Ch3_SteeleM

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 * Maddi's Physics Wiki - Ch3 **

**Homework** - Lesson 1
__//The author describes the characteristics of vectors and vector diagrams.//__ 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 include displacement, velocity, acceleration, and force.
 * a) Vectors and Direction**

Vector diagrams depict a vector by use of an arrow drawn to scale in a specific direction. Characteristics of a vector diagram:
 * 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.

__**Conventions for Describing Directions of Vectors**__ //The author explains one way of expressing the direction of a vector.// 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 author explains how the magnitude of a vector in a scaled vector diagram is depicted by the length of the arrow.//

//The author explains how 2 vectors can be added together to determine the result (or resultant)://
 * b) Vector Addition**

 Two methods to determine 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

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">__**Pythagorean Theorem**__ <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">//The author explains how the Pythagorean theorem is a method for finding the result of adding exactly two vectors __that make a right angle__ to each other://

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">__**Using Trigonometry to Represent a Vector's Direction**__ <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">//The author explains how the direction of a resultant vector can be determined by using trigonometry.//

<span style="display: block; font-family: Arial,Helvetica,sans-serif; font-size: 12px; height: 102px; text-align: left; width: 489px;">

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">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.

<span style="display: block; font-family: Arial,Helvetica,sans-serif; font-size: 12px; height: 191px; text-align: left; width: 341px;"> <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">__**Use of Scaled Vector Diagrams to Determine a Resultant**__ <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">//The author explains how the head-to tail method can determine the sum of two or more vectors.// 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 use the head-to-tail vector addition method.

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">A step-by-step method for applying the head-to-tail method to determine the sum of two or more vectors is given below.
 * 1) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">Choose a scale and indicate it on a sheet of paper.
 * 2) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">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: Arial,Helvetica,sans-serif; font-size: 12px;">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: Arial,Helvetica,sans-serif; font-size: 12px;">Repeat steps 2 and 3 for all vectors that are to be added
 * 5) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">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) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">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) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">Measure the direction of the resultant using the counterclockwise convention.

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">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.

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">

<span style="display: block; font-family: Arial,Helvetica,sans-serif; font-size: 12px; text-align: left;">(SCALE: 1 cm = 5 m)

**c) Resultant** __//The author explains how the **resultant** is the vector sum of two or more vectors.//__ It 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.



Vector R is the //resultant displacement// of displacement vectors A, B, and C. A person who walked with displacements A, then B, and then C would be displaced by the same amount as a person who walked with displacement R. Displacement vector R gives the same //result// as displacement vectors A + B + C. That is why it can be said that **A + B + C = R**.

__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. If the three forces are added together using methods of vector addition, then the resultant vector R can be determined. In this case, to experience the three forces A, B and C is the same as experiencing force R.



__//The author explains how the components of a vector represent the influence of that vector in a given direction.//__ 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.
 * d) Vector Components**

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.

__EXAMPLE 1:__

__EXAMPLE 2:__

__EXAMPLE 3:__

__//The author explains two methods of 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
 * e) Vector Resolution**
 * 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:
 * 1) Draw vector to scale in the indicated direction.
 * 2) Sketch a parallelogram around the vector.
 * 3) Draw the components of the vector. The components are the //sides// of the parallelogram.
 * 4) Meaningfully label the components of the vectors with symbols to indicate which component represents which side.
 * 5) Measure the length of the sides of the parallelogram an use the scale to determine the magnitude of the components in //real// units.

__EXAMPLE:__ (step-by-step parallelogram method)

__ **Trigonometric Method of Vector Resolution** __ The trigonometric method of vector resolution involves using trigonometric functions to determine the components of the vector. The method of employing trigonometric functions to determine the components of a vector are as follows:
 * 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.
 * 3) Draw the components of the vector.
 * 4) Meaningfully label the components of the vectors with symbols to indicate which component represents which side.
 * 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 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.

__EXAMPLE__: (step-by-step Pythagorean method)

//__<span style="background-color: white; font-family: Verdana,sans-serif; font-size: 9pt;">The author explains how to do vector addition by using the components of the vectors. __//
 * f) Component Addition**

<span style="background-color: white; font-family: Verdana,sans-serif; font-size: 9pt;">When the two vectors are added head-to-tail, the resultant is the hypotenuse of a right triangle. The resultant can be determined using the Pythagorean theorem.

