Ch3_ShermanB

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Vectors, 1A

A vector quantity is a quantity that is fully described by both magnitude and direction. On the other hand, a scalar quantity is a quantity that is fully described by its magnitude. Vector quantities are not fully described unless both magnitude and direction are listed.

Vector diagrams depict a vector by use of an arrow drawn to scale in a specific direction. Commonly called as [|free-body diagrams].


 * a scale is clearly listed
 * a vector arrow (with arrowhead) is drawn in a specified direction. The vector arrow has a //head// and a //tail//.
 * the magnitude and direction of the vector is clearly labeled. In this case, the diagram shows the magnitude is 20 m and the direction is (30 degrees West of North).


 * Conventions for Describing Directions of Vectors **
 * 1) The direction of a vector is often expressed as an angle of rotation of the vector about its " [|tail] " from east, west, north, or south
 * 2) The direction of a vector is often expressed as a counterclockwise angle of rotation of the vector about its " [|tail] " from due East. A vector with a direction of 270 degrees is a vector that has been rotated 270 degrees in a counterclockwise direction relative to due east.



A rotation of 240 degrees is equivalent to rotating the vector through two quadrants (180 degrees) and then an additional 60 degrees //into the// third quadrant. The magnitude of a vector in a scaled vector diagram is depicted by the length of the arrow. The arrow is drawn a precise length in accordance with a chosen scale. __**// A vector is a unit that has magnitude and direction. It can be displayed on a vector diagram. //**__

Vectors, 1B Two vectors can be added together to determine the result (or resultant). // Net force // experienced by an object was determined by computing the vector sum of all the individual forces acting upon that object.

The Pythagorean theorem is a useful method for determining the result of adding two (and only two) vectors __that make a right angle__ to each other. The method is not applicable for adding more than two vectors or for adding vectors that are __not__ at 90-degrees to each other.
 * The Pythagorean Theorem **

The direction of a // resultant // vector can often be determined by use of trigonometric functions. Sine, cosine, and tangent functions. These three functions relate an acute angle in a right triangle to the ratio of the lengths of two of the sides of the right triangle.
 * Using Trigonometry to Determine a Vector's Direction **



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

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

The head-to-tail method involves drawing a vector to scale on a sheet of paper beginning at a designated starting position. Where the head of this first vector ends, the tail of the second vector begins (thus, // head-to-tail // method). Once all the vectors have been added head-to-tail, the resultant is then drawn from the tail of the first vector to the head of the last vector; i.e., from start to finish. Once the resultant is drawn, its length can be measured and converted to // real // units using the given scale. The direction of the resultant can be determined by using a protractor and measuring its counterclockwise angle of rotation from due East. 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) Choose a scale and indicate it on a sheet of paper. The best choice of scale is one that will result in a diagram that is as large as possible, yet fits on the sheet of paper.
 * 2) Pick a starting location and draw the first vector // to scale //in the indicated direction. Label the magnitude and direction of the scale on the diagram (e.g., SCALE: 1 cm = 20 m).
 * 3) Starting from where the head of the first vector ends, draw the second vector // to scale // in the indicated direction. Label the magnitude and direction of this vector on the diagram.
 * 4) Repeat steps 2 and 3 for all vectors that are to be added
 * 5) Draw the resultant from the tail of the first vector to the head of the last vector. Label this vector as ** Resultant ** or simply ** R **.
 * 6) Using a ruler, measure the length of the resultant and determine its magnitude by converting to real units using the scale.
 * 7) Measure the direction of the resultant using the counterclockwise convention.

An example of the use of the head-to-tail method is illustrated below. The problem involves the addition of three vectors:

Interestingly enough, the order in which three vectors are added has no affect upon either the magnitude or the direction of the resultant. The resultant will still have the same magnitude and direction.


 * // When vectors make a right angle with one another, the Pythagorean theorem or the head to tail method can be used to find the resultant. If the vectors don’t form a right angle, a scaled vector diagram is necessary. //**

** -Vectors C, Resultants **

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. 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 ** Any two vectors can be added as long as they are the same vector quantity. In all such cases, the resultant vector (whether a displacement vector, force vector, velocity vector, etc.) is the result of adding the individual vectors. It is the same thing as adding A + B + C + ... . "To do A + B + C is the same as to do R."


