Chapter 3 Kinematics in Two Dimensions. Summary of Key Concepts. graphically:

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1 July :39 AM Chapter 3 Kinematics in Two Dimensions Summary of Key Concepts vectors (1.5 to 1.8) scalar and vector quantities (1.5) representing vectors numerically and graphically (1.5) converting from length + direction to components and vice versa (1.7) adding and subtracting vectors, using components, and graphically (1.6, 1.8) kinematical quantities in two dimensions (3.1) position and displacement vectors average velocity and average acceleration vectors instantaneous velocity and instantaneous acceleration vectors instantaneous speed is the length of the instantaneous velocity vector; average speed is total distance divided by total time 2-d kinematics equations for constant acceleration (3.2) projectile motion (3.3) relative motion (3.4) Vectors What is the distinction between scalar and vector quantities? How are vectors represented? graphically: numerically using length and direction: Ch3 Page 1

2 numerically using components If the vector's tail is not at the origin, you can determine its components as in the left diagram, or slide the vector parallel to itself until its tail is at the origin, as in the right diagram: Physical quantities that can be modelled using vectors: position, displacement, velocity, acceleration, force, momentum, angular momentum, torque, etc. Physical quantities that can be modelled using scalars: distance, speed, acceleration, energy, temperature, time, etc. How can we determine whether a physical quantity is represented mathematically by a vector or a scalar (or some other more Ch3 Page 2

3 sophisticated mathematical gadget, such as a tensor)? As always in science, experiments and observations decide: You do the experiments, record data from careful observations, analyze the data, and draw conclusions. Converting between vector representations: Example: The vector B has length 24 m and is inclined at an angle of 54 degrees north of east. Determine the components of the vector with respect to the standard axes. Solution: Draw a diagram! Ch3 Page 3

4 Example: The vector C has components 11 cm in the x-direction and 4 cm in the y-direction. Determine the magnitude and direction of C. Solution: Draw a diagram! The length C of the vector can be determined using Pythagoras's theorem: The direction of the vector can be determined using trigonometry: Ch3 Page 4

5 Thus, the length of the vector C is 11.7 cm and it is inclined 20 degrees below the positive x-axis. Adding vectors graphically Adding vectors using components Adding vectors using components yields the same result as adding vectors graphically, as you can see from the following diagram: Multiplying a vector by a scalar (graphically) Ch3 Page 5

6 Multiplying a vector by a scalar (in terms of components) Sketch the vectors in the calculations above to make sure this makes sense to you; i.e., to see that the numerical and graphical approaches lead to the same result. Subtracting vectors graphically Method 1: Ch3 Page 6

7 Method 2 (which leads to the same result): Subtracting vectors using components Sketch the vectors in the calculation above to make sure this makes sense to you; i.e., to see that the numerical and graphical approaches lead to the same result. Operations with vectors Example: Suppose that A = (-2, 3) and B = (1, 4). Determine 2A 3B Ch3 Page 7

8 and 5B 2.5A. Solution: Kinematical quantities in two dimensions Position vectors and displacement vectors Average velocity vectors Ch3 Page 8

9 Thus, the average velocity vector is in the same direction as the displacement vector, although its length may be different (it depends on the time interval), and it has different units. The average speed for two-dimensional motion is the distance travelled divided by the time interval, just as in one-dimensional motion. Example: Alice walks at a constant speed of 2 m/s counter-clockwise around a circular track that has radius 100 m. During the time interval in which she moves from the eastern-most point of the track to the northern-most point of the track, determine Alice's (a) displacement, (b) average velocity, and (c.) average speed. Solution: (a) Ch3 Page 9

10 (c.) We are told that Alice's speed is constant and has the value 2 m/s. This means that her average speed is 2 m/s. Note that Alice's average velocity is a vector. Also note that the magnitude of Alice's average velocity (calculate it!) is not the same as Alice's average speed. Instantaneous velocity vectors Ch3 Page 10

11 A good way to think about instantaneous velocity is to think about driving a car, as in the figure above. The direction of your headlights (or better, think about the single headlight on a motorcycle) is the direction of your instantaneous velocity. Thus, if you are driving on a curving road, the direction of your instantaneous velocity is changing continuously, even if your speed is constant. This is an important point, worth reflecting upon: Velocity can change even if speed is constant. Thus, the velocity changes if the direction of motion changes, if the speed changes, or both. Thus, for a motion in two (or three) dimensions, a non-zero acceleration might change the direction of motion, might change the speed, or both. There are some figures a little below that describe these changes more specifically. The instantaneous speed is the magnitude of the instantaneous velocity vector; the instantaneous speed is the reading on the car's speedometer. For example, for motion on a straight road at a constant speed, the acceleration is zero and the velocity is constant: Ch3 Page 11

