Ch3PolewayM

toc //Chapter 3//

Vectors Lesson 1
Vectors and Direction Quantities can be divided into two categories - __ [|vectors and scalars] __. A vector quantity is a quantity that is fully described by both magnitude and direction. A scalar quantity is a quantity that is fully described by its magnitude. Examples of vector quantities that have been __ [|previously discussed] __ include__ [|displacement] __, __ [|velocity] __, __ [|acceleration] __, and __ [|force] __. Vector quantities are not fully described unless both magnitude and direction are listed. Vector quantities are often represented by scaled __ [|vector diagrams] __. Vector diagrams depict a vector by use of an arrow drawn to scale in a specific direction. Such diagrams are commonly called as __ [|free-body diagrams] __. The vector diagram depicts a displacement vector.
 * a scale is clearly listed
 * a vector arrow (with arrowhead) is drawn in a specified direction. The vector arrow has a //head// and a //tail//.
 * the magnitude and direction of the vector is clearly labeled. In this case, the diagram shows the magnitude is 20 m and the direction is (30 degrees West of North).

**Conventions for Describing Directions of Vectors** Vectors can be directed due East, due West, due South, and due North. But some vectors are directed //northeast// (at a 45 degree angle); and some vectors are even directed //northeast//, yet more north than east. There are a variety of conventions for describing the direction of any vector. >
 * 1) The direction of a vector is often expressed as an angle of rotation of the vector about its "__ [|tail] __" from east, west, north, or south. For example, a vector can be said to have a direction of 40 degrees North of West (meaning a vector pointing West has been rotated 40 degrees towards the northerly direction) or 65 degrees East of South (meaning a vector pointing South has been rotated 65 degrees towards the easterly direction).
 * 2) The direction of a vector is often expressed as a counterclockwise angle of rotation of the vector about its "__ [|tail] __" from due East. Using this convention, a vector with a direction of 30 degrees is a vector that has been rotated 30 degrees in a counterclockwise direction relative to due east.

**Representing the Magnitude of a Vector** The magnitude of a vector in a scaled vector diagram is depicted by the length of the arrow. The arrow is drawn a precise length in accordance with a chosen scale.





Vector Addition Two vectors can be added together to determine the result (or resultant). Recall in our discussion of Newton's laws of motion, that the //net force// experienced by an object was determined by computing the vector sum of all the individual forces acting upon that object. That is the __ [|net force] __ was the result (or __ [|resultant] __) of adding up all the force vectors.

These rules for summing vectors were applied to __ [|free-body diagrams] __ in order to determine the net force (i.e., the vector sum of all the individual forces). Sample applications are shown in the diagram below. 

There are a variety of methods for determining the magnitude and direction of the result of adding two or more vectors.
 * the Pythagorean theorem and trigonometric methods
 * __ [|the head-to-tail method using a scaled vector diagram] __

**The Pythagorean Theorem** The Pythagorean theorem is a useful method for determining the result of adding two (and only two) vectors __that make a right angle__ to each other. The method is not applicable for adding more than two vectors or for adding vectors that are __not__ at 90-degrees to each other. To see how the method works, consider the following problem: >> Eric leaves the base camp and hikes 11 km, north and then hikes 11 km east. Determine Eric's resulting displacement. This problem asks to determine the result of adding two displacement vectors that are at right angles to each other. The result (or resultant) of walking 11 km north and 11 km east is a vector directed northeast as shown in the diagram to the right. Since the northward displacement and the eastward displacement are at right angles to each other, the Pythagorean theorem can be used to determine the resultant (i.e., the hypotenuse of the right triangle). The result of adding 11 km, north plus 11 km, east is a vector with a magnitude of 15.6 km

**Using Trigonometry to Determine a Vector's Direction** The direction of a //resultant// vector can often be determined by use of trigonometric functions. SOH CAH TOA is a mnemonic that helps one remember the meaning of the three common trigonometric functions - sine, cosine, and tangent functions. Sine= opposite over hypotenuse. Cosine= Adjacent over hypotenuse. Tangent= opposite over adjacent These three trigonometric functions can be applied to the __ [|hiker problem] __ in order to determine the direction of the hiker's overall displacement. The process begins by the selection of one of the two angles (other than the right angle) of the triangle. Once the angle is selected, any of the three functions can be used to find the measure of the angle. Once the measure of the angle is determined, the direction of the vector can be found. In this case the vector makes an angle of 45 degrees with due East. Thus, the direction of this vector is written as 45 degrees. The measure of an angle as determined through use of SOH CAH TOA is __not__ always the direction of the vector. The following vector addition diagram is an example of such a situation. Observe that the angle within the triangle is determined to be 26.6 degrees using SOH CAH TOA. This angle is the southward angle of rotation that the vector R makes with respect to West. Yet the direction of the vector as expressed with the CCW (counterclockwise from East) convention is 206.6 degrees.

When the two vectors that are to be added do not make right angles to one another, or when there are more than two vectors to add together, we will employ a method known as the head-to-tail vector addition method.

**Use of Scaled Vector Diagrams to Determine a Resultant** The magnitude and direction of the sum of two or more vectors can also be determined by use of an accurately drawn scaled vector diagram. Using a scaled diagram, the **head-to-tail method** is employed to determine the vector sum or resultant.

A step-by-step method for applying the head-to-tail method, as shown above, to determine the sum of two or more vectors is given below.
 * 1) Choose a scale and indicate it on a sheet of paper. The best choice of scale is one that will result in a diagram that is as large as possible, yet fits on the sheet of paper.
 * 2) Pick a starting location and draw the first vector //to scale// in the indicated direction. Label the magnitude and direction of the scale on the diagram (e.g., SCALE: 1 cm = 20 m).
 * 3) Starting from where the head of the first vector ends, draw the second vector //to scale// in the indicated direction. Label the magnitude and direction of this vector on the diagram.
 * 4) Repeat steps 2 and 3 for all vectors that are to be added
 * 5) Draw the resultant from the tail of the first vector to the head of the last vector. Label this vector as **Resultant** or simply **R**.
 * 6) Using a ruler, measure the length of the resultant and determine its magnitude by converting to real units using the scale (4.4 cm x 20 m/1 cm = 88 m).
 * 7) Measure the direction of the resultant using the counterclockwise convention discussed __ [|earlier in this lesson] __.

