Acceleration in Projectile Motion

10 Key excerpts on "Acceleration in Projectile Motion"

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  Physics

  • John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
  • 2018(Publication Date)
  • Wiley

    (Publisher)

  3.3 Projectile Motion Projectile motion is an idealized kind of motion that occurs when a moving object (the projectile) experiences only the ac- celeration due to gravity, which acts vertically downward. If the trajectory of the projectile is near the earth’s surface, the vertical component a y of the acceleration has a magnitude of 9.80 m/s 2 . The acceleration has no horizontal component (a x = 0 m/s 2 ), the effects of air resistance being negligible. There are several symmetries in projectile motion: (1) The time to reach maximum height from any vertical level is equal to the time spent returning from the maximum height to that level. (2) The speed of a projectile depends only on its height above its launch point, and not on whether it is moving upward or downward. 3.4 Relative Velocity The velocity of object A relative to object B is v → AB , and the velocity of object B relative to object C is v → BC . The velocity of A rel- ative to C is shown in Equation 3 (note the ordering of the subscripts). While the velocity of object A relative to object B is v → AB , the velocity of B relative to A is v → BA = −v → AB . v → AC = v → AB + v → BC (3) Note to Instructors: The numbering of the questions shown here reflects the fact that they are only a representative subset of the total number that are available online. However, all of the questions are available for assignment via WileyPLUS. Section 3.3 Projectile Motion 1. The drawing shows projectile motion at three points along the trajectory. The speeds at the points are υ 1 , υ 2 , and υ 3 . Assume there is no air resistance and rank the speeds, largest to smallest. (Note that the symbol > means “greater than.”) (a) υ 1 > υ 3 > υ 2 (b) υ 1 > υ 2 > υ 3 (c) υ 2 > υ 3 > υ 1 (d) υ 2 > υ 1 > υ 3 (e) υ 3 > υ 2 > υ 1 3. Two balls are thrown from the top of a building, as in the drawing. Ball 1 is thrown straight down, and ball 2 is thrown with the same speed, but upward at an angle θ with respect to the horizontal.

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  Physics

  • John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
  • 2021(Publication Date)
  • Wiley

    (Publisher)

  3.3 Projectile Motion Projectile motion is an idealized kind of motion that occurs when a moving object (the projectile) experiences only the acceleration due to gravity, which acts vertically downward. If the trajectory of the projectile is near the earth’s surface, the ver- tical component a y of the acceleration has a magnitude of 9.80 m/s 2 . The acceleration has no horizontal component (a x = 0 m/s 2 ), the effects of air resistance being negligible. There are several symmetries in projectile motion: (1) The time to reach maximum height from any vertical level is equal to the time spent returning from the maximum height to that level. (2) The speed of a projectile depends only on its height above its launch point, and not on whether it is moving upward or downward. 3.4 Relative Velocity The velocity of object A relative to object B is → v AB , and the velocity of object B relative to object C is → v BC . The velocity of A relative to C is shown in Equation 3 (note the ordering of the subscripts). While the velocity of object A relative to object B is → v AB , the velocity of B relative to A is → v BA = − → v AB . → v AC = → v AB + → v BC (3) Focus on Concepts Additional questions are available for assignment in WileyPLUS. Section 3.3 Projectile Motion 1. The drawing shows projectile motion at three points along the trajectory. The speeds at the points are υ 1 , υ 2 , and υ 3 . Assume there is no air resistance and rank the speeds, largest to smallest. (Note that the symbol > means “greater than.”) (a) υ 1 > υ 3 > υ 2 (b) υ 1 > υ 2 > υ 3 (c) υ 2 > υ 3 > υ 1 (d) υ 2 > υ 1 > υ 3 (e) υ 3 > υ 2 > υ 1 2. Two balls are thrown from the top of a building, as in the draw- ing. Ball 1 is thrown straight down, and ball 2 is thrown with the same speed, but upward at an angle θ with respect to the horizontal. Consider the motion of the balls after they are released.

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  Introduction to Physics

  • John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
  • 2015(Publication Date)
  • Wiley

    (Publisher)