12 For motion on a curved road, the acceleration is NOT zero and the velocity is NOT constant: Acceleration in two-dimensional motion In one-dimensional motion, if the acceleration is not zero, the velocity changes. Because velocity is a number for one-dimensional motion, a change in velocity simply means a change in a number. The situation is essentially different in two-dimensional (and three- Ch3 Page 12

13 dimensional) motion. Because velocity is a vector, it is possible for velocity to change without its magnitude changing. That is, even if the speed is constant, the velocity might change because the direction of the velocity changes. Average acceleration vectors Remember that For example, in the following figure a car is driving along a curved road. You can see that its velocity changes, because its direction changes. The direction of the change in velocity is shown in the figure on the right. The average acceleration between times 1 and 2 is in the same direction as the change in velocity. For a two-dimensional motion, we can separate the instantaneous acceleration into two components, one that is perpendicular to the instantaneous velocity, and the other that is parallel to the instantaneous velocity. The former is called the normal component of acceleration, and the latter is called the tangential component of acceleration. The normal component of acceleration changes the direction of the velocity but not its magnitude. The tangential component of acceleration changes the magnitude of the velocity but not its direction. The tangential acceleration increases the speed if it is in the same direction as the velocity, and decreases the speed if it is in the direction opposite to the velocity. Ch3 Page 13

14 In the following figure the speed is increasing (the tangential component of acceleration is in the same direction as the velocity). In the following figure the speed is decreasing (the tangential component of acceleration is in the direction opposite to the velocity). For motion at a constant speed in a curved path, the tangential component of acceleration is zero. The normal component of acceleration is non-zero, and it changes the direction of the velocity, but not its magnitude. One example of this is motion in a circle at a constant speed. Other examples are motion along a curved path at a constant speed. Ch3 Page 14

15 In the figure above, all the velocity vectors are meant to have equal magnitude (speed does not change), and all the acceleration vectors are meant to have equal magnitude and each is perpendicular to the corresponding velocity vector. Kinematics equations in two dimensions (for CONSTANT acceleration) If an object has non-zero acceleration, then it experiences a change in velocity. On the other hand, if the acceleration of an object is zero, then there is no change in velocity, which means the velocity is constant if acceleration is zero. This viewpoint was implicit in our work in Chapter 2, and it is embodied in the kinematics equations for constant acceleration developed in Chapter 2. The key fact of Chapter 3, which is supported by innumerable observations and precision experiments, is this: For an object moving in two (or three) dimensions, we can separate the kinematical vector quantities into components. Then the x- component of velocity is only affected by the x-component of acceleration, and the y-component of velocity is affected only by the y-component of acceleration. (A similar statement applies to the z- component, if the motion is in three dimensions.) Thus, we can analyze a two-dimensional (or three-dimensional) motion by separating it into components. This strategy fits in with the general strategy of reductionism, which is used frequently in mathematics and science. There are many other Ch3 Page 15

16 examples of reductionism, and you'll discover many if you look for them. Here are the kinematics equations for two-dimensional motion, as written in the textbook (see page 59 in Section 3.2): Projectile motion Basic simplifying assumption: no air resistance. What is a projectile? What is the path of a projectile? Road Runner and Coyote (0:16) Road Runner and Coyote (2:26) Ch3 Page 16

17 Road Runner and Coyote (0:58) If there is no air resistance, the horizontal component of velocity of a projectile remains constant. The acceleration due to gravity acts on the vertical component of velocity, but not on the horizontal component of velocity. Can you see these facts in the following figure? The following figure (and the next two video clips) illustrate the fact that gravity acts on a projectile in the same way whether it has a horizontal component of velocity or not. Once again, the main point is that the horizontal and vertical components of projectile motion are independent. Ch3 Page 17