<span style="font-family: Verdana,sans-serif; font-size: 9pt;">This //Pythagorean approach// is only for adding two vectors that are directed at right angles to one another. For this reason, the Pythagorean theorem can’t always be used to solve vector addition problems. Other methods must be used to do so

<span style="font-family: Verdana,sans-serif; font-size: 9pt;">Vector 1 = 6.0 km North <span style="font-family: Verdana,sans-serif; font-size: 9pt;">Vector 2 = 6.0 km East <span style="font-family: Verdana,sans-serif; font-size: 9pt;">Vector 3 = 2.0 km North
 * <span style="color: red; font-family: Verdana,sans-serif; font-size: 10.5pt;">Addition of Three or More Right Angle Vectors **
 * <span style="color: red; font-family: Verdana,sans-serif; font-size: 9pt;">Example 1: **

<span style="font-family: Verdana,sans-serif; font-size: 9pt;">What is the magnitude of the overall displacement of these vectors? <span style="background-color: white; font-family: Verdana,sans-serif; font-size: 9pt;">When these three vectors are added together in head-to-tail fashion, the resultant is a vector looks like this: <span style="background-color: white; font-family: Verdana,sans-serif; font-size: 9pt;">As can be seen in the diagram, the resultant vector (drawn in black) is not the hypotenuse of any right triangle. The vectors above were drawn in the order in which they were driven. However, rearranging the order in which the three vectors are added allows us to use the Pythagorean theorem.

<span style="background-color: white; font-family: Verdana,sans-serif; font-size: 9pt;">After rearranging the order in which the 3 vectors are added, the resultant vector is now the hypotenuse of a right triangle.

<span style="display: block; font-family: Verdana,sans-serif; font-size: 9pt; text-align: center;">R 2 = (8.0 km) 2 + (6.0 km) 2 <span style="font-family: Verdana,sans-serif; font-size: 9pt;">R 2 = 64.0 km 2 + 36.0 km 2 <span style="display: block; font-family: Verdana,sans-serif; font-size: 9pt; text-align: center;">R 2 = 100.0 km 2 <span style="display: block; font-family: Verdana,sans-serif; font-size: 9pt; text-align: center;">R = SQRT (100.0 km 2 ) **<span style="color: red; font-family: Verdana,sans-serif; font-size: 9pt;">R = 10.0 km ** <span style="background-color: white; font-family: Verdana,sans-serif; font-size: 9pt;">The resultant is independent by the order in which the vectors are added.

<span style="background-color: white; font-family: Verdana,sans-serif; font-size: 9pt;">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. In this picture, we wish to determine the angle measure of theta (Θ). The tangent function will be used to calculate the angle measure of theta (Θ). The work is shown below: <span style="font-family: Verdana,sans-serif; font-size: 9pt;">Tangent(Θ) = Opposite/Adjacent <span style="font-family: Verdana,sans-serif; font-size: 9pt;"> Tangent(Θ) = 6.0/8.0 <span style="font-family: Verdana,sans-serif; font-size: 9pt;"> Tangent(Θ) = 0.75 <span style="font-family: Verdana,sans-serif; font-size: 9pt;"> Θ = tan-1 (0.75) <span style="font-family: Verdana,sans-serif; font-size: 9pt;"> Θ = 36.869 …°
 * <span style="color: red; font-family: Verdana,sans-serif; font-size: 10.5pt;">SOH CAH TOA and the Direction of Vectors **
 * <span style="color: red; font-family: Verdana,sans-serif; font-size: 9pt;">Θ =37° **

<span style="background-color: white; font-family: Verdana,sans-serif; font-size: 9pt;">This angle measure must now be used to state the direction. Since the angle that the resultant makes with east is the complement of the angle that it makes with north, we could express the direction as 53° CCW.

