Ch3_CantorA

=toc = =Labs =

Cafeteria Vector Lab 10/19
Ali, Jake, Jessica, Kayla Walk the Path and Find Destination. Measure displacement.  **Graphical Results**  **Percent Error for Graphical Results**  **Analytical Results**  **Percent Error for Analytical Results** 

Shoot the Grade 10/26
Ali, Jessica, Kayla, Jake <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 110%;"> **Purpose** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Our objective is to launch a ball from the launcher at an angle of 10 degrees and speed (6.92 m/s) so that the ball passes consecutively through five rings and lands in a cup on the floor. We are trying to see if our calculations will correctly place the hoops in the right place so that the ball will effectively project through. We have to factor in the pull of gravity (when concerning the projectile) at a force of -9.8 m/s/s and air resistance. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 110%;">**Prediction/ Hypothesis** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">The ball will launch successfully through the hoops based on the heights that we had calculated. The rings will begin by increasing in height and then eventuate into a shorter height, replicating a parabolic trajectory. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 110%;">**Materials** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;"> launcher, ball, carbon paper, masking tape, measuring tape, right angle clamps, plumb bomb, string <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 110%;">**Procedure** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">By using our typical projectile equations, we were able to calculate where the hoops should exactly be placed. After calculating, we left the notebook and went to the real world scenario: setting up tape hoops via ceiling tiles and string We then measured the heights which were calculated, for both the y and x axises, and placed each hoop in their designated location. In order to ensure that our markings were correct, we launched the ball each time we set up a hoop. After each trial, we would maneuver the hoop slightly to make sure the ball would go through. Eventually, with a little manipulation, we got through setting up all hoops so that the ball would successfully project through all, technically landing in the cup.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">BELOW is the video of the ball shooting thru four hoops <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">media type="file" key="4 hoops lol.m4v" width="330" height="330" <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 110%;">**Observations and Data** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Finding initial velocity was necessary when determining where to place the hoops. Below are our calculations. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Chart for finding distance and time to place hoops, relative to our initial height of 1.202. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 110%;">**Analysis** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: small;">below is a chart showing percent error, with a sample underneath. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: small;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: small;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 110%;">**Conclusion** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">We thought that the ball would be able to go through all of the hoops, because we felt that our calculations were very accurate; however, it ended up that our calculations were slightly off. From our calculations, we got that the initial velocity would be about 6.92 m/s. Based on this calculated value, we found two distance components of where to place the hoops: vertical and horizontal. As shown in the data above, our values were the following- hoop 1: 1.2742 m vertically and .7185 meters horizontally, hoop 2: 1.2374 m vertically and 1.437 m horizontally, hoop 3: 1.0924 m vertically and 2.156 m horizontally, hoop4: .83350 m vertically and 2.874 m horizontally, and lastly hoop 5: .47440 m vertically and 3.593 m horizontally. It turned out that our horizontal values did not have to be altered at all, but the vertical heights differed. We were able to manipulate the hoops slightly so that the ball would go thru the hoops, with the heights recorded above. If we had time we most likely could have got the last hoop to the right height so that the ball would launch through it. As you can see above, our percent error in each case is very small, suggesting that our original calculations were somewhat accurate. Sources to account for our error were the following: Since there were many groups setting up the hoops, often, the strings were moved, and in some cases they fell, forcing our group to start setting it up again. Another source of error was the angle of the launcher. Often, after several trials, the angle would slightly increase from 10 degrees to around 15 degrees. Despite constant checking for this error, it kept occurring. We couldn't be as accurate as possible. One way to change the lab to address the first source of error is to make sure that everyone sets up their launcher in separate locations. Therefore, there will not be any issue with a group altering another's set up. Another way to change the lab to address the second source of error would be to have an item that can hold the launcher in place (regarding its angle). It seemed that the screws on the side that were designed to hold the launcher in place were not 100% effective. Thus, finding another object to keep it in place would be the best solution. This can be applied to many real life situations, ranging from some careers to simple observations. It is good to know the logistics behind a projectile and to understand how it functions because you never know when you may need the information; it may be something as simple as calculating how to score a field goal or how to hit the tennis ball into the right spot on the court. It's also beneficial that physics can help in these situations, all with the result of a logical solution. This lab showed us the importance of trial and error and the importance of consistency.