18 Projectile motion, Harvard University (1:34): Projectile motion, North Carolina School of Science and Mathematics (2:33): Projectile motion simulator, PhET, University of Colorado: The following three problems illustrate numerically the points made earlier qualitatively: The two components of projectile motion are independent. Note the solutions of the next two problems, and then note how the results of the third problem are related to the first two. The conclusion is that the motion in the third problem is a combination of the motions in the first two Ch3 Page 18

19 problems, so the results in the third problem are "the same" as the results in the first two problems (componentwise, anyway). Example: Consider a ball tossed vertically upwards from the ground with an initial speed of 15 m/s. (a) Determine the ball's maximum height above the ground. (b) Determine the time at which the ball reaches its maximum height. (c.) Determine the time at which the ball hits the ground. (d) Determine the ball's impact speed. Solution: Let's choose upwards to be positive, and we'll choose y = 0 at ground level. You might think about parts (a) and (b) together, as it's not clear which one we should start with. (With experience, I think you'll find that it's easier to begin with part (b).) Think about the condition that describes the ball when it reaches its peak height. The task is to convert this condition, which might initially be stated in words, into symbols that we can then use to help us solve kinematics equations. The condition for peak height is that the vertical component of velocity is zero; in our case, this translates to v = 0. We can use this condition to solve part (b): Now that we know the time at which the ball reaches its maximum height, we can determine the maximum height: Ch3 Page 19

20 Now that we know the time at which the ball reaches its maximum height, we can determine the maximum height: (c.) What is the condition when the ball hits the ground again? The condition is that y 3 = 0. (Note that the condition is NOT v 3 = 0, and understand why this is so.) The first solution (t = 0) reminds us that the ball is thrown upwards then (this represents position 1 in the diagram). The second solution is the one we are after, and tells us that the ball hits the ground again (position 3 in the diagram) after 3.06 s. (d) The impact speed is not zero! Sure, the ball eventually stops, but "impact speed" means the speed just an instant before contact with the ground. Ch3 Page 20

21 Thus, the impact speed is 15 m/s. Notice the symmetry in the values of the initial velocity and the final velocity. Will this always be true? Example: Consider a ball moving horizontally in a straight line on a frictionless table-top at a constant speed of 9 m/s. (a) Determine how far the ball travels after 1 s, 2 s, and 3 s. (b) Determine the ball's speed after 1 s, 2 s, and 3 s. Solution: The speed and direction of the ball are constant, which means that the acceleration is zero. (a) Thus the ball travels 9 m every second. It's displacements after 1 s, 2 s, and 3 s are, respectively, 9 m, 18 m, and 27 m. (b) The ball's speed is constant, so it is 9 m/s at each of the listed times. The motion in the next problem combines the motions in the previous two problems. Example: Consider a projectile with an initial velocity that has a y-component of 15 m/s upwards and an x-component of 9 m/s. (a) Determine the ball's maximum height above the ground. (b) Determine the time at which the ball reaches its maximum Ch3 Page 21

22 height. (c.) Determine the time at which the ball hits the ground. (d) Determine the ball's impact speed. (e.) Determine the horizontal distance travelled when the ball hits the ground. Solution: Once again, I'll take upwards to be positive. Consider parts (a) and (b). What is the condition that characterizes position (2) of the motion, which is where the projectile reaches its peak height? The condition is that the vertical component of velocity is zero. But then the solution is exactly the same as the solution to the earlier problem; the ball reaches its maximum height after 1.53 s and its maximum height above the ground is 11.5 m. (c.) Similarly, the ball reaches the ground again after 3.06 s. (d) When the ball hits the ground, the vertical component of its velocity is 15 m/s, just as in the earlier problem. The horizontal component of velocity is constant at 9 m/s. Therefore the impact speed can be calculated as follows: The impact speed is 17.5 m/s. Ch3 Page 22

23 (e.) The horizontal distance travelled is Remember, there is no acceleration in the horizontal direction! (We typically assume that there is no air resistance.) Compare the previous three problems to absorb the main point here: The components of the motion are independent. Example: A soccer ball is kicked from the ground with an initial speed of 25 m/s at an angle of 40 degrees above the ground. (a) Determine the time needed for the ball to reach its peak height. (b) Determine the time needed for the ball to return to the ground. (c) Determine the ball's impact speed and angle of impact. (d) Determine how far from the kicking point the ball lands. (e) Determine the ball's maximum height above the ground. Solution: As usual, we pretend that air resistance does not exist. (For all you advanced people, how would the results change if we assumed a small amount of air resistance?) Notice that in the diagram below we have chosen a coordinate system (which means we have chosen "upward" to be positive and "to the right" to be positive), and we have also labelled key positions of the ball's path as points 0, 1, and 2. Ch3 Page 23