A vector component describes the effect of a vector in a given direction. Any //<span style="font-family: Verdana,sans-serif; font-size: 12px; line-height: 18px;">angled // vector has a horizontal and a vertical component. Together, the effect of these two components is equal to the overall effect of the //<span style="font-family: Verdana,sans-serif; font-size: 12px; line-height: 18px;">angled //vector.
 * <span style="color: red; font-family: Verdana,sans-serif; font-size: 10.5pt;">Addition of Non-Perpendicular Vectors **

__<span style="background-color: white; font-family: Verdana,sans-serif; font-size: 9pt;">EXAMPLE 1 __<span style="background-color: white; font-family: Verdana,sans-serif; font-size: 9pt;">: <span style="background-color: white; font-family: Verdana,sans-serif; font-size: 9pt;">The situation is shown below:

<span style="background-color: white; font-family: Verdana,sans-serif; font-size: 9pt;">Vector **A** has two components - **Ax** and **Ay**. These two components together are equal to vector **A**.

<span style="background-color: white; font-family: Verdana,sans-serif; font-size: 9pt;">And since this is true, it makes since to say that **A + B = Ax + Ay + B.**

<span style="background-color: white; font-family: Verdana,sans-serif; font-size: 9pt;">The problem of A + B has been transformed into a problem in which all vectors are at right angles to each other. With all vectors being at right angles to one another, the Pythagorean theorem can be used to determine the magnitude of the resultant.

__<span style="font-family: Verdana,sans-serif; font-size: 9pt;">EXAMPLE 2 __<span style="font-family: Verdana,sans-serif; font-size: 9pt;">: <span style="font-family: Verdana,sans-serif; font-size: 9pt;">Vector A = 2.6 meters @ 270 degrees <span style="font-family: Verdana,sans-serif; font-size: 9pt;">Vector B = 2.2 meters @ 180 degrees <span style="font-family: Verdana,sans-serif; font-size: 9pt;">Vector C = 4.8 meters @ 240 degrees

<span style="font-family: Verdana,sans-serif; font-size: 9pt;">Determine the magnitude and direction of the resultant vector. <span style="background-color: white; font-family: Verdana,sans-serif; font-size: 9pt;">We need to determine the components of vector C in order to add the vectors. To determine the x-component, you multiply each vector quantity by the cosine of its angle. After doing this for each vector, you add the values. You do the same thing to determine the y-component, except you multiply each vector quantity by the sine of its angle instead of cosine.

<span style="background-color: white; font-family: Verdana,sans-serif; font-size: 9pt;">The triangle's perpendicular sides have lengths of 4.6 meters and 6.756 meters. The resultant's magnitude (R) can now be determined using the Pythagorean theorem.

<span style="display: block; font-family: Verdana,sans-serif; font-size: 9pt; text-align: center;">R2 = (6.756 m)2 + (4.6 m)2 <span style="display: block; font-family: Verdana,sans-serif; font-size: 9pt; text-align: center;">R2 = 45.655 m2 + 21.16 m2 <span style="display: block; font-family: Verdana,sans-serif; font-size: 9pt; text-align: center;">R2 = 66.815 m2 <span style="display: block; font-family: Verdana,sans-serif; font-size: 9pt; text-align: center;">R = SQRT(66.815 m2 ) <span style="display: block; font-family: Verdana,sans-serif; font-size: 9pt; text-align: center;">R = 8.174 m **<span style="color: red; font-family: Verdana,sans-serif; font-size: 9pt;">R = ~8.2 m ** <span style="background-color: white; font-family: Verdana,sans-serif; font-size: 9pt;">The direction of the resultant can be determined by finding the angle that the resultant makes with either the north-south or the east-west vector. Using the inverse tangent function, the angle theta (Θ) can be determined.

<span style="display: block; font-family: Verdana,sans-serif; font-size: 9pt; text-align: center;">Θ = tan-1 (1.46889) = 55.7536° **<span style="color: red; font-family: Verdana,sans-serif; font-size: 9pt;">Θ = ~56° ** <span style="background-color: white; font-family: Verdana,sans-serif; font-size: 9pt;">This 56° angle is the angle between the resultant vector and the westward direction. This makes the direction 56° south of west.

//The author explains that motion is relative to the observer.// Objects sometimes move within a medium that is moving with respect to an observer. For example, a boat on a river is moving amidst a current - water that is moving with respect to an observer on dry land. In such instances, the magnitude of the velocity of the moving object 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 observed speed of the boat must always be described relative to who the observer is.
 * g) Relative Velocity and Riverboat Problems**

__ EXAMPLE 1: __ Plane has a velocity of 100 km/hr. Tailwind increases velocity of plane by 25 km/hr. What is the velovity of the plane relative to an observer on the ground?