An example of the use of the head-to-tail method is illustrated below. The problem involves the addition of three vectors: 20 m, 45 deg. + 25 m, 300 deg. + 15 m, 210 deg.SCALE: 1 cm = 5 m The head-to-tail method is employed as described above and the resultant is determined (drawn in red). Its magnitude and direction is labeled on the diagram. SCALE: 1 cm = 5 m

Interestingly enough, the order in which three vectors are added has no affect upon either the magnitude or the direction of the resultant. The resultant will still have the same magnitude and direction. For example, consider the addition of the same three vectors in a different order.
 * 15 m, 210 deg. + 25 m, 300 deg. + 20 m, 45 deg.****SCALE: 1 cm = 5 m**

When added together in this different order, these same three vectors still produce a resultant with the same magnitude and direction as before (20. m, 312 degrees). SCALE: 1 cm = 5 m

Resultants The **resultant** is //the result// of adding two or more vectors together. If displacement vectors A, B, and C are added together, the result will be vector R. As shown in the diagram, vector R can be determined by the use of an __ [|accurately drawn, scaled, vector addition diagram] __. Displacement vector R gives the same //result// as displacement vectors A + B + C. **A + B + C = R**

The resultant vector (whether a displacement vector, force vector, velocity vector, etc.) is the result of adding the individual vectors. Vector R is the same result as vectors A + B + C!!



Vector Components We begin to see examples of vectors that are directed in //two dimensions// - upward and rightward, northward and westward, eastward and southward, etc.

In situations in which vectors are directed at angles to the customary coordinate axes, a useful mathematical trick will be employed to //transform// the vector into two parts with each part being directed along the coordinate axes.

Any vector directed in two dimensions can be thought of as having an influence in two different directions. Each part of a two-dimensional vector is known as a **component**. The components of a vector depict the influence of that vector in a given direction. The combined influence of the two components is equivalent to the influence of the single two-dimensional vector. The single two-dimensional vector could be replaced by the two components.

 Example:



Since each wire is stretched in two dimensions (both vertically and horizontally), the tension force of each wire has two components - a vertical component and a horizontal component. Focusing on the wire on the left, we could say that the wire has a leftward and an upward component. This is to say that the wire on the left could be replaced by two wires, one pulling leftward and the other pulling upward. If the single wire were replaced by two wires (each one having the magnitude and direction of the components), then there would be no affect upon the stability of the picture. The combined influence of the two components is equivalent to the influence of the single two-dimensional vector.





Another example:

Vector Resolution

The process of determining the magnitude of a vector is known as **vector resolution**. The two methods of vector resolution that we will examine are
 * the parallelogram method
 * __ [|the trigonometric method] __

**Parallelogram Method of Vector Resolution** The parallelogram method of vector resolution involves using an accurately drawn, scaled vector diagram to determine the components of the vector. A step-by-step procedure for using the parallelogram method of vector resolution is: The step-by-step procedure above is illustrated in the diagram below to show how a velocity vector with a magnitude of 50 m/s and a direction of 60 degrees above the horizontal may be resolved into two components.
 * 1) Select a scale and accurately draw the vector to scale in the indicated direction.
 * 2) Sketch a parallelogram around the vector: beginning at the __ [|tail] __ of the vector, sketch vertical and horizontal lines; then sketch horizontal and vertical lines at the __ [|head] __ of the vector; the sketched lines will meet to form a rectangle (a special case of a parallelogram).
 * 3) Draw the components of the vector. The components are the //sides// of the parallelogram. The tail of the components start at the tail of the vector and stretches along the axes to the nearest corner of the parallelogram. Be sure to place arrowheads on these components to indicate their direction (up, down, left, right).
 * 4) Meaningfully label the components of the vectors with symbols to indicate which component represents which side. A northward force component might be labeled Fnorth. A rightward velocity component might be labeled vx; etc.
 * 5) Measure the length of the sides of the parallelogram and __ [|use the scale to determine the magnitude] __ of the components in //real// units. Label the magnitude on the diagram.

**Trigonometric Method of Vector Resolution** The trigonometric method of vector resolution involves using trigonometric functions to determine the components of the vector. As such, trigonometric functions can be used to determine the length of the sides of a right triangle if an angle measure and the length of one side are known. The method of employing trigonometric functions to determine the components of a vector are as follows: The above method is illustrated below for determining the components of the force acting upon Fido. As the 60-Newton tension force acts upward and rightward on Fido at an angle of 40 degrees, the components of this force can be determined using trigonometric functions.
 * 1) Construct a //rough// sketch (no scale needed) of the vector in the indicated direction. Label its magnitude and the angle that it makes with the horizontal.
 * 2) Draw a rectangle about the vector such that the vector is the diagonal of the rectangle. Beginning at the __ [|tail] __ of the vector, sketch vertical and horizontal lines. Then sketch horizontal and vertical lines at the __ [|head] __ of the vector. The sketched lines will meet to form a rectangle.
 * 3) Draw the components of the vector. The components are the //sides// of the rectangle. The tail of each component begins at the tail of the vector and stretches along the axes to the nearest corner of the rectangle. Be sure to place arrowheads on these components to indicate their direction (up, down, left, right).
 * 4) Meaningfully label the components of the vectors with symbols to indicate which component represents which side. A northward force component might be labeled Fnorth. A rightward force velocity component might be labeled vx; etc.
 * 5) To determine the length of the side opposite the indicated angle, use the sine function. Substitute the magnitude of the vector for the length of the hypotenuse. Use some algebra to solve the equation for the length of the side opposite the indicated angle.
 * 6) Repeat the above step using the cosine function to determine the length of the side adjacent to the indicated angle.

Component Method of Vector Addition __ [|Earlier in this lesson] __, we learned that vectors oriented at right angles to one another can be added together using the Pythagorean theorem.