=<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Class Notes =

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Intro to Vectors
<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; line-height: 0px; overflow-x: hidden; overflow-y: hidden;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Magnitude: the value without the sign <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Vector: quantity with both magnitude and direction <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Resultant: single vector that replaces the components <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">collinear: same line <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">noncollinear: different line
 * <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">scalar is a quantitative value
 * <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">ie-speed, time, mass, distance
 * <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">components are the perpendicular pieces added to make the vector
 * <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">ie- velocity, acceleration, displacement, force

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">biggest possible magnitude is adding, and the smallest is subtracting. these are our max and min.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Vector Addition
<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Graphical Methods <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">The Analytical Method <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Horizontally Launched Projectiles
<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">We use the same equations we used with 1D kinematics > <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Sample: <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">
 * <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">a= Vf- Vi/ Δt
 * <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Vf= Vi +aΔt (no displacement)
 * <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Δd= 1/2( Vi + Vf)t
 * <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Δd= Vit + 1/2at(squared)
 * <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Vf(squared)=Vi(squared)+2aΔd

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Ground to Ground
<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">the Y displacement ALWAYS equals ZERO <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Systems Problem <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Off the Cliff
<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">

=<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">** Activities ** =

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: small;">Launching Ball Activity (10/21)
<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Ali, Jake, Kayla, Jessica <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">How fast does the launcher shoot the ball at medium range? <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Change the initial height, calculate where to place the cup on the floor so that the ball lands inside of it 3 times in a row. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Below is the procedure we used to get the ball in the cup. As you can see, the ball entered the cup, but due to a lack of strength of the tape, it fell over. However, it did enter the cup, but rolled out. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">media type="file" key="New Project - Medium.m4v" width="300" height="300" <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Percent Error <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Gourdarama Project
<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Ali and Nicole <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">**Conclusions:** Our velocity was 5.263 m/s and our acceleration was -2.77 m/s/s. Our gourd traveled 5 meters and weighed about 1.3 kg (a very light weight, not to mention!!) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">If Nicole and I were to change anything, we would have probably made sure that the axiles were lined up straighter so that the mobile would not crash into the wall. By doing this, the rate of "acceleration" would have decreased because it would have covered a longer distance. As you can tell, the high rate of acceleration was our biggest problem. By cutting this rate down, our cart would have been more successful. We could have also used three wheels instead of four to cut down on the weight a little bit more!

=<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Homework =

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Vectors Lesson 1 a +b (10/12)
<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Vectors can only be described by magnitude and direction. An arrow diagram is used to display vector form, and can even be used as a quantitative value, where you can multiply, divide, subtract, and add vectors. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">**Vectors and Direction** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;"> A vector quantity is a quantity that is fully described by both magnitude and direction. While a scalar quantity is a quantity that is dfully described by its magnitude. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Examples of vector quantities displacement, velocity, acceleration, and force. Vector quantities are not fully described unless both magnitude and direction are listed. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Vector quantities are often represented by scaled vector diagrams.They depict a vector by use of an arrow drawn to scale in a specific direction. They are sometimes called free-body diagrams. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Below are some characteristics: <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Below are some examples of what this will look like:
 * <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">a scale is clearly listed
 * <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">a vector arrow is drawn in a specified direction. The vector arrow has a //head// and a //tail//.
 * <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">the magnitude and direction of the vector is clearly labeled

<span style="color: #ff0000; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 12pt;">**Vector Addition** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Two vectors can be added together to determine the result. The //net force// experienced by an object is determined by computing the vector sum of all the individual forces acting upon that object. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Example:

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">** The Pythagorean Theorem ** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">The Pythagorean theorem determines the result of adding ONLY two vectors that make a RIGHT angle to each other. The Pythagorean theorem is a mathematical equation that relates the length of the sides of a right triangle to the length of the hypotenuse of a right triangle.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">To see how the method works, consider the following problem: <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Eric leaves the base camp and hikes 11 km, north and then hikes 11 km east. Determine Eric's resulting displacement.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">The result of adding 11 km, north plus 11 km, east is a vector with a magnitude of 15.6 km.

