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=Chapter 3 = toc

HW 10/12 Lesson 1 A&B
A vector quantity is a quantity that is fully described by both magnitude and direction. On the other hand, a scalar quantity is a quantity that is fully described by its magnitude. Vector quantities are not fully described unless both magnitude and direction are listed.  Vector diagrams depict a vector by use of an arrow drawn to scale in a specific direction. 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. The two conventions that will be discussed and used in this unit are described below:
 * 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).
 * 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) of 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. A vector with a direction of 160 degrees is a vector that has been rotated 160 degrees in a counterclockwise direction relative to due east.

The length of the arrow depicts the magnitude of a vector in a scaled vector diagram. The arrow is drawn a precise length in accordance with a chosen scale. The arrow points in the precise direction. Directions are described by the use of some convention. The most common convention is that the direction of a vector is the counterclockwise angle of rotation, which that vector makes with respect to due East. Two vectors can be added together to determine the result (or resultant). Observe the following summations of two force vectors:

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">There are a variety of methods for determining the magnitude and direction of the result of adding two or more vectors.
 * <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">the Pythagorean theorem and trigonometric methods
 * <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">head to tail method using a scaled vector diagram







<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">The direction of a //resultant// vector can often be determined by use of trigonometric functions. Most students recall the meaning of the useful mnemonic SOH CAH TOA from their course in trigonometry. SOH CAH TOA is the meaning of the three common trigonometric functions - sine, cosine, and tangent functions. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Once the measure of the angle is determined, the direction of the vector can be found. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">The measure of an angle as determined through use of SOH CAH TOA is __not__ always the direction of the vector. 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. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">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. This method is described below. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> 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. <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%; text-align: center;">



<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">The head-to-tail method involves d rawing a vector to scaleon a sheet of paper beginning at a designated starting position. Where the head of this first vector ends, the tail of the second vector begins (thus, //head-to-tail// method). Once all the vectors have been added head-to-tail, the resultant is then drawn from the tail of the first vector to the head of the last vector; i.e., from start to finish. Once the resultant is drawn, its length can be measured and converted to //real// units using the given scale. The direction of the resultant can be determined by using a protractor and measuring its counterclockwise angle of rotation from due East. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">A step-by-step method for applying the head-to-tail method to determine the sum of two or more vectors is given below. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">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.
 * 1) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> 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) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">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
 * 3) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">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) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">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**.
 * 5) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Measure the direction of the resultant using the counterclockwise convention.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">HW 10/13 Lesson 1 C&D
<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">The ** resultant ** is the vector sum of two or more vectors If displacement vectors A, B, and C are added together, the result will be vector R.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Displacement vector R gives the same //result// as displacement vectors A + B + C. That is why it can be said that <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%; text-align: center;">** A + B + C = R ** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">The above discussion pertains to the result of adding displacement vectors. When displacement vectors are added, the result is a //resultant displacement//. But any two vectors can be added as long as they are the same vector quantity. If two or more velocity vectors are added, then the result is a //resultant velocity//. If two or more force vectors are added, then the result is a //resultant force//.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">As an example, consider a football player who gets hit simultaneously by three players on the opposing team (players A, B, and C). The football player experiences three different applied forces. Each applied force contributes to a total or resulting force.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Vectors that are directed in //two dimensions.// <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">//Transform// the vector into two parts with each part being directed along the coordinate axes. A vector that is directed upward and rightward can be thought of as having two parts - an upward part and a rightward part. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Each part of a two-dimensional vector is known as a ** component **. The combined influence of the two components is **equivalent** to the influence of the single two-dimensional vector.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">HW 10/17 Lesson 1 E
<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">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
 * <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">the parallelogram method
 * <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">the trig method

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">The parallelogram method of vector resolution involves using an accurately drawn, scaled vector diagram to determine the components of the vector. Briefly put, the method involves drawing the vector to scale in the indicated direction, sketching a parallelogram around the vector such that the vector is the diagonal of the parallelogram, and determining the magnitude of the components (the sides of the parallelogram) using the scale.


 * 1) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Select a scale and accurately draw the vector to scale in the indicated direction.
 * 2) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">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
 * 3) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> 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) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">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.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">The trigonometric method of vector resolution involves using trigonometric functions to determine the components of the vector. 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.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">The method of employing trigonometric functions to determine the components of a vector are as follows:
 * 1) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">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) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">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) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> 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.
 * 4) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">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.