 Addition of Three or More Right Angle Vectors

As our first example, consider the following vector addition problem:

Example 1: > A student drives his car 6.0 km, North before making a right hand turn and driving 6.0 km to the East. Finally, the student makes a left hand turn and travels another 2.0 km to the north. What is the magnitude of the overall displacement of the student?



When these three vectors are added together in head-to-tail fashion, the resultant is a vector that extends from the tail of the first vector (6.0 km, North, shown in red) to the arrowhead of the third vector (2.0 km, North, shown in green). The head-to-tail vector addition diagram is shown below.

As can be seen in the diagram, the resultant vector (drawn in black) is not the hypotenuse of any right triangle - at least not of any immediately obvious right triangle. But would it be possible to force this resultant vector to be the hypotenuse of a right triangle? Yes. To do so, the order in which the three vectors are added must be changed.

After rearranging the order in which the three vectors are added, the resultant vector is now the hypotenuse of a right triangle.

R2 = (8.0 km)2 + (6.0 km)2

R2 = 64.0 km2+ 36.0 km2

R2 = 100.0 km2

R = SQRT (100.0 km2)

R = 10.0 km

 The resultant is independent by the order in which they are added. Adding vectors **A + B + C** gives the same resultant as adding vectors **B + A + C** or even **C + B + A**. As long as all three vectors are included with their specified magnitude and direction, the resultant will be the same.

 SOH CAH TOA and the Direction of Vectors  To begin our discussion, let's return to **Example 1** above where we made an effort to add three vectors: 6.0 km, N + 6.0 km, E + 2.0 km, N. Observe that the angle in the lower left of the triangle has been labeled as theta (Θ). Theta (Θ) represents the angle that the vector makes with the north axis. Theta (Θ) can be calculated using one of the three trigonometric functions introduced __ [|earlier in this lesson] __ - sine, cosine or tangent. In this problem, we use tangent (TOA).

Tangent(Θ) = Opposite/Adjacent

Tangent(Θ) = 6.0/8.0

Tangent(Θ) = 0.75

Θ = tan-1 (0.75)

Θ = 36.869 …°

Θ =37°

Example 2:



The tangent function will be used to calculate the angle measure of theta (Θ). The work is shown below. Tangent(Θ) = Opposite/Adjacent

Tangent(Θ) = 52.0/22.0

Tangent(Θ) = 2.3636 …

Θ = tan-1 (2.3636 …)

Θ = 67.067 …°

Θ =67.1°

 Addition of Non-Perpendicular Vectors

Now we will consider situations in which the two (or more) vectors that are being added are not at right angles to each other. It is possible to force two (or more) non-perpendicular vectors to be transformed into other vectors that do form a right triangle. The trick involves the __ [|concept of a vector component] __ and the __ [|process of vector resolution] __.

Now suppose that your task involves adding two non-perpendicular vectors together. We will call the vectors **A** and **B**. Vector **A** is a //nasty angled vector// that is neither horizontal nor vertical. And vector **B** is a nice, polite vector directed horizontally.







 To see how this process works with an actual vector addition problem, consider Example 3 shown below.

Example 3: > Max plays middle linebacker for South's football team. During one play in last Friday night's game against New Greer Academy, he made the following movements after the ball was snapped on third down. First, he back-pedaled in the southern direction for 2.6 meters. He then shuffled to his left (west) for a distance of 2.2 meters. Finally, he made a half-turn and ran downfield a distance of 4.8 meters in a direction of 240° counter-clockwise from east (30° W of S) before finally knocking the wind out of New Greer's wide receiver. Determine the magnitude and direction of Max's overall displacement.











R2 = (6.756… m)2 + (4.6 m)2

R2 = 45.655… m2 + 21.16 m2

R2 = 66.815… m2

R = SQRT(66.815… m2 )

R = 8.174 … m

R = ~8.2 m

tangent(Θ) = (6.756… m)/(4.6 m) = 1.46889…  Using the inverse tangent function, the angle theta (Θ) can be determined. On most calculators, this involves using the 2nd-Tangent buttons. Θ = tan-1 (1.46889…) = 55.7536… °

Θ = ~56°

Relative Velocity and Riverboat Problems On occasion objects move within a medium that is moving with respect to an observer. For example, a motorboat in a river is moving amidst a river current - water that is moving with respect to an observer on dry land. In such instances as this, the magnitude of the velocity of the moving object (whether it be a plane or a motorboat) with respect to the observer on land will not be the same as the speedometer reading of the vehicle. Motion is relative to the observer. The observer on land, often named (or misnamed) the "stationary observer" would measure the speed to be different than that of the person on the boat. The observed speed of the boat must always be described relative to who the observer is. To illustrate this principle, consider a plane flying amidst a **tailwind**. A tailwind is merely a wind that approaches the plane from behind, thus increasing its resulting velocity. If the plane is traveling at a velocity of 100 km/hr with respect to the air, and if the wind velocity is 25 km/hr, then what is the velocity of the plane relative to an observer on the ground below? The resultant velocity of the plane (that is, the result of the wind velocity contributing to the velocity due to the plane's motor) is the vector sum of the velocity of the plane and the velocity of the wind. If the plane encounters a headwind, the resulting velocity will be less than 100 km/hr. Since a headwind is a wind that approaches the plane from the front, such a wind would decrease the plane's resulting velocity. Suppose a plane traveling with a velocity of 100 km/hr with respect to the air meets a headwind with a velocity of 25 km/hr. Now consider a plane traveling with a velocity of 100 km/hr, South that encounters a **side wind** of 25 km/hr, West. The resulting velocity of the plane is the vector sum of the two individual velocities. The __ [|Pythagorean theorem] __ can be used.

(100 km/hr)2 + (25 km/hr)2 = R2 10 000 km2/hr2 + 625 km2/hr2 = R2  10 625 km2/hr2 = R2  SQRT(10 625 km2/hr2) = R  **103.1 km/hr = R**

The direction of the resulting velocity can be determined using a __ [|trigonometric function] __. The tangent function can be used

tan (theta) = (opposite/adjacent) tan (theta) = (25/100) theta = invtan (25/100) **theta = 14.0 degrees** If the resultant velocity of the plane makes a 14.0 degree angle with the southward direction (theta in the above diagram), then the direction of the resultant is 256 degrees.