<span style="color: #e72619; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 12pt;">**Using Trigonometry to Determine a Vector's Direction** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">The direction of a //resultant// vector can often be determined by use of trigonometric functions. Sine, cosine, and tangent relate an acute angle in a right triangle to the ratio of the lengths of two of the sides of the right triangle. Below are the equations.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">These three trigonometric functions can be applied to the hiker problem in order to determine the direction of the hiker's overall displacement.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Thus, trig functions can give us degree, but not always direction. We sometimes need to figure out direction ourselves.

<span style="color: #e72619; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 12pt;">**Use of Scaled Vector Diagrams to Determine a Resultant** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">The ** head-to-tail method ** is used to determine the vector sum. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">An example of the use of the head-to-tail method is illustrated below. The problem involves the addition of three vectors: <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%; text-align: center;">** 20 m, 45 deg. + 25 m, 300 deg. + 15 m, 210 deg. ** <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 90%; text-align: center;">** SCALE: 1 cm = 5 m **

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">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="color: #e72619; font-family: 'Trebuchet MS',Helvetica,sans-serif;">** SCALE: 1 cm = 5 m **

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">**FUN FACT:** The order in which three vectors are added has no affect upon either the magnitude or the direction of the resultant. The result will still have the same magnitude and direction.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 1.1em;">Vectors Lesson 1 c +d (10/13)
<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Vectors have resultants and components. Resultants are the *result* of the addition of two or more vectors. Components are the 2-dimensial breakdown of making up a vector.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">** Resultants **

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">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;">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; text-align: center;">** A + B + C = R **

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">The above discussion pertains to the result of adding displacement vectors. When displacement vectors are added, the result is a //resultant displacement//.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">** Vector Components **

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">A vector is a quantity that has both magnitude and direction. When there was a [|free-][|body][|diagram] depicting the forces acting upon an object, each individual force was directed in //one dimension// - either up or down or left or right. When an object had an acceleration and we described its 1-D direction. Now, we will have //two dimensions// - upward and rightward, northward and westward, eastward and southward, etc

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">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. For example, a vector that is directed northwest can be thought of as having two parts - a northward part and a westward part.



<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Each part of a two-dimensional vector is known as a ** component **.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">If Fido's dog chain is stretched upward and rightward and pulled tight by his master, then the tension force in the chain has two components - an upward component and a rightward component. To Fido, the influence of the chain on his body is equivalent to the influence of two chains on his body - one pulling upward and the other pulling rightward. If the single chain were replaced by two chains. with each chain having the magnitude and direction of the components, then Fido would not know the difference.



<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Any vector directed in two dimensions can be thought of as having two different components. The component of a single vector describes the influence of that vector in a given direction.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 120%;">Vectors Lesson 1 e (10/17)
<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">The resultants of vectors can be measured using two methods: the parallelogram method and the trigonomic method. Both require drawing diagrams and come out with similar results.

<span style="color: #2b87bd; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 12pt;">**Vector Resolution**

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">A vector directed in two dimensions can be thought of as having two components.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">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;">the parallelogram method
 * <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">the trigonometric method

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">** Parallelogram Method of Vector Resolution **

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">The parallelogram method of vector resolution involves using an accurately drawn, scaled vector diagram to determine the components of the vector. If one desires to determine the components as directed along the traditional x- and y-coordinate axes, then the parallelogram is a rectangle with sides that stretch vertically and horizontally.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Step by step breakdown:


 * 1) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Select a scale and accurately draw the vector to scale in the indicated direction.
 * 2) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">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) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">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
 * 4) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Meaningfully label the components of the vectors with symbols to indicate which component represents which side
 * 5) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">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;">**EXAMPLE**:



<span style="color: #e72619; font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 12pt;">**Trigonometric Method of Vector Resolution** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">The trigonometric method of vector resolution involves using trigonometric functions to determine the components of the vector. Trigonometric functions relate the ratio of the lengths of the sides of a right triangle to the measure of an acute angle within the right triangle. They 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;">Step by step breakdown:

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">**EXAMPLE**: <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">
 * 1) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">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;">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;"> 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) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Meaningfully label the components of the vectors with symbols to indicate which component represents which side
 * 5) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">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) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Repeat the above step using the cosine function to determine the length of the side adjacent to the indicated angle.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 120%;">Vectors Lesson 1f (10/18)
<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">** Component Method of Vector Addition ** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">When the two vectors are added head-to-tail as shown below, the resultant is the hypotenuse of a right triangle. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">A right triangle has two sides plus a hypotenuse; so the Pythagorean theorem is perfect for adding two right angle vectors.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">** Addition of Three or More Right Angle Vectors ** <span style="color: #000000; font-family: 'Trebuchet MS',Helvetica,sans-serif;">**Example 1:** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">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?