  7. Keep in mind that a kinematics problem may have two possible answers. Try to visualize the different physical situations to which the answers correspond. Check Your Understanding (The answer is given at the end of the book.) 2. A power boat, starting from rest, maintains a constant acceleration. After a certain time t, its displacement and velocity are r B and v B . At time 2t, what would be its displacement and veloc- ity, assuming the acceleration remains the same? (a) 2 r B and 2 v B (b) 2 r B and 4 v B (c) 4 r B and 2 v B (d) 4 r B and 4 v B 3.3 | Projectile Motion The biggest thrill in baseball is a home run. The motion of the ball on its curving path into the stands is a common type of two-dimensional motion called “projectile motion.” A good description of such motion can often be obtained with the assumption that air resistance is absent. Using the equations in Table 3.1, we consider the horizontal and vertical parts of the motion separately. In the horizontal or x direction, the moving object (the projectile) does not slow down in the absence of air resistance. Thus, the x component of the velocity re- mains constant at its initial value or v x 5 v 0x , and the x component of the acceleration is a x 5 0 m/s 2 . In the vertical or y direction, however, the projectile experiences the effect of gravity. As a result, the y component of the velocity v y is not constant and changes. The y component of the acceleration a y is the downward acceleration due to gravity. If the path or trajectory of the projectile is near the earth’s surface, a y has a magnitude of 9.80 m/s 2 . In this text, then, the phrase “projectile motion” means that a x 5 0 m/s 2 and a y equals the acceleration due to gravity. Example 3 and other examples in this section illustrate how the equations of kinematics are applied to projectile motion. EXAMPLE 3 | A Falling Care Package Figure 3.7 shows an airplane moving horizontally with a constant velocity of 1115 m/s at an altitude of 1050 m.

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  Physics

  • John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
  • 2015(Publication Date)
  • Wiley

    (Publisher)

  In the vertical or y direction, however, the projectile experiences the effect of gravity. As a result, the y component of the velocity v y is not constant and changes. The y component of the acceleration a y is the downward acceleration due to gravity. If the path or trajectory of the projectile is near the earth’s surface, a y has a magnitude of 9.80 m/s 2 . In this text, then, the phrase “projectile motion” means that a x 5 0 m/s 2 and a y equals the acceleration due to gravity. Example 3 and other examples in this section illustrate how the equations of kinematics are applied to projectile motion. EXAMPLE 3 | A Falling Care Package Figure 3.7 shows an airplane moving horizontally with a constant velocity of 1115 m/s at an altitude of 1050 m. The directions to the right and upward have been chosen as the positive directions. The plane releases a “care package” that falls to the ground along a curved trajectory. Ignoring air resistance, determine the time required for the package to hit the ground. 60 Chapter 3 | Kinematics in Two Dimensions Reasoning The time required for the package to hit the ground is the time it takes for the package to fall through a vertical distance of 1050 m. In falling, it moves to the right, as well as downward, but these two parts of the motion occur independently. Therefore, we can focus solely on the vertical part. We note that the package is moving initially in the horizontal or x direction, not in the y direction, so that v 0y 5 0 m/s. Furthermore, when the package hits the ground, the y component of its displacement is y 5 21050 m, as the drawing shows. The accel- eration is that due to gravity, so a y 5 29.80 m/s 2 . These data are summarized as follows: y = –1050 m +x +y  y x = 115 m/s  x = 115 m/s  x = 115 m/s    y  Figure 3.7 The package falling from the plane is an example of projectile motion, as Examples 3 and 4 discuss.

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  Physics, Volume 1

  • Robert Resnick, David Halliday, Kenneth S. Krane(Authors)
  • 2016(Publication Date)
  • Wiley

    (Publisher)

  In Fig. 4-16 you can see that the direction of is the same as the direction of just as the vector rela- tionship of Eq. 2-14 requires. Both in free fall and in projectile motion is constant in direction and magnitude, and we can use the equations developed for constant acceleration. We cannot use these equations for uniform circular motion because varies in direction and is therefore not constant. The units of centripetal acceleration are the same as those of an acceleration resulting from a change in the mag- nitude of a velocity. Dimensionally, we have which are the usual dimensions of acceleration. The units therefore may be m/s 2 , km/h 2 , or similar units of dimension L/T 2 . The acceleration resulting from a change in direction of a velocity is just as real and just as much an acceleration in every sense as that arising from a change in magnitude of a velocity. By definition, acceleration is the time rate of change of velocity, and velocity, being a vector, can change in direction as well as magnitude. If a physical quantity is a vector, its directional aspects cannot be ignored, for their effects will prove to be every bit as important and real as those produced by changes in magnitude. According to Newton’s second law in its vector form the acceleration and the net force must have the same direction. In the case of circular motion at con- stant speed, the net force must thus point toward the center of the circle. For now we will write this result in terms of magnitudes: For uniform circular motion, and so (4-30) The quantity on the left side of Eq. 4-30 is sometimes called the “centripetal force.” The centripetal force is not a new kind of force. When a particle moves in a circular path   F B   mv 2 r . a  a c  v 2 /r   F B   ma. ( F B  m a B ), [a]  [v 2 ] [r]  (L/T) 2 L  L T 2 , a B a B  v B , a B a c  v 2 r . a y  v 2 /r,    at constant speed, several forces may act on it.