 * //__ A resultant is the sum of two or more vectors. __//**

**-5d, Vector Components** In situations in which vectors are directed at angles to the customary coordinate axes, a useful mathematical trick will be employed to //transform// the vector into two parts with each part being directed along the coordinate axes.

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



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.


 * //__Two dimensional vectors have one combination component which can be broken into two individual components.__//**

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1e-Vector Resolution The process of determining the magnitude of a vector is known as ** vector resolution **. The method involves drawing the vector to scale in the indicated direction, sketching a parallelogram around the vector such that the vector is the diagonal of the parallelogram, and determining the magnitude of the components (the sides of the parallelogram) using the scale. Select a scale and accurately draw the vector to scale in the indicated direction.
 * Parallelogram Method of Vector Resolution **
 * 1) 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).
 * 2) Draw the components of the vector. The components are the //sides// of the parallelogram. The tail of the components start at the tail of the vector and stretches along the axes to the nearest corner of the parallelogram. Be sure to place arrowheads on these components to indicate their direction (up, down, left, right).
 * 3) Meaningfully label the components of the vectors with symbols to indicate which component represents which side. A northward force component might be labeled Fnorth. A rightward velocity component might be labeled vx; etc.
 * 4) 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.

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


 * //__ Vectors resolutions can be measured graphically, or algebraically. Graphically requires the vector’s components to be drawn to scale and then have the measurments found by hand, and algebraically requires the Pythagorean theorem and trig. __//**

**Vectors G,** ** Relative Velocity and Riverboat Problems ** Motion is relative to the observer. The observer on land, often named (or misnamed) the "stationary observer" would measure the speed to be different than that of the person on the boat. The observed speed of the boat must always be described relative to who the observer is. To illustrate this principle, consider a plane flying amidst a ** tailwind **. A tailwind is merely a wind that approaches the plane from behind, thus increasing its resulting velocity.

If the plane encounters a headwind, the resulting velocity will be less than 100 km/hr.

http://www.physicsclassroom.com/Class/vectors/u3l1f3.gif

Now consider a plane traveling with a velocity of 100 km/hr, South that encounters a ** side wind ** of 25 km/hr, West. 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 theorem can be used. This is illustrated in the diagram below.

tan (theta) = (opposite/adjacent) tan (theta) = (25/100) theta = invtan (25/100) ** theta = 14.0 degrees ** 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 (that is, if the boat were to point its bow straight towards the other side), it would not reach the shore directly across from its starting point. The river current influences the motion of the boat and carries it downstream. The motorboat may be moving with a velocity of 4 m/s directly across the river, yet the resultant velocity of the boat will be greater than 4 m/s and at an angle in the downstream direction. While the speedometer of the boat may read 4 m/s, its speed with respect to an observer on the shore will be greater than 4 m/s.
 * Analysis of a Riverboat's Motion **

(4.0 m/s)2 + (3.0 m/s)2 = R2 16 m2/s2 + 9 m2/s2 = R2 25 m2/s2 = R2 SQRT (25 m2/s2) = R ** 5.0 m/s = R **

tan (theta) = (opposite/adjacent) tan (theta) = (3/4) theta = invtan (3/4) ** theta = 36.9 degrees ** Given a boat velocity of 4 m/s, East and a river velocity of 3 m/s, North, the resultant velocity of the boat will be 5 m/s at 36.9 degrees Motorboat problems such as these are typically accompanied by three separate questions: ** ave. speed = distance/time ** A motorboat traveling 4 m/s, East encounters a current traveling 3.0 m/s, North. The solution to the first question has already been shown in the [|above discussion]. The resultant velocity of the boat is 5 m/s at 36.9 degrees. We will start in on the second question. The river is 80-meters wide. That is, the distance from shore to shore as measured straight across the river is 80 meters. The time to cross this 80-meter wide river can be determined by rearranging and substituting into the average speed equation. ** time = distance /(ave. speed) ** The distance of 80 m can be substituted into the numerator. But what about the denominator? What value should be used for average speed? Should 3 m/s (the current velocity), 4 m/s (the boat velocity), or 5 m/s (the resultant velocity) be used as the average speed value for covering the 80 meters? With what average speed is the boat traversing the 80 meter wide river? Yet the value of 5 m/s is the speed at which the boat covers the diagonal dimension of the river. And the diagonal distance across the river is not known in this case. If one knew the ** distance C ** in the diagram below, then the ** average speed C ** could be used to calculate the time to reach the opposite shore. Similarly, if one knew the ** distance B ** in the diagram below, then the ** average speed B ** could be used to calculate the time to reach the opposite shore. And finally, if one knew the ** distance A ** in the diagram below, then the ** average speed A ** could be used to calculate the time to reach the opposite shore.
 * 1) What is the resultant velocity (both magnitude and direction) of the boat?
 * 2) If the width of the river is //X// meters wide, then how much time does it take the boat to travel shore to shore?
 * 3) What distance downstream does the boat reach the opposite shore?
 * ** Example 1 **
 * 1) What is the resultant velocity of the motorboat?
 * 2) If the width of the river is 80 meters wide, then how much time does it take the boat to travel shore to shore?
 * 3) What distance downstream does the boat reach the opposite shore? ||