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26 The following calculation verifies that the initial and final speeds are the same: Let's try another basic projectile motion problem: A ball is kicked with an initial speed of 20 m/s at an initial angle of 25 degrees above the horizontal. (a) When does the ball hit the ground? Ch3 Page 26

27 (b) What if instead the ball is caught when it is 1 m above the ground; when is it caught? (c.) When does the ball reach its peak height? (d) Determine the components of the ball's velocity 1 s after it is kicked. (e.) Determine the components of the ball's velocity when it is caught 1 m above the ground (when the ball is on the way down). (f) Determine the ball's peak height. (g) Determine the ball's horizontal displacement when it hits the ground. (h) Determine the ball's horizontal displacement when it is caught 1 m above the ground. Solution: Ch3 Page 27

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30 Page 80, Problem 14: A tennis ball is struck and departs from the racket horizontally with a speed of 28.0 m/s. The ball hits the court at a horizontal distance of 19.6 m from the racket. How far above the court is the tennis ball when it leaves the racket? Solution: If we knew the time that the ball was in the air, we could calculate the Ch3 Page 30

31 initial height of the ball using the y-component equations, because we know that the initial y-component of velocity is zero. But we can calculate the time of flight because we know the x-component of the displacement and we know that the x-component of velocity is constant (remember, the x-component of acceleration is zero). This figure strikes me as a little high, but I suppose it is reasonable for an overhead shot. (A serve is certainly angled down, but perhaps an overhead shot in the middle of a rally might be hit so that the ball's initial velocity is horizontal.) Page 81, Problem 30: A quarterback claims that he can throw the football a horizontal distance of 183 m (200 yards). Furthermore, he claims that he can do this by launching the ball at the relatively low angle of 30 degrees above the horizontal. To evaluate this claim, determine the speed with which this quarterback must throw the ball. Assume that the ball is launched and caught at the same vertical level and that air resistance can be ignored. For comparison a baseball pitcher who can accurately throw a fastball at 45 m/s (100 mph) would be considered exceptional. Ch3 Page 31

32 Solution: The claim is completely unrealistic; even the best professional quarterbacks would have extreme difficulty throwing 100 m, let alone nearly 200 m. But the point of the problem is to calculate the initial speed, so let's do it. Strategy: First use an x-component equation to relate the initial speed to the time of flight. Then use a y-component equation to relate the total y-component of displacement (which is zero) to the initial speed and the time of flight. This leads to two equations in two unknowns; then we can eliminate time to determine the initial speed. Ch3 Page 32

33 So the initial speed is not unrealistic, and yet the range of the football is unrealistic. Why? Perhaps looking up baseball records will be helpful. The longest baseball throw was by a Canadian fellow by the name of Glen Gorbous, and had a range of over 135 m. According to our calculations, a hard-throwing pitcher (let's say an "exceptional" one) should be able to throw a baseball 183 m. How can we reconcile these facts? And further, how can we reconcile the calculations with the fact that a strong-armed professional football quarterback can throw a football perhaps 60 m or so? Page 81, Problem 38: A marble is thrown horizontally with a speed of 15 m/s from the top of a building. When it strikes the ground, the marble has a velocity that makes an angle of 65 degrees with the horizontal. From what height above the ground was the marble thrown? Solution: What does it mean that the impact angle is 65 degrees? This calls for a diagram: Ah, OK; the impact angle is the angle of the final velocity vector relative to the ground. From the little triangle above, this means that Ch3 Page 33

34 So we have just calculated the y-component of the final velocity, and we are asked about the y-component of the initial position, so we should turn to y-component equations to make further progress. One way to solve the problem is as follows: I like using this method because we were not given any time information, and we were not asked for time. However, the final result is rather high (about 17 storeys high), and so I double-checked the solution by determining how long the marble was in the air, and then used the other displacement equation (the one that includes time) to calculate the initial height. You might like to do the same. Page 82, Problem 50: In the annual battle of the dorms, students gather on the roofs of Jackson and Walton dorms to launch water balloons at each other with slingshots. The horizontal distance between the buildings is 35.0 m, and the heights of the Jackson and Walton buildings are, respectively, 15.0 m and 22.0 m. The first balloon launched by the Jackson team hits Walton dorm 2.0 s after launch, striking it halfway between the ground and the roof. Ignore air resistance. (a) Find the direction of the balloon's initial velocity. Give your result as an angle (in degrees) measured above the horizontal. (b) A second balloon launched at the same angle hits the edge of Walton's roof. Find the initial speed of this second balloon. Ch3 Page 34