__ EXAMPLE 2: __ Plane has a velocity of 100 km/hr, South. Plane encounters a side wind of 25 km/hr, West. What would the resulting velocity of the plane be? *Use Pythagorean theorem to solve.

The algebraic steps are as follows: (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** tan (theta) = (opposite/adjacent) tan (theta) = (25/100) theta = invtan (25/100) **theta = 14.0 degrees**

//The author explains how the components of a vector are independent of each other, therefore a change in one component does not affect the other.// 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.
 * h) Independence of Perpendicular Components of Motion**

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. An alteration in one of the components will not affect the other component.

**Homework** - Lesson 2
__Main idea__: //A projectile is an object upon which the only force acting is gravity. The two components of a projectile are vertical and horizontal, and they are independent of each other.//
 * a) What is a Projectile?**
 * b) Characteristics of a Projectile's Trajectory**

1. What is a projectile?
 * An object upon which the only force acting is gravity.

2. Are there different types of projectile motion?
 * Yes.
 * 1) Object dropped from rest
 * 2) Object thrown vertically upward
 * 3) Object thrown upward at an angle

3. What is Newton's first law of motion?
 * An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

4. What are the two components of a projectile's trajectory?
 * Horizontal and vertical motion. Since they are perpendicular components of motion, they are independent of each other.

5. Do projectiles experience horizontal acceleration?
 * No. The presence of gravity does not affect the horizontal motion of the projectile. The force of gravity acts downward, and therefore cannot alter the horizontal motion.

__Main idea__: //The horizontal displacement of a projectile remains constant, while vertical displacement of a projectile changes because it is influenced by gravity.//
 * c) Describing Projectiles with Numbers (a & b)**

1. What are the characteristics of a projectile?
 * 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.

2. How are projectiles measured?
 * By how the numerical values of the x- and y-components of the velocity and displacement change with time (or remain constant).

3. What are important equations to know for solving projectile equations?
 * y = 0.5*g*t^2 (vertical displacement of a projectile)
 * x = vix*t (horizontal displacement of a projectile)
 * y = viy*t + 0.5*g*t^2 (vertical displacement of an angled-launched projectile)

4. What would happen if a projectile was launched horizontally in the absence of gravity?
 * The projectile would travel in horizontal line, traveling a constant value per second.
 * [[image:Untitled.png width="299" height="231"]]

5. What would happen if a projectile was launched at an angle 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 (straight line).

Perpendicular components of motion are independent of each other. The two components of motion are to be analyzed independently of each other. It is important not to //mix// horizontal motion information with vertical motion information. It is for this reason that one of the initial steps of a projectile motion problem is to determine the components of the initial velocity.
 * d) Initial Velocity Components**

If a projectile is launched at an angle to the horizontal, then the initial velocity of the projectile has both a horizontal and a vertical component. The analysis of projectile motion problems begins by using trigonometric methods to determine the horizontal and vertical components of the initial velocity.

There are two basic types of projectile problems:
 * e) Horizontally Launched Projectile Problems - Problem Solving**

A projectile is launched with an initial horizontal velocity from an elevated position and follows a parabolic path to the ground. 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.
 * Problem Type 1: **** Horizontal Launches **

Ground-to-ground. 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.
 * Problem Type 2: **** Ground-to-Ground Launches **

Because the two components of a projectile’s 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. Three common kinematic equations that will be used for both type of problems include the following:
 * Equations for the Horizontal and Vertical Motion of a Projectile **



Summary of problem-solving approach:
 * Solving Projectile Problems **
 * 1) List known and unknown information in terms of the symbols of the kinematic equations. Make a table with horizontal information on one side and vertical information on the other side.
 * 2) Identify the unknown quantity that the problem requests you to solve for.
 * 3) Select either a horizontal or vertical equation to solve for the time of flight of the projectile.
 * 4) With the time determined, use one of the other equations to solve for the unknown.