**Analysis of a Riverboat's Motion** T If a motorboat were to head straight across a river (that is, if the boat were to point its bow straight towards the other side), it would not reach the shore directly across from its starting point. The river current influences the motion of the boat and carries it downstream. The motorboat may be moving with a velocity of 4 m/s directly across the river, yet the resultant velocity of the boat will be greater than 4 m/s and at an angle in the downstream direction. While the speedometer of the boat may read 4 m/s, its speed with respect to an observer on the shore will be greater than 4 m/s. The resultant velocity of the boat is the vector sum of the boat velocity and the river velocity. The __ [|Pythagorean theorem] __ can be used to determine the resultant velocity. Suppose that the river was moving with a velocity of 3 m/s, North and the motorboat was moving with a velocity of 4 m/s, East. (4.0 m/s)2 + (3.0 m/s)2 = R2 16 m2/s2 + 9 m2/s2 = R2  25 m2/s2 = R2  SQRT (25 m2/s2) = R  **5.0 m/s = R**

This angle can be determined using a trigonometric function as shown below. tan (theta) = (opposite/adjacent) tan (theta) = (3/4) theta = invtan (3/4) **theta = 36.9 degrees**

Motorboat problems such as these are typically accompanied by three separate questions:
 * 1) What is the resultant velocity (both magnitude and direction) of the boat?
 * 2) If the width of the river is //X// meters wide, then how much time does it take the boat to travel shore to shore?
 * 3) What distance downstream does the boat reach the opposite shore?

The second and third of these questions can be answered using the __ [|average speed equation] __ (and a lot of logic). **ave. speed = distance/time**


 * < Example 1 A motorboat traveling 4 m/s, East encounters a current traveling 3.0 m/s, North.
 * 1) What is the resultant velocity of the motorboat?
 * 2) If the width of the river is 80 meters wide, then how much time does it take the boat to travel shore to shore?
 * 3) What distance downstream does the boat reach the opposite shore? ||

We will start in on the second question. The distance from shore to shore as measured straight across the river is 80 meters. The time to cross this 80-meter wide river can be determined by rearranging and substituting into the average speed equation. **time = distance /(ave. speed)** The distance of 80 m can be substituted into the numerator. Most students want to use the resultant velocity in the equation since that is the actual velocity of the boat with respect to the shore. Yet the value of 5 m/s is the speed at which the boat covers the diagonal dimension of the river. And the diagonal distance across the river is not known in this case. If one knew the **distance C** in the diagram below, then the **average speed C** could be used to calculate the time to reach the opposite shore. Similarly, if one knew the **distance B** in the diagram below, then the **average speed B** could be used to calculate the time to reach the opposite shore. And finally, if one knew the **distance A** in the diagram below, then the **average speed A** could be used to calculate the time to reach the opposite shore. In our problem, the 80 m corresponds to the distance A, and so the average speed of 4 m/s (average speed in the direction straight across the river) should be substituted into the equation to determine the time. **time = (80 m)/(4 m/s) = 20 s** It requires 20 s for the boat to travel across the river. During this 20 s of crossing the river, the boat also drifts downstream. __ [|Part c of the problem] __ asks "What distance downstream does the boat reach the opposite shore?" The same equation must be used to calculate this //downstream distance//. And once more, the question arises, which one of the three average speed values must be used in the equation to calculate the distance downstream? The distance downstream corresponds to **Distance B** on the above diagram. The speed at which the boat covers this distance corresponds to **Average Speed B** on the diagram above (i.e., the speed at which the current moves - 3 m/s). And so the average speed of 3 m/s (average speed in the downstream direction) should be substituted into the equation to determine the distance. **distance = ave. speed * time = (3 m/s) * (20 s)** **distance = 60 m** The boat is carried 60 meters downstream during the 20 seconds it takes to cross the river.

Independence of Perpendicular Components of Motion

Any vector - whether it is a force vector, displacement vector, velocity vector, etc. - directed at an angle can be thought of as being composed of two perpendicular components. These two components can be represented as legs of a right triangle formed by projecting the vector onto the x- and y-axis. The two perpendicular parts or components of a vector are independent of each other. A change in the horizontal component does not affect the vertical component. Changing a component will affect the motion in that specific direction.

Example:

A boat on a river often heads straight across the river, perpendicular to its banks. Yet because of the flow of water (i.e., the current) moving parallel to the river banks, the boat does not land on the bank directly across from the starting location. The resulting motion of the boat is the combination (i.e., the vector sum) of these two simultaneous and independent velocity vectors - the boat velocity plus the river velocity. In the diagram at the right, the boat is depicted as moving eastward across the river while the river flows southward. The boat starts at Point A and heads itself towards Point B. But because of the flow of the river southward, the boat reaches the opposite bank of the river at Point C. The time required for the boat to cross the river from one side to the other side is dependent upon the boat velocity and the width of the river. Only an eastward component of motion could affect the time to move eastward across a river. Suppose that the boat velocity is 4 m/s and the river velocity is 3 m/s. The magnitude of the resultant velocity could be determined to be 5 m/s using the Pythagorean Theorem. The time required for the boat to cross a 60-meter wide river would be dependent upon the boat velocity of 4 m/s. It would require 15 seconds to cross the 60-meter wide river. **d = v • t** So **t = d / v** (60 m) / (4 m/s) **15 seconds** The southward river velocity will not affect the time required for the boat to travel in the eastward direction. If the current increased such that the river velocity became 5 m/s, then it would still require 15 seconds to cross the river. Perpendicular components of motion are independent of each other. An increase in the river velocity would simply cause the boat to travel further in the southward direction during these 15 seconds of motion. An alteration in a southward component of motion only affects the southward motion.



Vectors Lesson 2
What is a Projectile? The most common example of an object that is moving in //two dimensions// is a projectile. A projectile is an object upon which the only force acting is gravity. There are a variety of examples of projectiles. An object dropped from rest is a projectile (provided that the influence of air resistance is negligible). An object that is thrown vertically upward is also a projectile (provided that the influence of air resistance is negligible). And an object which is thrown upward at an angle to the horizontal is also a projectile (provided that the influence of air resistance is negligible). A projectile is any object that once //projected// or dropped continues in motion by its own __ [|inertia] __ and is influenced only by the downward force of gravity.