35 Solution: OK, the book is definitely American, as we would use the term "res" instead of "dorm." In either country I believe the dons would take exception to residents being on the roof for shenanigans, but it leads to a worthy problem. Discussion: In the previous problem we learned that the impact angle is related to the relative sizes of the components of the final velocity vector. Here we are asked to calculate the projection angle, and so perhaps we can get a handle on the angle by working with the components of the initial velocity. It may be hard to see how this is going to work out, but if you can't think of anything better to do, it's good to try something. Just write down expressions for the components of the initial velocity; then by examining them you might get an idea about how to proceed. OK, so let's start with the x-component of the initial velocity. We can use the equation for the x-component of displacement: OK, it's still not clear where this is going, but we have one equation for two unknowns, the initial speed and the angle. Let's Ch3 Page 35

36 write down a similar equation for the y-component. If we are in luck, we may end up with two equations in two unknowns that we can solve for the angle. OK, so we have two equations in two unknowns, and we wish to eliminate the initial speed and solve for the angle. An easy way to do this is to divide one equation by the other: We can use a strategy similar to the one in Part (a), but in this case we don't know the time of flight. So, in the first step we'll solve for time and substitute the resulting expression into the second step: Ch3 Page 36

37 We calculated the angle in Part (a), but we'll wait to substitute its specific value until the end of the problem. Now we'll write the y-component equation and substitute the expression for time: This is working out: We have just a single equation for one unknown, the initial speed. (Remember that we know the angle; it's the same as in Part (a).) Solving for the initial speed, we get: Relative velocity and relative motion problems Suppose you are on an airplane that is moving at a speed of 800 km/h. You get up from your seat and walk down the aisle towards the front of the plane. You figure you must be walking at a speed of about 3 km/h. But you are on a plane travelling at 800 km/h. So what is your real speed, 800 km/h or 3 km/h? Ch3 Page 37

38 The key point of this section of the textbook is that there is no such thing as "real speed." After all, the airplane moves together with the entire Earth, which is rotating on its axis and revolving around the Sun at high speed. But the whole solar system, which includes the Earth and all its contents, is orbiting the centre of the galaxy. And the galaxy, with all its contents, is also moving. There is no such thing as "real speed;" there is only speed relative to some reference point. We can make this idea precise by speaking about relative velocity; that is, there is a nice, simple basic formula that deals with how you compare relative velocities that are based on different reference points. You can get a sense for what this formula must be by imagining that the airplane we've been discussing is transparent and you are on the ground observing someone sitting in their seat; let's call the person Alice. Suppose it is night time and all is dark except that Alice, the person on the transparent airplane, has a bright light attached to her clothing. Then when Alice is sitting in her seat, all we can see is the light on her clothing, moving forward at 800 km/h. When Alice gets up from her seat and walks forward in the airplane, at a speed that she reckons is about 3 km/h, you see her light moving forward at a speed of 803 km/h. On the other hand, if Alice gets up from her seat and walks backward in the airplane, at a speed that she reckons is about 3 km/h, you see her light moving forward at a speed of 797 km/h. Ch3 Page 38

39 This is very reminiscent of vector addition, isn't it? You can test it out for yourself by walking on a moving walkway (or an escalator) such as is found in most airports. If you and your friend walk at the "same speed" and then you happen to get onto a moving walkway, you will move ahead of your friend. On the other hand, if you walk in the opposite direction on the moving walkway, if you do so at the right speed you'll stay even with someone who is standing still to the side of the walkway. (You might have done this sort of thing by walking up a downwardmoving escalator when you were a child.) We can embody the example of Alice walking within the airplane using the basic equation for relative motion: V AB = V AC + V CB The meaning of the symbols: V AB is the velocity of Alice (A) relative to the ground (B); V AC is the velocity of Alice (A) relative to the airplane(c); V CB is the velocity of the airplane (C) relative to the ground(b). The basic relative velocity formula that we just wrote is a formula involving vectors. It's valid in two (and three) dimensions as well, with no modifications. For example, consider a boat crossing a river: Ch3 Page 39