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  Guide to Mechanics

  • Philip Dyke, Roger Whitworth(Authors)
  • 2017(Publication Date)
  • Red Globe Press

    (Publisher)

  If the velocity of projection is V at an angle to Ox , as illustrated in Figure 5.12, then the velocity of projection can be written as the vector: u ˆ V cos i ‡ V sin j The horizontal motion is unaffected by the acceleration and the component of velocity in that direction is constant. The upwards vertical motion is subject to an acceleration of g . The acceleration vector is written in the form: a ˆ g j For a given point r relative to O at time t on the trajectory, we can write: r ˆ u t ‡ 1 2 a t 2 The vector diagram in Figure 5.13 shows the dependence of the position vector r on the vectors u and a . This gives, for a given point on the path with coordinates ( x , y ): x i ‡ y j ˆ V cos † t i ‡ V sin † t j † ‡ 1 2 g j † t 2 O y V x Figure 5.12 The trajectory 124 Guide to Mechanics O r u t ½ a t 2 Figure 5.13 A triangle of vectors illustrating the equation r ˆ u t ‡ 1 2 a t 2 The set of equations: x ˆ V cos † t and y ˆ V sin † t 1 2 gt 2 represents the parametric equations of the trajectory. Making t the subject of the first of these formulae, we obtain: t ˆ x V cos This enables us to derive: y ˆ x tan gx 2 sec 2 2 V 2 5 : 26 † as the Cartesian equation for the trajectory. From this equation, we can see that, for a given V and , the trajectory is a parabola. Having once determined this we can use any of the properties of a parabola to discuss projectile motion. The most important of these is the symmetry of the curve about a vertical line through its maximum value, which is of course its greatest height. Using this symmetry property, we can make the following observations about projectile motion: (a) The greatest height is the maximum value of y . (b) The range is the value of x for which y ˆ 0. (c) The range is twice the x value to the greatest height. (The time to the range is twice the time to the greatest height, which follows directly.) (d) All heights are symmetrical about the greatest height's horizontal position.

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  Workshop Physics Activity Guide Module 1

  Mechanics I

  • Priscilla W. Laws, David P. Jackson, Brett J. Pearson(Authors)
  • 2023(Publication Date)
  • Wiley

    (Publisher)

  It’s worth noting that the gravitational force, which acts vertically, has no effect on our horizontal motion experiments like the cart on a track. 12 In other words, motions in the horizontal and vertical directions are independent of one another. We will observe this direction independence shortly, when we inves- tigate the motion of an object that is launched near the surface of Earth. Such motion is commonly referred to as projectile motion. Before studying motion in two dimensions, we would like to do an experiment that demonstrates the physical nature of a constant, continuous force in one dimension. For example, a dropped ball experiences a constant downward force due to gravity. We can create a horizontal analog of this motion by continuously tapping a ball in one direction on a flat surface. The similarity between a falling ball and a tapped ball will help us study projectile motion, in which an object falls vertically while simultaneously moving horizontally. For the measurements described below, you will use a twirling baton with a rubber tip to tap a bowling ball and watch its motion. You should have the fol- lowing equipment available: • 1 bowling ball (or other heavy ball) • 1 twirling baton (or other stick used to tap the ball) 11 Clearly, Earth cannot be seen as flat from their perspective. 12 We did need to be careful to keep the motion only in the horizontal direction—any “tilt” to the cart track would have led to an acceleration “downhill” caused by the gravitational force. UNIT 6: GRAVITY AND PROJECTILE MOTION 183 Find a stretch of smooth, level floor over which the ball can roll for some dis- tance (a hallway can be used if the classroom is not large enough).

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  An Introduction to Physical Science

  • James Shipman, Jerry Wilson, Charles Higgins, Bo Lou, James Shipman(Authors)
  • 2020(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 2.3 Acceleration 35 When any of these changes occur, an object is accelerating. Examples are 1. a car speeding up (or slowing down) while traveling in a straight line, 2. a car rounding a curve at a constant speed, and 3. a car speeding up (or slowing down) while rounding a curve. Acceleration is defined as the time rate of change of velocity. Taking the symbol Δ (delta) to mean “change in,” the equation for average acceleration (a) can be written as average acceleration 5 change in velocity time for change to occur a 5 Dv Dt 5 v f 2 v o t 2.2 The change in velocity (Δv) is the final velocity v f minus the original velocity v o . Also, the interval Δt is commonly written as t (Δt 5 t 2 t o 5 t), with t o taken to be zero (t is understood to be an interval). v o is not zero if the car is initially moving. The units of acceleration in the SI are meters per second per second, (m/s)/s, or meters per second squared, m/s 2 . These units may be confusing at first. Keep in mind that an acceleration is a measure of a change in velocity during a given time period. Consider a constant acceleration of 9.8 m/s 2 . This value means that the velocity changes by 9.8 m/s each second. Thus, for straight-line motion, as the number of seconds increases, the velocity goes from 0 to 9.8 m/s during the first second, to 19.6 m/s (that is, 9.8 m/s 1 9.8 m/s) during the second second, to 29.4 m/s (that is, 19.6 m/s 1 9.8 m/s) during the third second, and so forth, adding 9.8 m/s each second. This sequence is illustrated in ● Fig. 2.7 for an object that falls with a constant acceleration due to gravity of 9.8 m/s 2 . Because the velocity increases, the distance traveled by the falling object each second also increases, but not uniformly.