<span style="color: #000000; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">HW 10/18 Lesson 1 G&H
<span style="color: #000000; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">On occasion objects move within a medium that is moving with respect to an observer. 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. <span style="color: #000000; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">A tailwind is merely a wind that approaches the plane from behind, thus increasing its resulting velocity. 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. <span style="color: #000000; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Since a headwind is a wind that approaches the plane from the front, such a wind would decrease the plane's resulting velocity. In this case, the resultant velocity would be 75 km/hr; this is the velocity of the plane relative to an observer on the ground. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> 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. To determine the resultant velocity, the plane velocity (relative to the air) must be added to the wind velocity. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> In this situation of a side wind, the southward vector can be added to the westward vector using the usual methods of vector addition. The magnitude of the resultant velocity is determined using Pythagorean theorem. <span style="color: #ff0000; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">(100 km/hr)2 + (25 km/hr)2 = R2 <span style="color: #ff0000; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">10 000 km2/hr2 + 625 km2/hr2 = R2 <span style="color: #ff0000; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">10 625 km2/hr2 = R2 <span style="color: #ff0000; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">SQRT(10 625 km2/hr2) = R <span style="color: #ff0000; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">**103.1 km/hr = R**

<span style="color: #000000; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">

<span style="color: #ff0000; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">tan (theta) = (opposite/adjacent) <span style="color: #ff0000; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">tan (theta) = (25/100) <span style="color: #ff0000; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">theta = invtan (25/100) <span style="color: #ff0000; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">**theta = 14.0 degrees** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> 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. Like any vector, the resultant's direction is measured as a counterclockwise angle of rotation from due East.

<span style="color: #000000; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">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. <span style="color: #000000; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">The resultant velocity of the motorboat can be determined in the same manner as was done for the plane. The resultant velocity of the boat is the vector sum of the boat velocity and the river velocity. Since the boat heads straight across the river and since the current is always directed straight downstream, the two vectors are at right angles to each other.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> The direction of the resultant is the counterclockwise angle of rotation that the resultant vector makes with due East. This angle can be determined using a trigonometric function as shown below.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> Finding the width of the river, the amount of the time it took for the boat to travel, and the distance downstream the boat reaches opposite shore can be answered using the average speed equation. <span style="color: #ff0000; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">**ave. speed = distance/time** <span style="color: #000000; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">The river is 80-meters wide. That is, 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. <span style="color: #ff0000; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">**time = distance /(ave. speed)** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> The distance of 80 m can be substituted into the numerator. 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. <span style="color: #000000; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">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) <span style="color: #ff0000; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">**time = (80 m)/(4 m/s) = 20 s** <span style="color: #000000; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">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. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> "What distance downstream does the boat reach the opposite shore?" The same equation must be used to calculate this //downstream distance//. 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). <span style="color: #ff0000; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">**distance = ave. speed * time = (3 m/s) * (20 s)** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">The boat is carried 60 meters downstream during the 20 seconds it takes to cross the river.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> A **component** describes the affect of a single vector in a given direction. Any force vector that is exerted at an angle to the horizontal can be considered as having two parts or components. The vector sum of these two components is always equal to the force at the given angle.

<span style="color: #000000; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">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. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> <span style="color: #000000; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">The two perpendicular parts or components of a vector are independent of each other. Consider the pull upon Fido as an example. If the horizontal pull upon Fido increases, then Fido would be accelerated at a greater rate to the right; yet this greater horizontal pull would not exert any vertical influence upon Fido. Pulling horizontally with more force does not lift Fido vertically off the ground. A change in the horizontal component does not affect the vertical component. A change in one component does not affect the other component. Changing a component will affect the motion in that specific direction. While the change in one of the components will alter the magnitude of the resulting force, it does not alter the magnitude of the other component. <span style="color: #000000; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">The resulting motion of a plane flying in the presence of a crosswind is the combination (or sum) of two simultaneous velocity vectors that are perpendicular to each other. If the plane encounters a crosswind directed towards the west, then the resulting velocity of the plane would be northwest. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">

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. <span style="color: #ff0000; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">**d = v • t** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">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. 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. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Lab 10/19 Vector Addition Activity/Mapping
<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">We started at the Northeast corner and ended at the 3rd pole in the lunch room from the east. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">This was the direction of our steps and the amount we moved for each direction. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">This is the drawing of our map and the scale is 1cm= 2cm. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">The graphical resultant is 20.4 cm at 32 degrees SW or 212 degrees. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">The analytical resultant was 20.5 cm at 33.3 degrees or 213.3 degrees Southwest <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">The graphical percent error= 6.47% <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">The analytical percent error= 6.93 %