A non-horizontally launched projectile is a projectile that begins its motion with an initial velocity that is both horizontal and vertical. The solution of any non-horizontally launched projectile problem (in which vi and Theta are given) should begin by first resolving the initial velocity into horizontal and vertical components using the trigonometric functions.
 * f) Non-Horizontally Launched Projectile Problems - Problem Solving**

The following procedure summarizes the above problem-solving approach:
 * 1) Use the given values of the initial velocity (the magnitude and the angle) to determine the horizontal and vertical components of the velocity.
 * 2) Carefully read the problem and list known and unknown information in terms of the symbols of the kinematic equations.
 * 3) Identify the unknown quantity that the problem requests you to solve for.
 * 4) Select either a horizontal or vertical equation to solve for the time of flight of the projectile.
 * 5) With the time determined, use a horizontal equation to determine range of the projectile.
 * 6) The peak height of the projectile can be found using a time value that is half the total flight time.

__Activity__: Orienteering
__Displacement by measuring__: 25.30 m
 * Data:**


 * Analytical Vector Addition:**


 * Graphical Vector Addition:**


 * Percent Error Calculation:**

The measured resultant vector was 25.30 meters, the "analytical" resultant vector was 23.40 meters, and the "graphical" resultant vector was 23.80 meters. The analytical method yielded a 4.12% error, and the graphical method yielded a 6.30% error. Evidently, the analytical method was slightly more accurate than the graphical method, as it yielded a resultant vector closest to the actual measured resultant vector. One source of error could have occurred when doing the graphical method. My scale was very small, which made it difficult to get exact measurements. To fix this problem in the future, I would make my scale larger. Another source of error could have occurred while doing the analytical method. I rounded only two decimal places, which could have skewed my resultant vector value. To fix this problem, I would round my numbers to 4 decimal places, in order to yield a more accurate resultant vector.
 * Conclusion:**

__Lab__: Target Practice

 * Objectives:**
 * 1) Measure the initial velocity of a ball.
 * 2) Apply concepts from two-dimensional kinematics to predict the impact point of a ball in projectile motion.
 * 3) Take into account trial-to-trial variations in the velocity measurement when calculating the impact point.


 * Pre-Lab Questions:**
 * 1) If you were to drop a ball, releasing it from rest, what information would be needed to predict how much time it would take for the ball to hit the floor? What assumptions must you make?
 * In order to solve for time, I need the distance from the starting to the ending point. I would assume that the initial velocity is 0 m/s and that the acceleration is -9.8 m/s^2.
 * 1) If the ball in Question 1 is traveling at a known horizontal velocity when it starts to fall, explain how you would calculate how far it will travel before it hits the ground.
 * Because I know 3 out of the 4 y-component variables, I would solve for time using these variables in the equation d = (vi)(t) + ½(a)(t^2). Since the ball is traveling horizontally (theta = 0 degrees), the initial y-velocity would be zero. The y-acceleration would be -9.8 m/s^2 (acceleration of gravity), and the y-change in distance would be the distance from the shooter to the floor. I would solve for time, and use my result to solve for the x-change in distance (range). This is possible since the initial velocity is known, the acceleration is 0 m/s, and the time is known.
 * 1) A single Photogate can be used to accurately measure the time interval for an object to break the beam of the Photogate. If you wanted to know the velocity of the object, what additional information would you need?
 * To solve for velocity, it is necessary to know the hang time (which is calculated by the Photogate timer) and the object’s range (x-change in distance). We can assume that the horizontal acceleration is zero.
 * 1) What data will you need to collect? Remember that you must run multiple trials. Keep in mind your end goal!
 * We will need to collect the distance from the shooter (point where ball is launched) to the ground, and the distance from the shooter to each dot on the carbon paper. This is information is necessary to solve for the shooter's initial velocity at "medium range."
 * 1) How will you analyze your results in terms of precision and/or in terms of accuracy?
 * For the carbon paper test, I will analyze my results in terms of precision. To do so, I will shoot the ball many times and see where it lands on the carbon paper each time. If the dots are close together, it means that my results are precise. If the dots are very spread out and random, it means that my results are not precise.
 * For the "ball in the cup" experiment, I will analyze my results in terms of precision and accuracy. To analyze my results in terms of accuracy, I will compare where the ball actually lands to the theoretical position (theoretical position = avg. distance of all dots on carbon paper). I will do this using the percent error formula. To analyze my results in terms of precision, I will see if the ball landed in the same area each time. I will do so using the percent difference formula.