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  Engineering Mechanics

  Dynamics

  • L. G. Kraige, J. N. Bolton(Authors)
  • 2018(Publication Date)
  • Wiley

    (Publisher)

  Therefore, every- thing covered in Art. 2 ∕ 2 on rectilinear motion can be applied separately to the x-motion and to the y-motion. Projectile Motion An important application of two-dimensional kinematic theory is the problem of projectile motion. For a first treatment of the subject, we neglect aerodynamic drag and the curvature and rotation of the earth, and we assume that the altitude change is small enough so that the acceleration due to gravity can be considered constant. With these assumptions, rectangular coordinates are useful for the tra- jectory analysis. For the axes shown in Fig. 2 ∕ 8, the acceleration components are a x = 0 a y = −g r = x i + y j y v = r ˙ = x ˙ i + y ˙ j y a = v ˙ = r ¨ = x ¨ i + y ¨ j y ¨ Path j i x i y j r A x y A v a v y v x a x a y  FIGURE 2/7 Article 2/4 Rectangular Coordinates (x-y) 29 Integration of these accelerations follows the results obtained previously in Art. 2 ∕ 2a for constant acceleration and yields v x = ( v x ) 0 v y = ( v y ) 0 − gt x = x 0 + ( v x ) 0 t y = y 0 + ( v y ) 0 t − 1 2 gt 2 v y 2 = ( v y ) 0 2 − 2 g( y − y 0 ) In all these expressions, the subscript zero denotes initial conditions, frequently taken as those at launch where, for the case illustrated, x 0 = y 0 = 0. Note that the quantity g is taken to be positive throughout this text. We can see that the x- and y-motions are independent for the simple projectile conditions under consideration. Elimination of the time t between the x- and y- displacement equations shows the path to be parabolic (see Sample Problem 2 ∕ 6). If we were to introduce a drag force which depends on the speed squared (for exam- ple), then the x- and y-motions would be coupled (interdependent), and the trajec- tory would be nonparabolic.

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  Principles of Physics: Extended, International Adaptation

  • David Halliday, Robert Resnick, Jearl Walker(Authors)
  • 2023(Publication Date)
  • Wiley

    (Publisher)

  (4.4.6) The trajectory (path) of a particle in projectile motion is parabolic and is given by y = (tan θ 0 )x − gx 2 ____________ 2( v 0 cos θ 0 ) 2 , (4.4.7) if x 0 and y 0 of Eqs. 4.4.3 to 4.4.6 are zero. The particle’s horizontal range R, which is the horizontal distance from the launch point to the point at which the particle returns to the launch height, is R = v 0 2 __ g sin 2θ 0 . (4.4.8) Uniform Circular Motion If a particle travels along a circle or circular arc of radius r at constant speed v, it is said to be in uniform circular motion and has an acceleration a → of constant magnitude a = v 2 ___ r . (4.5.1) The direction of a → is toward the center of the circle or circular arc, and a → is said to be centripetal. The time for the particle to complete a circle is T = 2πr ____ v . (4.5.2) T is called the period of revolution, or simply the period, of the motion. Relative Motion When two frames of reference A and B are moving relative to each other at constant velocity, the velocity of a particle P as measured by an observer in frame A usually differs from that measured from frame B. The two measured velocities are related by v → PA = v → PB + v → BA , (4.7.2) where v → BA is the velocity of B with respect to A. Both observers measure the same acceleration for the particle: a → PA = a → PB . (4.7.3) 1 Figure 4.1 shows the path taken by a skunk foraging for trash food, from initial point i. The skunk took the same time T to go from each labeled point to the next along its path. Rank points a, b, and c according to the magnitude of the average velocity of the skunk to reach them from initial point i, greatest first. FIGURE 4.1 Question 1. a i b c 2 Figure 4.2 shows the initial position i and the final position f of a particle. What are the (a) initial position vector r → i and (b) final position vector r → f , both in unit-vector notation? (c) What is the x component of displacement Δ r → ? z x i f y 4 m 4 m 1 m 2 m 3 m 3 m 3 m 5 m FIGURE 4.2 Question 2.

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