The __ [|free-body diagram] __ of a projectile would show a single force acting downwards and labeled force of gravity (or simply Fgrav). Regardless of whether a projectile is moving downwards, upwards, upwards and rightwards, or downwards and leftwards, the free-body diagram of the projectile is still as depicted in the diagram at the right.

**Projectile Motion and Inertia** Newton's laws suggest that forces are only required to cause an acceleration (not a motion). And in the case of a projectile that is moving upward, there is a downward force and a downward acceleration. That is, the object is moving upward and slowing down. To further ponder this concept of the downward force and a downward acceleration for a projectile, consider a cannonball shot horizontally from a very high cliff at a high speed. And suppose for a moment that the //gravity switch// could be //turned off// such that the cannonball would travel in the absence of gravity? According to __ [|Newton's first law of motion] __, such a cannonball would continue in motion in a straight line at constant speed. If not acted upon by an unbalanced force, "an object in motion will ...". This is __ [|Newton's law of inertia] __.

Now suppose that the //gravity switch// is turned on and that the cannonball is projected horizontally from the top of the same cliff. Gravity will act downwards upon the cannonball to affect its vertical motion. Gravity causes a vertical acceleration. The ball will drop vertically below its otherwise straight-line, inertial path. Gravity is the downward force upon a projectile that influences its vertical motion and causes the parabolic trajectory that is characteristic of projectiles.

Characteristics of a Projectile's Trajectory Many projectiles not only undergo a vertical motion, but also undergo a horizontal motion. There are the two components of the projectile's motion - horizontal and vertical motion. And since __ [|perpendicular components of motion are independent of each other,] __these two components of motion can (and must) be discussed separately.

**Horizontally Launched Projectiles** Consider a cannonball projected horizontally by a cannon from the top of a very high cliff. In the absence of gravity, the cannonball would continue its horizontal motion at a constant velocity. This is consistent with the __ [|law of inertia] __. And furthermore, if merely dropped from rest in the presence of gravity, the cannonball would accelerate downward, gaining speed at a rate of 9.8 m/s every second. If our thought experiment continues and we project the cannonball horizontally in the presence of gravity, then the cannonball would maintain the same horizontal motion as before - a constant horizontal velocity. Furthermore, the force of gravity will act upon the cannonball to cause the same vertical motion as before - a downward acceleration. The cannonball falls the same amount of distance as it did when it was merely dropped from rest (refer to diagram below). However, the presence of gravity does not affect the horizontal motion of the projectile.




 * <  ||< **Horizontal****Motion** ||< **Vertical****Motion** ||
 * < Forces (Present? - Yes or No)(If present, what dir'n?) ||< No ||< YesThe force of gravity acts downward ||
 * < Acceleration (Present? - Yes or No)(If present, what dir'n?) ||< No ||< Yes"g" is downward at 9.8 m/s/s ||
 * < **Velocity** (Constant or Changing?) ||< Constant ||< Changing(by 9.8 m/s each second) ||

**Non-Horizontally Launched Projectiles** Now suppose that our cannon is aimed upward and shot at an angle to the horizontal from the same cliff. In the absence of gravity (i.e., supposing that the //gravity switch//could be //turned off//) the projectile would again travel along a straight-line, inertial path. An object in motion would continue in motion at a constant speed in the same direction if there is no unbalanced force. This is the case for an object moving through space in the absence of gravity. However, if the //gravity switch// could be //turned on// such that the cannonball is truly a projectile, then the object would once more //free-fall// below this straight-line, inertial path. In fact, the projectile would travel with a //parabolic// //trajectory//. The downward force of gravity would act upon the cannonball to cause the same vertical motion as before - a downward acceleration. The cannonball falls the same amount of distance in every second as it did when it was merely dropped from rest (refer to diagram below). The projectile still moves the same horizontal distance in each second of travel as it did when the //gravity switch// was turned off.

Describing Projectiles With Numbers:

(Horizontal and Vertical Velocity) Consider again the cannonball launched by a cannon from the top of a very high cliff. Suppose that the cannonball is launched horizontally with no upward angle whatsoever and with an initial speed of 20 m/s. If there were no gravity, the cannonball would continue in motion at 20 m/s in the horizontal direction. Yet in actuality, gravity causes the cannonball to accelerate downwards at a rate of 9.8 m/s/s. This means that the vertical velocity is changing by 9.8 m/s every second. If a__ [|vector diagram] __ (showing the velocity of the cannonball at 1-second intervals of time) is used to represent how the x- and y-components of the velocity of the cannonball is changing with time, then x- and y- velocity vectors could be drawn and their magnitudes labeled. 


 * < **Time** ||< **Horizontal****Velocity** ||< **Vertical****Velocity** ||
 * < 0 s ||< 20 m/s, right ||< 0 ||
 * < 1 s ||< 20 m/s, right ||< 9.8 m/s, down ||
 * < 2 s ||< 20 m/s, right ||< 19.6 m/s, down ||
 * < 3 s ||< 20 m/s, right ||< 29.4 m/s, down ||
 * < 4 s ||< 20 m/s, right ||< 39.2 m/s, down ||
 * < 5 s ||< 20 m/s, right ||< 49.0 m/s, down ||

The diagram depicts an object launched upward with a velocity of 75.7 m/s at an angle of 15 degrees above the horizontal. For such an initial velocity, the object would initially be moving 19.6 m/s, upward and 73.1 m/s, rightward. These values are x- and y-__ [|components] __ of the initial velocity and will be discussed in more detail in __ [|the next part of this lesson] __.


