In projectile motion, the horizontal and the vertical components of the motion are treated separately. A projectile moves both horizontally and vertically. Its horizontal motion is constant. Its vertical motion is affected by the acceleration due to gravity. The only variable shared by both types of motion is time. Every point on the trajectory is the vector sum of the horizontal and the vertical components of the velocity.

See if you can score a "hole-in-one."

An object projected horizontally (projected perfectly parallel to the surface) will reach the ground in the same time as an object dropped vertically. Since speed at any point in a trajectory is the vector sum of the horizontal and vertical velocity components at that point, the projected object will have a greater speed when it strikes.

The maximum range for a given initial velocity is obtained when the angle of projection is 45°.

Equations that are used to describe the horizontal and vertical motion:

• horizontal motion (constant velocity):
  d = v t
  where v is the horizontal velocity component and d is the horizontal distance (range)

  vertical motion (acceleration):

  d = v_it + ½ a t²

  v_f = v_i + a t

  v_f² = v_i² + 2 a d

  where d is the vertical distance, v_i is the initial vertical velocity, and v_f is the final vertical velocity

Speed

Directions

AP assignment of an origin

Advanced calculations

• If horizontal distance is known: Use the horizontal velocity component to calculate time in the air to that point in the trajectory. Substitute that value for time in d = v_it + ½ a t² and solve for vertical displacement, knowing the initial vertical velocity component.
• If vertical displacement is known (remember to assign your origin system and use the appropriate positive and negative signs for vertical displacement and initial vertical velocity): Use d = v_it + ½ a t² to solve for time to that point in the trajectory knowing the initial vertical velocity component.
• Calculate the total time the object is in the air by using d = v_it + ½ a t² and the initial vertical velocity component. Remember, to set your origin and use the appropriate positive and negative signs for the initial vertical velocity component and for vertical displacement.
• Suppose an object is launched from the ground with speed v and at an angle q. You can use d = v_it + ½ a t² and the initial vertical velocity component to find the total time in the air. Remember, when it hits the ground, the vertical displacement would be zero.

We will be working two types of problems.

• The first type of problem involves an object that is thrown or projected at the top (or apex) of its trajectory (at point B in the picture). At this point, the vertical velocity is zero. Since it starts at this point, we let v_i = zero. It accelerates vertically downward at a rate of -9.8 m/s². We usually use the first acceleration formula listed to find the time the projectile is in the air. The time that it travels vertically is the same as it travels horizontally. We can use this time to determine its horizontal range. Its vertical displacement is given by distance BD and it is negative. Its horizontal range is given by distance DC.
  vertical: v_i = v sin q
  horizontal: v_i = v cos q

  In this case, q = 0°. Thus, cos q = 1 and sin q = 0 which makes the initial vertical velocity component equal to zero and the horizontal velocity component equal the the horizontally projected velocity.

  Advanced calculations

  Consider this problem: An object rolls down an incline and then off the edge of a cliff. Motion information is given allowing one to calculate to speed and angle with which the object leaves the edge of the cliff. Find the initial vertical and horizontal velocity components at the edge of the cliff. Remember, set your origin at the point t=0. The object is going down, so the sign of the initial vertical velocity component is negative. If the height of the cliff is known, the time to fall can be calculated using d = v_it + ½ a t², the initial vertical velocity component, and the acceleration due to gravity. Once time to fall is known, the horizontal velocity component can be used to calculate the distance the object strikes from the base of the cliff. One can also calculate the vertical velocity component the instant before it strikes the ground. It can be used to determine the speed and angle with which the object strikes the ground (remember, it is the vector sum of the horizontal and vertical velocity components at that point).

  

  The second type of problem involes an object that is projected from the ground (point A) and whose path is parabolic. At this point, the speed at which it is projected, v, is the vector sum of its horizontal and its vertical components. At point A, its initial vertical velocity component and its horizontal velocity component are given by:

  vertical: v_i = v sin q

  horizontal: v_i = v cos q

  At pointB, its vertical velocity component is zero. The acceleration at all points is -9.8 m/s². At point B, the object is accelerating (even though its vertical velocity is zero), because its direction is changing. At all points in the trajectory, the horizontal velocity component remains the same (in the absence of air friction). We will work these problems by finding total time in the air using the second acceleration formula. v_f at point C is numerically equal to v_i at point A but of the opposite sign. Remember, velocity is a vector quantity! Once we have total time, we can use d = v t to find the horizontal range, knowing that this uses the horizontal velocity componet. We can use the last acceleration formula to predict the maximum vertical displacement (working from points B and D).

A formula can be derived for the horizontal range:

Advanced calculations

A virtual lab in which the user can control the angle of the cannon. Cursor interrogation allows maximum vertical height and total horizontal range to be determined. A target is provided for amusement.

