walkway moves at a speed of 1.50 m/s, what is the velocity of the ball relative to the ground? 3. Critical Thinking Describe the motion of the following objects if they are observed from the stated frames of reference: a. a person standing on a platform viewed from a train traveling north b. a train traveling north viewed by a person standing on a platform c. a ball dropped by a boy walking at a speed of 1 m/s viewed by the boy d. a ball dropped by a boy walking 1 m/s as seen by a nearby viewer who is stationary Two-Dimensional Motion and Vectors 105 Use the information below to answer questions 9–12. 14. A roller coaster travels 41.1 m at an angle of 40.0° A girl riding a bicycle at 2.0 m/s throws a tennis ball above the horizontal. How far does it move hori- horizontally forward at a speed of 1.0 m/s from a zontally and vertically? height of 1.5 m. At the same moment, a boy standing 15. A ball is thrown straight upward and returns to the on the sidewalk drops a tennis ball straight down from thrower’s hand after 3.00 s in the air. A second ball a height of 1.5 m. is thrown at an angle of 30.0° with the horizontal. 9. What is the initial speed of the girl’s ball relative to At what speed must the second ball be thrown to the boy? reach the same height as the one thrown vertically? A. 1.0 m/s B. 1.5 m/s C. 2.0 m/s EXTENDED RESPONSE D. 3.0 m/s 16. A human cannonball is shot out of a cannon at 10. If air resistance is disregarded, which ball will hit 45.0° to the horizontal with an initial speed of 25.0 the ground first? m/s. A net is positioned at a horizontal distance of F. the boy’s ball 50.0 m from the cannon. At what height above the G. the girl’s ball cannon should the net be placed in order to catch H. neither the human cannonball? Show your work. J. The answer cannot be determined from the Read the following passage to answer question 17. given information. Three airline executives are discussing ideas for devel- 11. If air resistance is disregarded, which ball will have oping flights that are more energy efficient. a greater speed (relative to the ground) when it hits the ground? Executive A: Because the Earth rotates from west to east, we could operate “static flights”—a helicopter or A. the boy’s ball airship could begin by rising straight up from New B. the girl’s ball York City and then descend straight down four hours C. neither later when San Francisco arrives below. D. The answer cannot be determined from the given information. Executive B: This approach could work for one-way flights, but the return trip would take 20 hours. 12. What is the speed of the girl’s ball when it hits the ground? Executive C: That approach will never work. Think F. 1.0 m/s about it. When you throw a ball straight up in the air, it G. 3.0 m/s comes straight back down to the same point. H. 6.2 m/s Executive A: The ball returns to the same point because J. 8.4 m/s Earth’s motion is not significant during such a short time. SHORT RESPONSE 17. In a paragraph, state which of the executives is correct, and explain why. 13. If one of the components of one vector along the direction of another vector is zero, what can you conclude about these two vectors? If you get stuck answering a question, move on. You can return to the question later if you have time. Two-Dimensional Motion and Vectors 115