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">The head-to-tail vector addition diagram is shown below.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">The order in which the three vectors are added must be changed in order to make a right triangle. If the three vectors are added in the order 6.0 km, N + 2.0 km, N + 6.0 km, E, then the diagram will look like this:

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">After rearranging the order in which the three vectors are added, the resultant vector is now the hypotenuse of a right triangle. The lengths of the perpendicular sides of the right triangle are 8.0 m, North (6.0 km + 2.0 km) and 6.0 km, East. The magnitude of the resultant vector (R) can be determined using the Pythagorean theorem. <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">R 2 <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;"> (8.0 km) 2 + (6.0 km) 2  R 2 <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;"> 64.0 km 2 + 36.0 km 2  R 2 <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;"> 100.0 km 2  R <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;"> SQRT (100.0 km2) ** R = 10.0 km ** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">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.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">The resultant of the addition of three or more right angle vectors can be easily determined using the Pythagorean theorem. Doing so involves the adding of the vectors in a different order.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">** SOH CAH TOA and the Direction of Vectors ** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">The convention is known as the counter-clockwise from east convention, often abbreviated as the **CCW** convention. Using this convention, the direction of a vector is often expressed as a counter-clockwise angle of rotation of the vector about its //tail// from due East.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">**Example:** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;"> 6.0 km, N + 6.0 km, E + 2.0 km, N. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">In this problem, we know the length of the side opposite theta (Θ) - 6.0 km - and the length of the side adjacent the angle theta (Θ) - 8.0 km. The tangent function will be used to calculate the angle measure of theta (Θ). <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">Tangent(Θ) = Opposite/Adjacent Tangent(Θ) = 6.0/8.0 Tangent(Θ) = 0.75 Θ = tan -1 (0.75) Θ = 36.869 …° ** Θ =37° ** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">The **result** is 37° east of north. 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.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">In summary, the direction of a vector can be determined by finding the angle of rotation counter-clockwise from due east.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">** Addition of Non-Perpendicular Vectors ** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">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. ** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">**Example:** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">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.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">As is the usual case, the solution begins with a diagram of the vectors being added.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">The resultant is the vector sum of these three vectors; a head-to-tail vector addition diagram reveals that the resultant is directed southwest. Of the three vectors being added, vector C is clearly the //nasty// vector. Its direction is neither due south nor due west. The solution involves resolving this vector into its components.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Vector C makes a 30° angle with the southern direction. By sketching a right triangle with horizontal and vertical legs and C as the hypotenuse, it becomes possible to determine the components of vector C. This is shown in the diagram below. The side adjacent this 30° angle in the triangle is the vertical side; the vertical side represents the vertical (southward) component of C - C y. So to determine C y, the cosine function is used. The side opposite the 30° angle is the horizontal side; the horizontal side represents the horizontal (westward) component of C – C. The cosine function is used to determine the southward component since the southward component is adjacent to the 30° angle. The sine function is used to determine the westward component since the westward component is the side opposite to the 30° angle.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Now our vector addition problem has been transformed from the addition of two nice vectors and one nasty vector into the addition of four nice vectors.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">With all vectors oriented along are customary north-south and east-west axes, they can be added head-to-tail in any order to produce a right triangle whose the hypotenuse is the resultant.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">The triangle's perpendicular sides have lengths of 4.6 meters and 6.756 meters. The length of the horizontal side (4.6 m) was determined by adding the values of B (2.2 m) and C x (2.4 m). The length of the vertical side (6.756… m) was determined by adding the values of A (2.6 m) and C y (4.156… m). The resultant's magnitude (R) can now be determined using the Pythagorean theorem. <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">R 2 <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;"> (6.756… m) 2 + (4.6 m) 2  R 2 <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;"> 45.655… m 2 + 21.16 m 2  R 2 <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;"> 66.815… m 2  R <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;"> SQRT(66.815… m 2 ) R = 8.174 … m ** R = ~8.2 m **