 * < **Time** ||< **Horizontal****Velocity** ||< **Vertical****Velocity** ||
 * < 0 s ||< 73.1 m/s, right ||< 19.6 m/s, up ||
 * < 1 s ||< 73.1 m/s, right ||< 9.8 m/s, up ||
 * < 2 s ||< 73.1 m/s, right ||< 0 m/s ||
 * < 3 s ||< 73.1 m/s, right ||< 9.8 m/s, down ||
 * < 4 s ||< 73.1 m/s, right ||< 19.6 m/s, down ||
 * < 5 s ||< 73.1 m/s, right ||< 29.4 m/s, down ||
 * < 6 s ||< 73.1 m/s, right ||< 39.2 m/s, down ||
 * < 7 s ||< 73.1 m/s, right ||< 49.0 m/s, down ||

There is a horizontal velocity that is constant and a vertical velocity that changes by 9.8 m/s each second. As the projectile rises towards its peak, it is slowing down (19.6 m/s to 9.8 m/s to 0 m/s); and as it falls from its peak, it is speeding up (0 m/s to 9.8 m/s to 19.6 m/s to ...). Finally, the //symmetrical// nature of the projectile's motion can be seen in the diagram above: the vertical __ [|speed] __ one second before reaching its peak is the same as the vertical __ [|speed] __ one second after falling from its peak. The vertical __ [|speed] __ two seconds before reaching its peak is the same as the vertical __ [|speed] __ two seconds after falling from its peak. For non-horizontally launched projectiles, the direction of the velocity vector is sometimes considered + on the way up and - on the way down; yet the magnitude of the vertical velocity (i.e., vertical __ [|speed] __) is the same an equal interval of time on either side of its peak. At the peak itself, the vertical velocity is 0 m/s; the velocity vector is entirely horizontal at this point in the trajectory.

Describing Projectiles With Numbers:

(Horizontal and Vertical Displacement) The vertical displacement ( **//y//** ) of a projectile can be predicted using the same equation used to find the displacement of a free-falling object undergoing one-dimensional motion. The equation can be written as follows. **y = 0.5 • g • t2**(equation for vertical displacement for a horizontally launched projectile) where **g** is -9.8 m/s/s and **t** is the time in seconds. The above equation pertains to a projectile with no initial vertical velocity and as such predicts the vertical distance that a projectile falls if dropped from rest. The horizontal displacement of a projectile is only influenced by the speed at which it moves horizontally ( **vix** ) and the amount of time ( **t** ) that it has been moving horizontally. Thus, if the horizontal displacement ( **x** ) of a projectile were represented by an equation, then that equation would be written as **x = vix • t** The diagram below shows the trajectory of a projectile (in red), the path of a projectile released from rest with no horizontal velocity (in blue) and the path of the same object when gravity is turned off (in green). The position of the object at 1-second intervals is shown. In this example, the initial horizontal velocity is 20 m/s and there is no initial vertical velocity (i.e., a case of a horizontally launched projectile). 
 * < **Time** ||< HorizontalDisplacement ||< VerticalDisplacement ||
 * < 0 s ||< 0 m ||< 0 m ||
 * < 1 s ||< 20 m ||< -4.9 m ||
 * < 2 s ||< 40 m ||< -19.6 m ||
 * < 3 s ||< 60 m ||< -44.1 m ||
 * < 4 s ||< 80m ||< -78.4 m ||
 * < 5 s ||< 100 m ||< -122.5 m ||

The diagram below depicts the position of a projectile launched at an angle to the horizontal.

In the absence of gravity, a projectile would rise a vertical distance equivalent to the time multiplied by the vertical component of the initial velocity (viy• t). In the presence of gravity, it will fall a distance of 0.5 • g • t2. Combining these two influences upon the vertical displacement yields the following equation**.** **y = viy • t + 0.5 • g • t2**(equation for vertical displacement for an angled-launched projectile) where **viy** is the initial vertical velocity in m/s, **t** is the time in seconds, and **g** = -9.8 m/s/s (an approximate value of the acceleration of gravity).

y = (19.6 m/s) * (1 s) + 0.5*(-9.8 m/s/s)*(1 s)2 y = 19.6 m + (-4.9 m) y = 14.7 m (approximately) ||< x = vix * twhere vix = 33.9 m/s x = (33.9 m/s) * (1 s) x = 33.9 m ||   || Vertical displacement here shows symmetry.
 * < Calculations for t = 1 second ||< y = viy * t + 0.5*g*t2where viy = 19.6 m/s
 * < **Time** ||< HorizontalDisplacement ||< VerticalDisplacement ||
 * < 0 s ||< 0 m ||< 0 m ||
 * < 1 s ||< 33.9 m ||< 14.7 m ||
 * < 2 s ||< 67.8 m ||< 19.6 m ||
 * < 3 s ||< 101.7 m ||< 14.7 m ||
 * < 4 s ||< 135.6 m ||< 0 m ||

Initial Velocity Components If analyzing the motion to determine the vertical displacement, one would use kinematic equations with vertical motion parameters.

The analysis of projectile motion problems begins by using __ [|the trigonometric methods discussed earlier] __ to determine the horizontal and vertical components of the initial velocity. Consider a projectile launched with an initial velocity of 50 m/s at an angle of 60 degrees above the horizontal. Such a projectile begins its motion with a horizontal velocity of 25 m/s and a vertical velocity of 43 m/s. These are known as the horizontal and vertical components of the initial velocity. These numerical values were determined by constructing a sketch of the velocity vector with the given direction and then using trigonometric functions to determine the sides of the //velocity// triangle. The sketch is shown at the right and the use of trigonometric functions to determine the magnitudes is shown below.



**Determination of the Time of Flight** The time for a projectile to rise vertically to its peak (as well as the time to fall from the peak) is dependent upon vertical motion parameters. The process of determining the time to rise to the peak is an easy process. It is evident that the time for a projectile to rise to its peak is a matter of dividing the vertical component of the initial velocity (viy) by the acceleration of gravity.