Projetile Motion applet These two types of problems can also be solved graphically using parametric equations on the graphing calculator. If you are interested in learning how to do this, it is explained in the link below.

Graphing calculator solution to problems

One of the classic questions in physics is this: What should a monkey in a tree do when a gun pointed at him fires -- jump down or stay where it is? A graphing calculator solution to this problem is found at Monkey and Hunter Problem

An interactive "shoot the monkey" activity"

AP Multiple Choice Questions -- Motion in Two Dimensions

1. Be prepared to perform simple calculations. For example, an object is rolled off a table a height h above the ground. The object has initial horizontal speed v. Calculate how long the object is in the air. Calculate how far it lands from the edge of the table. Predict how time in the air and distance it lands from the edge of the table would change if the horizontal velocity of the object were increased or decreased.
2. Be prepared to perform simple calculations. For example, and projectile is launched with velocity v from the ground at an angle q. Points are labeled on its trajectory, one being at its apex. Be able to identify the acceleration vector for each point. Be able to compare the speeds at each point. Be able to compare the velocities at each point. Be able to identify graphs of its vertical velocity, its horizontal velocity, and its acceleration for each point in its trajectory.
3. Important - know that the horizontal velocity is constant for each point in the trajectory. Know that the vertical velocity is accelerating. At the apex of the trajectory, the object has a vertical velocity of zero. It still has acceleration (gravity) and horizontal velocity. The speed of the object at any point is the vector sum of the horizontal and vertical velocities at that point. Remember -- going up, vertical velocity is positive and acceleration is negative; going down, vertical velocity is negative and acceleration is negative.
4. Remember -- acceleration involves a change in speed and/or direction.

AP Free Response Questions -- Motion in Two Dimensions

1. These questions occur over and over in free response questions! They all involve an object that is some height h above the ground. Something has given it horizontal velocity v at this point. Its initial vertical velocity is zero. You will be asked to describe its motion as it falls to the ground. Remember -- describe its horizontal motion (it moves with constant horizontal speed v); describe its vertical motion (it accelerates downward at -9.8 m/s/s); and, describe the trajectory that you see as the vector sum of these two motions.
2. Be able to recognize this type of problem which is asked over and over. They all involve an object that is some height h above the ground. Something has given it horizontal velocity v at this point. Its initial vertical velocity is zero. You have to calculate the time it takes for the object to reach the ground. You have to calculate how far it lands (horizontally). You have to calculate its speed when it hits the floor (remember, it is the vector sum of its final vertical velocity and its horizontal veloctiy).
3. Be able to recognize an Atwood machine or its variation. Two objects with different masses are hung by a cord across a frictionless pulley (in its variation, one object is on a table and the other hangs off the edge or the object is on an incline and the other object hangs off its edge). Calculate the acceleration of the system. You have to calculate the time it takes for the object to reach the ground. You have to calculate how far it lands (horizontally). You have to calculate its speed when it hits the floor (remember, it is the vector sum of its final vertical velocity and its horizontal veloctiy).
4. Trajectory problems and their variations (common variation is to launch the object and have it clear a fence, etc.). Be able to calculate how high the object goes and how long it is in the air. Be able to calculate its height at a specific point above the ground.
5. Be able to draw horizontal velocity vs time, vertical velocity vs time, and acceleration vs time graphs for projectile motion.

simple harmonic motion
   periodic motion where the unbalanced force varies directly with the displacement from the equilibrium point; this motion is described by the period, the frequency, and the amplitude of the motion

period (T)
    the time in seconds needed to complete one cycle of motion

amplitude
    the distance from the equilibrium point to the point of greatest displacement

frequency
    the number of vibrations in a time interval; its SI unit is hertz (Hz)

I like to use the Greek letter, n, as the symbol for frequency. This can be confusing to students since its appearance is similar to the letter v. It can also be represented using the symbol f

1 Hz = 1 sec^-1

T = 1/n  or T = 1/f

n = 1/T  or f = 1/T

characteristics of a simple pendulum:

1. period is independent of mass
2. period is directly proportional to the square root of its length
3. period is indirectly proportional to the square root of the acceleration due to gravity
4. period is independent of amplitude if the arc is less than 10°

For a pendulum, speed is zero and acceleration is a maximum at the point of maximum displacement (point A). For a pendulum, speed is a maximum and acceleration is zero at the equilibrium point (point B).

This Pendulum Applet shows the variation of displacement, velocity, acceleration, and force in a pendulum's swing. You can choose whichever graph display that you would like.

The same thing as above, but for a Spring Pendulum.