Our results were very close to accuracy but we still had a kind of big percent error. This may have had to do with the fact that while we measured the distance of the vector segments we may not have been exact with our distances when we moved the measuring tape for each segment. Next time, I would try and be as accurate as possible to find the distance for each segment and the resultant.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">HW 10/19 Lesson 2 A&B
<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Lesson 2A <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">What is a projectile? Examples? <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> An object thats only force is gravity provided that the air resistance is insignificant. 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 <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">What is the force on a projectile? <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> The force of gravity no matter what direction <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Why is there a force? <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> A force is only required to maintain an acceleration. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">What would happen in the absence of gravity? <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> The object would continue at the same speed in the same direction <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">What does gravity do to the projectile? <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> Gravity causes a vertical acceleration. Gravity is the downward force on a projectile which influences the vertical motion and causes the parabolic motion. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Main Idea: <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Gravity is the only force that is acting on a projectile!

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Lesson 2B <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">What two components make up a projectile? <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> Vertical and horizontal motions <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Do the two components affect one another? <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> The presence of gravity does not affect the horiztonal motion but it causes a vertical motion for downward acceleration. The vertical forces are perpendicular to the horizontal motion which shows that they are independent from one another <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> Horizontal Motion Vertical Motion <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">What shape does a projectile make when gravity is present? <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> With gravity, it makes a parabolic shape <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">What shape does a projectile make when gravity is not present? <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> Without gravity, a straight path. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Why does this happen with gravity? <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> The downward force of gravity accelerates the object down.
 * < <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> Forces (Present? - Yes or No)(If present, what dir'n?) ||< <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">No ||< <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">YesThe force of gravity acts downward ||
 * < <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> Acceleration (Present? - Yes or No)(If present, what dir'n?) ||< <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">No ||< <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Yes"g" is downward at 9.8 m/s/s ||
 * < <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;"> **Velocity** (Constant or Changing?) ||< <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Constant ||< <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Changing(by 9.8 m/s each second) ||

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">Main Idea: <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">The two components in a projectile are independent from one another and gravity is the force that makes the object go downward.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">HW 10/20 Lesson 2 C
<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%;">What are rules of projectiles? How is gravity affecting the projectile? It causes the object to accelerate downward at a rate of 9.8 m/s/s. The horizontal shift remains a constant velocity which means there is no horizontal acceleration.
 * A projectile is any object upon which **the only force is gravity**,
 * Projectiles travel with a **parabolic** trajectory __due to the influence of gravity__,
 * There are __no horizontal forces__ acting upon projectiles and thus **no horizontal acceleration**,
 * The //horizontal velocity// of a projectile is **constant** (a never changing in value),
 * There is a __**vertical acceleration caused by gravity**__; its value is 9.8 m/s/s, down,
 * The **vertical velocity of a projectile __changes__ by 9.8 m/s __each second__,**
 * The horizontal motion of a projectile is **independent** of its vertical motion.

What if the projectile is launched into the air? The initial velocity is found and then goes to 9.8 m/s to 0 than back to 9.8. Horizontal never changes. What is a symmetrical projectile? Magnitude of the velocity is the same at equal intervals of time on either side of its peak. What is vertical displacement? vertical displacement (denoted by the symbol **//y//** in the discussion below) of a projectile is dependent only upon the acceleration of gravity and not dependent upon the horizontal velocity. What is horizontal displacement? Influenced by the speed at which it moves horizontally. What happens when an object is launched in the air at an angle?
 * y = 0.5 • g • t2**
 * x = vix • t**
 * y = viy • t + 0.5 • g • t2**

Main Idea: The numerical value shows that their is a symmetrical nature for projectile trajectory.

10/24 Projectile/ Ball in Cup Part 1
Our percent error was 10.98%; the reason it was so off was because our angle measure may have been different for our trials so our theoretical may have been wrong because the projectile may have been launched at an angle other than 0 degrees therefore everything else we calculated like initial velocity, time, and range would be off. The lateral movement was also a source of error. Every time the ball was launched it may not have been the exact same tug on the rope there the results may have varied. To improve the results, we would have to be more careful and and consistent.

media type="file" key="Movie on 2011-10-25 at 09.15.mov" width="300" height="300"

Gordorama Project:
Partner: Ali Cantor For the most part, I think that our project worked very well the main problem that was it would not go straight. I would make sure there was little room for the wheels to move on our axles. I would want to make it have a softer landing when it went past the drop of the ramp and the cardboard to the floor. I'm not sure what I could do to improve the little jump but that was a big cause to making the cart change directions.