Once the time to rise to the peak of the trajectory is known, the total time of flight can be determined. For a projectile that lands at the same height which it started, the total time of flight is twice the time to rise to the peak. Remember symmetry

**Determination of Horizontal Displacement** The horizontal displacement of a projectile is dependent upon the horizontal component of the initial velocity. As discussed in the previous part of this lesson, the horizontal displacement of a projectile can be determined using the equation x = vix • t

**Determination of the Peak Height** The process of rising to the peak is a vertical motion and is again dependent upon vertical motion parameters (the initial vertical velocity and the vertical acceleration). The height of the projectile at this peak position can be determined using the equation y = viy • t + 0.5 • g • t2 where **viy** is the initial vertical velocity in m/s, **g** is the acceleration of gravity (-9.8 m/s/s) and **t** is the time in seconds it takes to reach the peak.

Horizontally Launched Projectile Problems **Problem Type 1:** A projectile is launched with an initial horizontal velocity from an elevated position and follows a parabolic path to the ground. Predictable unknowns include the initial speed of the projectile, the initial height of the projectile, the time of flight, and the horizontal distance of the projectile. Examples of this type of problem are **Problem Type 2:** A projectile is launched at an angle to the horizontal and rises upwards to a peak while moving horizontally. Upon reaching the peak, the projectile falls with a motion that is symmetrical to its path upwards to the peak. Predictable unknowns include the time of flight, the horizontal range, and the height of the projectile when it is at its peak. Examples of this type of problem are
 * 1) A pool ball leaves a 0.60-meter high table with an initial horizontal velocity of 2.4 m/s. Predict the time required for the pool ball to fall to the ground and the horizontal distance between the table's edge and the ball's landing location.
 * 2) A soccer ball is kicked horizontally off a 22.0-meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.
 * 1) A football is kicked with an initial velocity of 25 m/s at an angle of 45-degrees with the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the football.
 * 2) A long jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28-degrees above the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the long-jumper.

Three common kinematic equations that will be used for both type of problems include the following:

**Equations for the Horizontal Motion of a Projectile** Since these two components of motion are independent of each other, two distinctly separate sets of equations are needed - one for the projectile's horizontal motion and one for its vertical motion. For the horizontal components of motion, the equations are

**Equations for the Vertical Motion of a Projectile** For the vertical components of motion, the three equations are In each of the above equations, __ [|the vertical acceleration of a projectile is known to be -9.8 m/s/s] __ (the acceleration of gravity). Furthermore, for the special case of __ [|the first type of problem] __ (horizontally launched projectile problems), viy = 0 m/s. Thus, any term with viy in it will cancel out of the equation.

**Solving Projectile Problems**

The solution of this problem begins by equating the known or given values with the symbols of the kinematic equations - x, y, vix, viy, ax, ay, and t. Because horizontal and vertical information is used separately, it is a wise idea to organized the given information in two columns - one column for horizontal information and one column for vertical information. 0 m/s/s ||< y -0.60 mviy = 0 m/s ay = -9.8 m/s/s || The __ [|first vertical equation] __ (y = viy•t +0.5•ay•t2) will allow for the determination of the time. Once the time has been determined, a __ [|horizontal equation] __ can be used to determine the horizontal displacement of the pool ball. The first horizontal equation (x = vix•t + 0.5•ax•t2) can then be used to solve for "x."
 * < **Example** A pool ball leaves a 0.60-meter high table with an initial horizontal velocity of 2.4 m/s. Predict the time required for the pool ball to fall to the ground and the horizontal distance between the table's edge and the ball's landing location. ||
 * < Horizontal Information ||< **Vertical Information** ||
 * < x = ???**vix** = 2.4 m/s
 * ax**
 * -0.60 m = (0 m/s)•** **t** **+ 0.5•(-9.8 m/s/s)•** **t2** **-0.60 m = (-4.9 m/s/s)•** **t2** **0.122 s2 =** **t2****t = 0.350 s** **(rounded from 0.3499 s)**

Non-Horizontally Launched Projectile Problems The vix and viy values in kinematic equations can be determined by the use of trigonometric functions. The initial x-velocity (vix) can be found using the equation vix = vi•cosine(Theta) where Theta is the angle that the velocity vector makes with the horizontal. The initial y-velocity (viy) can be found using the equation viy = vi•sine(Theta) where Theta is the angle that the velocity vector makes with the horizontal. 
 * x = (**2.4 m/s**)•(0.3499 s) + 0.5•(0 m/s/s)•(0.3499 s)2****x = (**2.4 m/s**)•(0.3499 s)**, **x = 0.84 m** **(rounded from 0.8398 m)**
 * < **__Example__** A football is kicked with an initial velocity of 25 m/s at an angle of 45-degrees with the horizontal. Determine the time of flight, the horizontal displacement, and the peak height of the football. ||

vix = 17.7 m/s ||< viy = vi•sin(Theta)viy = 25 m/s•sin(45 deg) viy = 17.7 m/s || In this case, it happens that the vix and the viy values are the same as will always be the case when the angle is 45-degrees. ???**vix** 17.7 m/s 0 m/s/s ||< **y** ???**viy** = 17.7 m/s As indicated in the table, the final x-velocity (vfx) is the same as the initial x-velocity (vix). This is due to the fact that the__ [|horizontal velocity of a projectile is constant] __; there is __ [|no horizontal acceleration] __. The table also indicates that the final y-velocity (vfy) has the same magnitude and the opposite direction as the initial y-velocity (viy). This is due to __ [|the symmetrical nature of a projectile's trajectory] __.
 * < Horizontal Component ||< **Vertical** Component ||
 * < vix = vi•cos(Theta)vix = 25 m/s•cos(45 deg)
 * < Horizontal Information ||< **Vertical Information** ||
 * < **x**
 * vfx** = 17.7 m/s
 * ax**
 * vfy** = -17.7 m/s
 * ay** = -9.8 m/s/s ||

From the vertical information in the table above and the second equation listed among the __ [|vertical kinematic equations] __ (vfy= viy + ay*t), it becomes obvious that the time of flight of the projectile can be determined. -17.7 m/s = 17.7 m/s + (-9.8 m/s/s)• t -35.4 m/s = (-9.8 m/s/s)• t 3.61 s = t (rounded from 3.6077 s)

The total time of flight of the football is 3.61 seconds. With the time determined, information in the table and the horizontal kinematic equations can be used to determine the horizontal displacement (x) of the projectile.

x = (17.7 m/s)•(3.6077 s) + 0.5•(0 m/s/s)•(3.6077 s)2x = (17.7 m/s)•(3.6077 s) x = 63.8 m

The horizontal displacement of the projectile is 63.8 m. Finally, the __ [|problem statement] __ asks for the height of the projectile at is peak. y (17.7 m/s)• (1.80 s) + 0.5*(-10 m/s/s)• (1.80 s)2 y 31.9 m + (-15.9 m) y = 15.9 m  The time of flight of the football is 3.61 s, the horizontal displacement of the football is 63.8 m , and the peak height of the football 15.9 m.