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;"> <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">tangent(Θ) = (6.756… m)/(4.6 m) = 1.46889… <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">so… <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">Θ = tan -1 (1.46889…) = 55.7536… ° ** Θ = ~56° ** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">This 56° SW angle is the angle between the resultant vector. The (CCW) can be determined by adding 180° to the 56°. So the CCW direction is 236°.

<span style="color: #000000; font-family: 'Trebuchet MS',Helvetica,sans-serif;">**Example:** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Cameron Per (his friends call him Cam) and Baxter Nature are on a hike. Starting from home base, they make the following movements. <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">**A: 2.65 km, 140° CCW B: 4.77 km, 252° CCW C: 3.18 km, 332° CCW**

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">This is the angle measure diagrams.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">A table to organize values.

<span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">**A** <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">**2.65 km** <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">**140° CCW** || <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">(2.65 km)•cos(40°) <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">= 2.030… km, West || <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">(2.65 km)•sin(40°) <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">= 1.703… km, North || <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">**B** <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">**4.77 km** <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">**252° CCW** || <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">(4.77 km)•sin(18°) <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">= 1.474… km, West || <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">(4.77 km)•cos(18°) <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">= 4.536… km, South || <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">**C** <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">**3.18 km** <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">**332° CCW** || <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">(3.18 km)•cos(28°) <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">= 2.808… km, East || <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">(3.18 km)•sin(28°) <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">= 1.493… km, South || <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">**Sum of** <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">**A + B + C** || <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">0.696 km, West  || <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">4.326 km, South  || <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">R 2 <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;"> (0.696 km) 2 + (4.326 km) 2  R 2 <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;"> 0.484 km 2 + 18.714 km 2  R 2 <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;"> 19.199 km 2  R <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;"> SQRT(19.199 km 2 ) ** R = ~4.38 km ** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">
 * <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">** Vector ** || <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">** East-West Component **  || <span style="display: block; font-family: 'Trebuchet MS',Helvetica,sans-serif; text-align: center;">** North-South Component **  ||

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Tangent(Θ) = opposite/adjacent <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Tangent(Θ) = (4.326 km)/(0.696 km) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Tangent(Θ) = 6.216 <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Θ = tan -1 (6.216) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">** Θ = 80.9° ** <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;"> <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">It would be worded as 80.9° SW. Since west is 180° counterclockwise from east, the direction could also be expressed in the counterclockwise (CCW) from east convention as 260.9°.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">**FINAL RESULTANT:** 4.38 km with a direction of 260.9° (CCW).

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 17px;">Vectors Lesson 1 g+h (10/18)
<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">On occasion objects move within a medium that is moving with respect to an observer. In such instances, 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. 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. This resultant velocity is quite easily determined if the wind approaches the plane directly from behind. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;"> 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. 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. This is the same procedure that was used above for the headwind and the tailwind situations; only now, the resultant is not as easily computed. Since the two vectors to be added - the southward plane velocity and the westward wind velocity - are at right angles to each other, the Pythagorean can be used. 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. The direction of the resulting velocity can be determined using a trigonometric function. Since the plane velocity and the wind velocity form a right triangle when added together in head-to-tail fashion, the angle between the resultant vector and the southward vector can be determined using the sine, cosine, or tangent functions. The tangent function can be used. Like any vector, the resultant's direction is measured as a counterclockwise angle of rotation from due East. <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">The affect of the wind upon the plane is similar to the affect of the river current upon the motorboat. If a motorboat were to head straight across a river, 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. Motorboat problems such as these are typically accompanied by three separate questions:


 * 1) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">What is the resultant velocity (both magnitude and direction) of the boat?
 * 2) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">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) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">What distance downstream does the boat reach the opposite shore?