An Advanced Look at Simple Harmonic Oscillation

A block is attached to a spring. The block is pulled to position A, stretching the spring from its equilibrium position x=0. This amount of stretch represents the displacement or the amplitude of the oscillating block on the spring. At this point (maximum displacement or amplitude of the oscillation), the speed of the block is zero (v=0) and the spring experiences a restoring force F. An external force must be applied to stretch the spring. This force is given by Hooke's Law (F =-kx, where k is the spring constant and x is the displacement). (see image 1) When the block is released, the spring exerts a force (given by Hooke's Law) that restores the spring/block to the equilibrium position. As the block moves closer to the equilibrium position, the speed increases and the force decreases. The speed is its maximum and the restoring force is zero at the equilibrium position (see image 2). As the block continues to move past its equilibrium position, the force acting on it tends to slow it down until the speed is again zero at the point of maximum displacement (see image 3).

Work must be done to stretch the spring. Potential energy is stored in a stretched or compressed spring. This elastic potential energy is given by

The total mechanical energy of an oscillating spring/block system is the sum of the kinetic and potential energies at that point. The total mechanical energy is equal to the elastic potential energy at maximum displacement.

The total mechanical energy is proportional to the amplitude. If the spring is stretched twice as far, the force is twice as great. Since the energy is proportional to the square of the amplitude, stretching it twice as far quadruples the energy. Since the force is twice as great, the acceleration is twice as great.

The period of a simple harmonic oscillator is dependent upon the spring constant and the mass that is oscillating.

AP Multiple Choice Questions on Pendulums and Springs in Simple Harmonic Motion

1. Be able to predict how the period of a pendulum would change if its length were doubled.
2. Be able to perform simple calculations to predict the period of a pendulum.
3. Be able to predict what would happen to the period of a pendulum if the mass of its bob were doubled.
4. Be able to predict what would happen to the period of a mass on a spring in SHM if the mass were doubled.
5. Be able to predict what the period and the amplitude for an object on a spring in SHM would be for each point in its oscillation. In other words, what would they be after 1/4 the period has elapsed? 1/2 the period? 3/4 the period?
6. Be able to identify when the acceleration and/or the velocity is the greatest (the least) for an object in SHM.

AP Free Response Questions on Pendulums and Springs in Simple Harmonic Motion

1. Not commonly asked unless it is combined with energy concepts. This will be discussed in the next chapter.

acceleration involves a change in speed and/or direction; it is caused by an unbalanced force

in circular motion, the object moves at constant speed but is accelerating because its direction is constantly changing

Uniform Circular Motion

centripetal acceleration:

centripetal force:

Newspaper article about sky diving accident

You can use your graphing calculator to determine how the magnitude of the centripetal force varies the speed with which the object is swung in the horizontal circle, the mass of the object, or the radius of the horizontal circle.

Remember: when a mass moves in a horizontal circle, it completes one revolution in its period. It goes a distance of one revolution or the circumference of the circle, given by 2pr. Since it is moving at constant speed in a horizontal circle, the speed can be found by

Usually, one takes frequency data. In other words, one counts revolutions for a given time interval. Remember, the period is the inverse of the frequency.

Graphing Calculator

Horizontal Circles The image below shows the forces acting on a ball attached to a cord swung in a perfectly horizontal circle. There are two forces acting on the ball, the tension in the string and the weight of the ball. In reality, the ball's weight makes it impossible to swing the ball in a perfectly horizontal circle. We will assume that the ball's weight is small enough that we can ignore it. The tension (labeled T in the image) is the unbalanced center-seeking force and provides the centripetal force on the ball.

Horizontal Circles The image below shows the forces acting on a ball attached to a cord swung in a horizontal circle. The cord is attached to a pole at angle q to the pole. There are two forces (shown in red) acting on the ball, the tension in the string and the weight of the ball. The diagram to the right shows the x and y components of the tension. The y-component of the tension is equal in magnitude and opposite in direction to the weight of the ball. The x-component of the tension is the unbalanced center-seeking force which provides the centripetal force on the ball. If q is measured between the cord and the pole, (A handy trig identity to remember is that tanq=sinq /cosq)

Vertical Circles The image shows a picture of a ball attached to a cord with tension T swung in a vertical circle. We will look at three points on the circle, labeled A, B, and C. There are always two forces acting on the ball, the tension in the cord T and the weight mg. At the top of the circle (point A), both the weight and the tension act down. Their sum (T + mg) is the unbalanced force which is the source of the centripetal force. The minimum speed that the ball must have for the ball to just clear the top of the circle occurs when T=0 (thus, mg = mv,sup>2/r). At the bottom of the circle (point C), the tension is acting up and the weight is acting down. The unbalanced force which is the center-seeking centripetal force is T - mg. At point B, only the tension is center-seeking (the weight is acting perpendicular to the tension) and is the sole source of the centripetal force.