 * Procedure:**
 * a) Carbon paper measurement (no video)
 * 1. Shoot ball at "medium range"
 * 2. Put carbon paper on floor where ball seems to be landing
 * 3. Shoot the ball onto the carbon paper many times.
 * 4. Measure distance from shooter to each dot on carbon paper.
 * 5. Take the average of these distances.
 * b) Ball in cup
 * media type="file" key="Ball in cup movie.mov" width="300" height="300"

a) How fast does the launcher shoot the ball at "medium range"? b) Change initial height, calculate where to place cup on floor so that ball lands inside.
 * Calculations:**
 * [[image:vi_table.png]]
 * __Answer__: 6.85 m/s
 * [[image:111.png width="494" height="371"]]
 * __Calculated position__: 2.80 m away from shooter
 * [[image:1234.png width="497" height="238"]]


 * Analysis:**
 * __Theoretical position of cup__: 2.80 m away from shooter
 * __Actual position of cup where ball landed inside__: 2.55 m away from shooter
 * [[image:perror.png width="499" height="216"]]

The purpose of this experiment was to show students how it is possible to figure unknown information about a projectile based on other known variables. Ultimately, I applied my knowledge of projectiles to solve for the shooter's initial velocity at "medium range", which was 6.85 m/s. Knowing the initial velocity, I was able to solve for the ball's range. If we solved correctly, the ball would land in a cup placed at the end of the ball's range. Because our ball did not make it in the cup at the theoretical position, we must have miscalculated the shooter's initial velocity at "medium range." We had to move the cup much closer to the shooter for the ball to make it in. Because our experimental value was different than our theoretical value, we had a high percent error.
 * Conclusion:**

__Lab__: Shoot Your Grade
**Hypothesis:** If the 5 suspended hoops are correctly aligned, the ball should go through all of them and land in the cup at the end.


 * Purpose with Rationale: **
 * For this lab, my goal is to shoot a ball through 5 hoops and have it land in a cup at the end. My procedure involves solving mathematically for the exact position of each hoop and the cup. This procedure is logical because it ensures (theoretically) that the ball won’t hit the suspended hoops and veer off course.


 * Materials and Methods: **
 * Materials:
 * To shoot the ball, I used a shooter set at “medium range”. I used rings of masking tape to serve as hoops, and black string to hang them from the ceiling. I used a tape measurer to correctly mark the position of each hoop in relation to the shooter and ground. To take accurate measurements, I used a string with a pointy weight at the end. When dangled under a suspended hoop, this device helped determine the exact horizontal position of the hoop.
 * Procedure:
 * 1) Verify initial velocity
 * 2) Choose a horizontal position (in relation to shooter) for each hoop.
 * 3) For each hoop, use known x-variables to solve for time with d=(vi)(t)+(1/2)(a)(t^2) equation.
 * 4) For each hoop, use time variable to solve for y-distance with d=(vi)(t)+(1/2)(a)(t^2) equation.
 * 5) Hang each hoop at its calculated horizontal and vertical position.
 * 6) Place cup at ending position.
 * 7) Shoot ball from shooter at “medium range”.
 * [[image:diagram_hoops.png width="464" height="303"]]


 * Observations and Data from Initial Velocity: **
 * To verify the initial velocity, I shot the ball at 20 degrees at “medium range” onto carbon paper many times. I then measured the distance from the shooter to each dot, and took the average of these measurements.
 * [[image:verify_initial_velocity_chart_1.png]]
 * Using my known y-variables (vi = 0, a = -9.8m, and d = -1.1582m), I solved for time. I then used the time (0.45s) and the other x-variables (a = 0, d = 4.931m) to solve for initial velocity. My initial velocity was 6.759 m/s, which differed from my previously determined initial velocity (6.85 m/s). Verifying the initial velocity was important, because all subsequent calculations involved the initial velocity.