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">The first of these three questions: the resultant velocity of the boat can be determined using the Pythagorean theorem (magnitude) and a trigonometric function (direction). The second and third of these questions can be answered using the average speed equation.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">** ave. speed = distance/time. **A force vector that is directed upward and rightward has two parts - an upward part and a rightward part. That is to say, if you pull upon an object in an upward and rightward direction, then you are exerting an influence upon the object in two separate directions - an upward direction and a rightward direction. These two parts of the two-dimensional vector are referred to as components. 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. 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. All vectors can be thought of as having perpendicular components. In fact, any motion that is at an angle to the horizontal or the vertical can be thought of as having two perpendicular motions occurring simultaneously. These perpendicular components of motion occur independently of each other. Any component of motion occurring strictly in the horizontal direction will have no affect upon the motion in the vertical direction. Any alteration in one set of these components will have no affect on the other set.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 120%;">Vectors Lesson 2 a+b (10/19)
<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 120%;">Lesson A <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Central idea: A projectile is an object upon which the only force is gravity. Gravity causes a vertical motion, which causes a vertical acceleration.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 120%;">Lesson B <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Central idea: A projectile has two components: a vertical motion and horizontal motion. Vertical motion is effected by the downward force of gravity, and therefore is always changing. However, horizontal motion is unaffected by gravity, and therefore stays at a constant rate.
 * 1) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">What is a projectile?
 * 2) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">It is an object dropped from rest or thrown vertically upwards at an angle to the horizontal
 * 3) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">It is any object projected or dropped that continues in motion by its own inertia
 * 4) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">What forces affect upon a projectile?
 * 5) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">The only force is of downward gravity
 * 6) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">What is inertia?
 * 7) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">The resistance an object has to a change in a state of motion
 * 8) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">How do we represent projectiles?
 * 9) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">A free-body diagram is a good representation of projectiles.
 * 10) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">What type of motion will we see from this diagram?
 * 11) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">We will see a Parabolic Trajectory, similar to a diagram in free fall.
 * 1) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Can projectiles move in a direction other than vertical motion?
 * 2) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Yes, projectiles can move in a horizontal motion as well. Horizontal and vertical are both COMPONENTS of the projectile.
 * 3) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Does gravity affect the horizontal motion of a projectile?
 * 4) No, it does not. This is because gravity acts as a downward force, therefore unable to affect the horizontal motion.
 * 5) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">How does a projectile travel?
 * 6) <span style="font-family: arial,helvetica,sans-serif;">With constant horizontal velocity and a downward vertical acceleration.
 * 7) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Is the velocity of downward vertical motion changing or constant?
 * 8) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">It is ALWAYS changing, due to the account of gravity's acceleration of -9.8 m/s/s.
 * 9) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Is acceleration present in horizontal motion?
 * 10) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">No. It is always moving at a constant rate because it is not affected by gravity.

<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: 120%;">Vectors Lesson 2 c (10/20)
<span style="font-family: 'Trebuchet MS',Helvetica,sans-serif; font-size: small;">Central Idea: Depending on the projectile, it can either be moving against time solely, or considering the property of displacement.Both of these properties can be addressed numerically.
 * 1) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">What is the velocity of horizontal trajectory? vertical?
 * 2) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Constant and changing, respectively.
 * 3) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">What is the difference between free fall and a projectile? What are the equations for projectiles?
 * 4) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Free fall is an example of projectiles. However, in many cases, a projectile will have an initial velocity (as opposed to zero velocity) due to a launch or upward motion.
 * 5) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">What are the equations for certain projectiles?
 * 6) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">vertical displacement for a 1D object
 * 7) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">y= 1/2gt 2
 * 8) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">vertical displacement for a horizontally launched projectile
 * 9) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">X= V ix t
 * 10) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">vertical displacement for an angled-launched projectile
 * 11) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">v iy t+ 1/2gt 2
 * 12) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Can vertical motion increase? decrease?
 * 13) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">Yes, the diagram should most of the time look like a parabola. Therefore, the velocities will increase and decrease.
 * 14) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">What is the difference between a launched projectile over time versus one dealing with displacement?
 * 15) <span style="font-family: 'Trebuchet MS',Helvetica,sans-serif;">If displacement is a factor, the numbers will be increasing or decreasing (uniformed rate) for horizontal components.