A common example of centripetal acceleration is a car rounding a curve. If the road surface is flat, friction is the source of the centripetal force. If the curve is banked, a component of the normal force provides the centripetal force. For a given banking angle, there would be one speed for which no friction is required.

Banking Angle

AP Force Examples

AP Multiple Choice Questions on Centripetal Motion

1. Be able to predict how the speed of an object twirled in a horizontal circle would change if the radius were doubled or halved. If the force supplying the centripetal force were doubled or halved.
2. Be able to perform simple ratio calculations in which you set the Universal Law of Gravitation equal to mg. For example, this could be used to determine the acceleration due to gravity on Planet X comparing everything to earth data. It also could be used to find the weight of an object (mg) on Planet X comparing everything to earth data.
3. Be able to identify the direction of the force providing the centripetal force for an object in circular motion. For example, the frictional force acts toward the center for a car rounding a flat curve.
4. Be able to perform simple calculations at the bottom (or at the top) of the swing of an object swung in a vertical circle. You should be able to calculate an expression for the centripetal force, knowing weight and tension. Once you know this, you could also determine centripetal acceleration.
5. Know that the work done on an object in circular motion in one period (or one revolution) is zero.

AP Free Response Questions on Centripetal Motion

1. This is frequently asked!
2. Remember how to calculate speed when an object moves in a circle. Remember, it goes a distance of one circumference in a time of one period. Its speed is constant.
3. You may be given a problem in which an object on a string is twirled in a horizontal circle. The object makes an angle theta with respect to the vertical. Represent all forces using a FBD. Calculate the tension in the string (remember: sin q/cos q = tan q). They also combine motion concepts with this type of problem by asking you to describe the motion of the object when the string breaks. You will also be asked to calculate how long it takes the object to hit the ground, hwo far it travels horizontally, and its speed the instant before it strikes.
4. You may be given a problem in which an object on a string is twirled in a vertical circle. Represent all forces using FBDs at the top of the swing, the bottom of the swing, and a radial distance above the bottom. Express the tension in the string at each one of these points. Calculate the minimum speed needed for the object to clear the top of the swing. They also combine motion concepts with this type of problem by asking you to describe the motion of the object when the string breaks at each point. You will also be asked to calculate how long it takes the object to hit the ground, hwo far it travels horizontally, and its speed the instant before it strikes.
5. An object can roll down a frictionless arc of a circle onto a plane. Calculate the speed of the object at the bottom of the arc. Calculate the centripetal force acting on the object while on the arc. They also combine motion with this type of problem.
6. Another common type of problems combines centripetal motion with electrostatics and magnetism concepts. These will be described in the spring.

Kepler's Laws of Planetary MotionThis is not asked on the AP B test.

Kepler's First Law: The path of each planet about the sun is an ellipse with the sun at one focus.

Kepler's Second Law: Each planet moves so that an imaginary line drawn from the sun to the planet sweeps out equal areas in equal periods of time.

Kepler's Third Law: The ratio of the squares of the periods (the time needed for one revolution about the sun) of any two planets revolving about the sun is equal to the ratio of the cubes of their mean distances from the sun.

How are measurements made in different reference frames related to one another?

1. Velocities along a straight line:
   • Simple addition or subtraction is used.
   • Example: Car A moving at 20 m/s passes Car B moving at 15 m/s. The velocity of the Car A relative to Car B is 20 m/s - 15 m/s = 5 m/s.
   • Example: Car A moving at 20 m/s approaches Car B moving at 15 m/s in the opposite direction. Observers in either car will measure a speed of 20 m/s + 15 m/s = 35 m/s relative to their position for the approaching car.
2. When determining relative velocities, we will draw a diagram and use a labeling process in which each velocity is labeled by two subscripts. The first subscript refers to the object and the second subscript refers to the reference frame in which it has this velocity.
   • Example: A boat heads at angle q across a river. The boat has velocity v measured with respect to the shore (This is the velocity the boat would have if the water were still). Using our labeling convention, we will call the boat's velocity v_bs. A current is flowing perpendicular to the boat. The velocity of the current with repsect to the shore is v_ws. To find the velocity of the boat with respect to the shore, we perform the necessary vector addition. (This would be the velocity that is perpendicular to the shore.)
     v_bs = v_bw + v_ws

     Note: using our labeling method, the two outer subscripts for the sum of the vectors on the right are the same as the subscripts of the velocity vector on the left.

AP Objectives-Motion in Two Dimensions

AP Objectives-Circular Motion

Motion in Two Dimensions Homework

Periodic Motion Homework

Motion in Two Dimensions Sample Problems

AP Motion in Two Dimensions Sample Problems

AP Dynamics of Circular Motion Sample Problems