Lab: Shoot Your Grade
Partners: Michael Solimano, Sammy Caspert, and George Souflis Purpose: To get the ball launched at a 15 degree angle to go through all of five hoops and then the cup on the floor. We calculated the distance the ball would move by find the height of the launcher and desk like an off the cliff problem which assumes that the horizontal acceleration is 0 and the vertical acceleration is -9.8 m/s. We divided the range by 6 and found intervals at which we would place the hoops.

Materials and Methods: We used measuring tape, duck tape, string, and carbon paper. We first found the range with the carbon paper. We also found the distance from the launcher to the floor by measuring. We used the horizontally launched table and found the time it took for the ball to hit the ground. We divided the range by 6 and then found at what time the ball would be at this interval and at distance the ball would be from the height of the launcher. We placed the tape hoops at the right range intervals and then would practice shooting the ball to see the height of it where it would go through the hoop. It was then just a trial and error to find the perfect spot for each hoop by moving the strings in certain directions to line up with the launcher at the correct position.

Data: Distances are referred to the point of launch. Interval 1 .78 m horizontally .139 m vertically Interval 2 1.56m horizontally .138m vertically Interval 3 2.34m horizontally -.01m vertically

Interval 4 3.12 m horizontally -.29m vertically

Interval 5 3.9m horizontally -.72m vertically Interval 6 cup height .092m cup width .075m 4.66m horizontally -1.23 vertically w/o the cup factoring into it

The data is broken down into intervals. We found the range intervals by finding the total range and dividing it by 6 which gave us that it will be in increments of .78 so each of the hoops would be .78m apart. Each interval represents where the ball would be at that certain location and time on the range as it is travel in a parabolic motion. We found the velocity for the x and y components and used them for each interval.
 * due to a miscommunication and availability to school email; our groups heights are not the actual heights but a close guess to what seem to be attained. The height is in reference to the point of projectile launcher.

Percent Error: Average Percent Error for the Vertical Height: 2.18% The vertical placement had to be maneuvered due to the inconsistency of the launcher.

Final Performance: Here is a video of the ball being launched at 15 degrees and going through 4 hoops.

media type="file" key="Movie on 2011-11-09 at 09.26.mov" width="300" height="300" Conclusion: Our hypothesis was correct because we did get the ball at a 15 degree angle to go through 4 out of 5 of the hoops. Even though we had to estimate the vertical heights each hoops height had a small percent error and the overall average of the vertical percent error was 2.18% which is very small. We would only tamper with the vertical height and leave the hoop at the same exact range from the launcher each time. This small percent error proves that our calculations were very accurate but due to some sources of error they were not 0 % error.

There are numerous sources of error in this lab. The main sources of error would be hanging the hoops and shooting the ball consistently. Each time we hung the hoop, the string would either fall a little or it would move. Sometimes one of us would have to hold the string so the hoops position would not alter. The strings were not easy to work with so next time I would try using a thicker string to make sure that it was able to hold the tape hoop without falling as well as being able to maneuver easily. I would have also got a stronger tape like duct tape to keep the hoops and strings in place. The launcher was not consistently shooting. The angle measure could have changed when pushing the ball into the shooter. I would have tried to find a way to keep the launcher from changing angles as well as moving horizontally. Also we tried to finish it quickly so we would shoot the ball many times in a short time and that would heat up the launcher which would make the ball go in different directions so we kept changing the position of the tape to fit the needs of the ball's different launches so it would go through the hole.

There are multiple real life applications to off the cliff problems. They are used everywhere. This lab also taught me a lot about teamwork and patience. We all had to be on the same page for this lab because it had many steps. Due to the many steps as well as the trial and error, we had to have patience for finding the perfect place for the ball to go through the hoop. There was a lot of frustrating times where the ball would go in a different direction every time but we had to remain calm and take a break to give the launcher time to recover. When we kept using the launcher with no breaks, the heat would make the launcher shoot the ball in many different directions but when we gave it a rest; the launcher would work more consistently.