Best trial: media type="file" key="Shooter_FINAL.m4v" width="300" height="300"
 * Observations and Data from Performance: **


 * Physics Calculations: **
 * Assumptions and justifications:
 * I can assume that the vertical acceleration is always -9.8 m/s 2, because this is the acceleration of gravity on earth.
 * I can also assume that horizontal acceleration is always zero, because this is a known property of projectiles.
 * Calculation table for each hoop / cup:
 * [[image:hoop_position_xy_chart1.png]]
 * Sample calculations:
 * [[image:hoop_calculations_REAL.png width="448" height="235"]]
 * [[image:velocity_REAL.png width="513" height="330"]]


 * Error analysis: **
 * Data table
 * [[image:good_percent_error.png]]
 * Sample calculations:
 * [[image:percent_error_calc.png width="520" height="288"]]
 * __Note__: Percent error calculations were only done for the y-component of the hoops. This is because we didn't solve for a theoretical x-component. Instead, we based the horizontal distance of each hoop on where it was convenient to hang it.
 * As you can see, even though there was % error, the hoops still went through. This was a fairly accurate experiment.


 * Conclusion ** **:** Was the purpose satisfied?
 * Was your hypothesis correct? Provide __specific__ evidence from the experiment to justify your claims.
 * My hypothesis was correct, even though my group did not get the ball through all 5 hoops. Because my group was rushing to finish, we didn’t carefully align the fourth hoop. For this reason, the ball only went through the first, second, third, and fifth hoops. If we had taken the time to carefully align the fourth hoop, the ball would have went through all of them, therefore verifying my hypothesis.


 * Conclusion: ** Experimental Errors
 * How much? Where did the error occur? Why did the error occur? 2-3 sources of error should be described thoroughly.
 * The first source of error occurred while hanging up the hoops. Because the only way to adjust each hoop was by pulling the two attached strings, there was no way to accurately adjust it to the perfect position (both vertically and horizontally). When I tried pulling both strings evenly, one string always pulled the hoop up more than the other. Thankfully, this source of error did not greatly affect the ball’s trajectory. Because the hoop was much larger than the ball, the ball could still make it through even if the hoop’s position was slightly off. The second source of error occurred while shooting the ball. Sometimes after shooting the ball, the shooter shifted or the angle changed. This happened due to the force of the spring pushing the ball out of the shooter. Even though this may sound insignificant, it greatly affected our results. After every few trials, the ball would start veering way off track. When this happened, we knew that the angle and/or horizontal position of the shooter had changed. The third source of error occurred when the ball made contact with the hoop. One hoop in particular had the tendency to slip from its designated vertical position when grazed by the ball. This was a major source of error, as the hoop kept changing vertical position and needed constant adjustment.


 * Conclusion: ** Implications for further discussion
 * How would you change the lab to address the error? What is a relevant real-life application of this concept? (Why is this important to know/understand?)
 * To fix the problem of not being able to accurately adjust both strings, I would attach a tape measure to the strings. This would allow me to see exactly how much I adjusted each side of the hoop. To address the problem of the shooter shifting position, I would use two clamps to hold it down. I would also draw lines around the base of the shooter at its correct position. This would allow me to easily fix the shooter’s position if it shifted. To address the problem of the shooter changing angle, I would check the angle after each trial and make sure that it is at exactly 20 degrees. Last, to fix the hoop that kept changing position upon contact with the ball, I would tie a knot around the ceiling bar with the dangling end of each string. I would fasten this knot to the ceiling with masking tape. This would ensure that the hoop does not move vertically.
 * This concept can be applied to warfare. When dropping a bomb, it is very important to know information such as the bomb's initial velocity upon release, its horizontal and vertical distance from the target, and the time it takes to reach the target. Knowing how to figure out these things can be a matter of success or failure during a bomb raid.

__Project__: Gourd-O-Rama
Madison Steele & Andrea Aronsky


 * Pictures of final project:**

Calculations:


 * Results:**
 * __Mass of vehicle__: Mrs. Burns has this information.
 * __Time__: 4.21 seconds
 * __Distance traveled__: 11.0 meters
 * __Initial velocity__: 5.23 m/s
 * __Acceleration__: -1.24 m/s** 2 **

Our cart went extremely far because we used skateboard wheels with low friction bearings. These wheels carried our pumpkin 11.0 meters! However, if I were to redo this project, I would make sure to perfectly align the axles. Because our axles were a little crooked, our vehicle veered off and hit the wall. If they were aligned straight, our vehicle would have traveled a longer distance and had a lower deceleration. Also, I would have used lighter wheels to decrease the vehicle's mass.
 * Conclusion:**