40 Example: You paddle across a 1-km wide river at a constant speed, with the front of your canoe pointing directly west, which is perpendicular to the banks of the river. If you were in still water, your constant speed would have been 5 km/h, but the current in the river drags you north. If you were not paddling, your speed due to the current would be a constant 2 km/h. a. Determine your velocity relative to the ground. b. Determine how long it takes you to cross the river. c. Determine how far up the river you end up. d. In which direction would you have to paddle in order to cross the river to a point directly across your initial position? Solution: Ch3 Page 40

41 Solution: If you wish to specify the velocity vector using its magnitude (i.e., the speed) and direction, then use Pythagoras's theorem to determine the magnitude and use trigonometry to determine the angle. The result is: Thus, the speed is 5.4 km/h, and the direction of the velocity is 22 degrees North of West. (b) Now that we know the speed, which is constant, we could determine the time needed to cross the river if only we knew the distance travelled by the canoe along the slanting path. This can be done by noting that the triangle in the figure below, which shows the displacement of the canoe, is similar (that is, has the same angles) as the triangle above that shows the velocity and its components. Ch3 Page 41

42 However, an alternative solution for this problem is to observe that the canoe's motion can be separated into two independent components. The paddling effort (and its associated westward component of velocity) pushes the canoe across the river; the river current (which is directed northward) pushes the canoe up the river. To determine how long it takes for the canoe to cross the river, we can focus our attention on just the westward components of displacement and velocity: (c) Because the displacement triangle is similar to the velocity triangle, Ch3 Page 42

43 (d) Experienced paddlers will know that you have to point your canoe up river when you wish to go directly across. The velocities have to look something like this: Knowing the magnitudes of two of the velocities in the triangle enables us to determine the angle, which gives us the direction in which we have to point the canoe. Thus, to go directly across the river, we must point the canoe at an angle of 24 degrees south of west. Similar problems arise in many other types of motion problems. In airplane navigation, one speaks of "air speed" and "ground speed." In situations where one of the motions is not at a constant velocity, additional complications arise. For example, the Earth rotates, which means its surface does not move at a constant velocity (though the speed may be approximately constant). This means that air or water currents that move north or south tend to be deflected, which leads to the well-known cyclone patterns of air movements in the northern and southern hemispheres, and to similar patterns of water currents. Think back to the last time you were moving about on a merry-go-round or carousel and you'll get a sense for the complications. Example: A small airplane has an airspeed of 130 km/h and a heading of 25 degrees North of East. The wind blows at a steady speed of 50 Ch3 Page 43

44 km/h and is directed 40 degrees West of North. Determine the velocity of the airplane with respect to the ground. Solution: Ch3 Page 44

45 This is a perfectly good answer, and so we've satisfied the requirements of the problem by determining the velocity of the plane with respect to the ground. However, the pilot will undoubtedly prefer to express the velocity of the plane in terms of speed and direction; let's do this now: Thus, we can also say that the velocity of the plane is 127 km/h [E 47.4 N]. You can see from the calculations and the diagram that the wind doesn't change the speed of the plane very much, but it does change the plane's heading significantly. Pilots of small planes are constantly monitoring their heading to make sure they arrive at their intended destination with a minimum of meandering, because they carry very limited fuel. Ch3 Page 45

46 The same observations that we made about walking in an airplane were made about four centuries ago by Galileo during heated arguments about whether the Earth moves. Some thinkers, who were certain that the Earth does not move, argued that if the Earth did move then if you stand still and drop an object it will fall behind you as the Earth moves out from under it. Since this doesn't happen, they used this as an argument that the Earth does not move. However, Galileo countered that if a sailor is at the top of a mast of a moving ship, and he drops an object, the object falls at the base of the mast, not behind it as the ship moves out from under it. (Surely many people had experienced travelling in carriages by then, and could have made similar observations.) Galileo argued that the object is being carried along by the ship, and it continues to move forward along with the ship as it falls. The idea of relative velocity is embedded in this old discussion. Galileo won the argument, and many people were convinced that maybe the Earth does move. Thanks to Galileo's work, and the work of other great thinkers (Newton, Kepler, Copernicus, and others), our understanding of the universe and our place in it was utterly transformed. We'll learn more about this in a later chapter. Ch3 Page 46