Vector Addition Lab


Our percent error was very high in this experiment, as it is always unacceptable to go over 10 % for an error. In both our graphical and analytical methods, we got a really high percent error. This is most likely because we didn't exactly start from the origin when measuring. We started on the opposite side of the pillar and didn't think about it right. That definitely hurt our results, and if we did it again, this error would not happen.



Ball in Cup Activity


We never got the ball actually in the cup, as it always hit the bottom of it. Our percent error shows this, as it is very high for a percent error. We possibly went somewhere wrong in our calculations and this caused the ball to be short. We even measured the height of the cup of the distance, so we definitely had it in the right place. When we do this experiment again with the rings we need to be spot on to make sure it goes in the cup. There can be no more room for error, and it shows that we have a lot of error with a percent error over 10.

Gordorama Project


To make this project, we had to get a shoebox, two metal rods, wheels from a roller skate, four stoppers, and four washers. We used the two metal rods for axels and then put the wheels on. Once the wheels were on we put on washers and stoppers to stop the wheels from moving. Furthermore, we had to cut the rods with a saw to fit on the ramp. Then we simply taped the cardboard shoe box onto the axels.



These are our calculations for initial velocity and acceleration. Our distance was 18.4 m in 7.58 seconds and from there we could plug those into our equation. Our initial velocity was 4.85 m/s and our acceleration was -.604 m/s^2.

All in all, our project did very well and traveled the most distance in our class at 18.4 m. If it didn't crash into the wall at the end, it definitely would've reached the end of the hallway. If we had measured the axels more carefully instead of giving a good guess, then our project would have went more straight. It wouldn't have hit the wall and our car would've made it to the end. Furthermore, our project wasn't able to hold much weight, as the pumpkin broke it after the first run. We were only good for one run. If we were to do this project again we would measure for the axels to make sure they were straight. We would also make sure that we used a stronger base then a tiny shoebox. If we were to fix both of these things, our project would've been flawless.

=Lab: Shoot Your Grade=

Purpose: In this lab, we are required to shoot a ball from the launcher above the cabinet at a 25 degree angle. We have to set up 5 rings and a cup at the end and attempt to get it through them all and into the cup. The more rings we get the ball through, the better our grade is.

Hypothesis: The ball should make it through all our rings and land in our cup if calculations are correct.

Procedure: To calculate the initial velocity, we had to set up the equation d=vit + 1/2at^2 and find vit. Once we had vit, we could find the time by plugging vit in. Finally, once time was calculated, we could then plug that in and find the initial velocity. Then, we had to subtract the height of the cup from the total vertical distance, or it would hit the bottom of the cup every time. By doing this, we could the ball right in the cup. Then, we calculated the distances each ring had to be from the launcher and how high up they had to be for the ball to go through. We used all our previous information to find where to put them.

This is the amount of distance the ball traveled from the launcher and it helped us get our initial velocity
 * Trials || Distance (m) ||
 * Trial 1 || 3.25 ||
 * Trial 2 || 3.20 ||
 * Trial 3 || 3.22 ||
 * Trial 4 || 3.17 ||
 * Trial 5 || 3.22 ||





media type="file" key="Shoot+your+grade+lab+video.mov" width="300" height="300"


 * Rings/Cup || Did it make it through? ||
 * Ring #1 || Yes ||
 * Ring #2 || Yes ||
 * Ring #3 || Yes ||
 * Ring #4 || Yes ||
 * Ring #5 || Yes ||
 * Cup || YES ||


 * || X (cm) || Time (s) || Theoretical Y in cm || Experimental Y in cm || % error ||
 * Ring #1 || 50 || .1147 || 16.9 || 21.5 || 27.2% ||
 * Ring #2 || 100 || .2294 || 20.85 || 24.0 || 15.1% ||
 * Ring #3 || 151 || .346 || 11.65 || - || - ||
 * Ring #4 || 208 || .477 || -14.6 || -15.5 || 6.16% ||
 * Ring #5 || 258 || .592 || -51.73 || -56.0 || 8.25% ||
 * Cup || 313 || .737 || 0 || 0 || 0% ||

Percent error:

We got our ball through all the rings and into the cup but our percent error was still very high. It was so high because the angle kept changing on the launcher every time we shot the ball, and we were always altering the rings according to that.

Conclusion: In conclusion, we were able to shoot the ball through all 5 rings and eventually get it into the cup. As we thought it would, this ball followed a parabolic path like all projectiles do. We shot the ball a few times to see where it landed and we were able to measure the horizontal distance with that. Then, we measured the vertical distance. Our group then found the vi which was 4.8 cm/s. After finding the initial velocity, we could use it to find all the ring placements or the respective area they were in. There were several errors with our ring placement though, mostly because the launcher angle changed every time we put the ball in. There was a loose screw, which hurt our results. This is why our percent error was so high. In addition, the launcher was inconsistent and every new trial the ball would be shot differently. Furthermore, whenever you shot the ball and pulled the string, the ball would tend to veer one direction depending on which direction you pulled the string. There are also other forces than gravity acting upon this like air resistance so that contributes to the errors. To fix the launcher, we would have to replace the broken screw that keeps changing the angle and be careful when pulling the string. There's nothing we can really do about the inconsistency of the launcher because all of them are like that. Doing more trials though would give you better results. The only way to avoid the air resistance would be to put the experiment in a vacuum, which is unrealistic. All in all, our ball made it through 5 rings and into the cup which shows that we were successful. After all our equations and results, we were able to figure out exactly where to place the rings, which helped us get it through them all.