In our problem, the 80 m corresponds to the distance A, and so the average speed of 4 m/s (average speed in the direction straight across the river) should be substituted into the equation to determine the time. ** time = (80 m)/(4 m/s) = 20 s ** It requires 20 s for the boat to travel across the river. During this 20 s of crossing the river, the boat also drifts downstream. The same equation must be used to calculate this //downstream distance//. And once more, the question arises, which one of the three average speed values must be used in the equation to calculate the distance downstream? The distance downstream corresponds to ** Distance B ** on the above diagram. The speed at which the boat covers this distance corresponds to ** Average Speed B ** on the diagram above (i.e., the speed at which the current moves - 3 m/s). And so the average speed of 3 m/s (average speed in the downstream direction) should be substituted into the equation to determine the distance. ** distance = ave. speed * time = (3 m/s) * (20 s) ** ** distance = 60 m ** The boat is carried 60 meters downstream during the 20 seconds it takes to cross the river.
 * //__ Motion is relative to the observer – headwinds, tailwinds, and side winds all change the motion of an object to the observer, but not to the object itself. __//**

**Vectors H** Independence of Perpendicular Components of Motion Any vector - whether it is a force vector, displacement vector, velocity vector, etc. - directed at an angle can be thought of as being composed of two perpendicular components. These two components can be represented as legs of a right triangle formed by projecting the vector onto the x- and y-axis.

The two perpendicular parts or components of a vector are independent of each other. A change in one component does not affect the other component. Changing a component will affect the motion in that specific direction. While the change in one of the components will alter the magnitude of the resulting force, it does not alter the magnitude of the other component. Perpendicular components of motion do not affect 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.
 * //__ X and Y components can be altered, but changing one won’t impact the other. __//**

** __Projectile Motion, A__ ** Main Idea: A projectile is an object that is acted on solely by gravity, and it only influences the vertical motion of the projectile.

Questions: 1.) What is a projectile? A - A projectile is an object where the only force acting upon it is gravity. It is an object that when projected or dropped continues in motion by its own inertia and is influenced only by the downward force of gravity.

2.) How do we display a projectile visually? A – A projectile can be displayed visually with a free body diagram

3.) How does Newtonian physics tie in with projectiles? A – Newton’s laws imply that forces are only required to cause acceleration. Ultimately, a force is not needed to keep an object in motion. Newton said that without gravity, objects would continue to travel with the same speed in the same direction.

4.) Are there any other factors that have an effect on a projectile? A – No. An object can’t be a projectile if other forces are acting on it.

5.) Does gravity impact the speed of a projectile? A – No. Gravity only changes the course of a projectile, but doesn’t slow it down or speed it up.

** __Projectile Motion B__ ** Main Idea: Projectiles have a parabolic trajectory because gravity causes the object to lose it’s straight trajectory and ultimately fall back to earth.

Questions: 1.) How does acceleration of gravity affect a projectile? A – The acceleration of gravity, 9.8m/s^2, acts on a projectile and changes its trajectory.

2.) What happens when an object is launched upwards, instead of horizontally? A – When not launched horizontally, a projectile would travel at an angle upwards and then start to free fall.

3.) Does parabolic trajectory mean that gravity is influencing the horizontal motion of the projectile? A – No. Parabolic trajectory means there is a change in vertical motion rather than horizontal motion.

4.) Why is the trajectory of a projectile a parabola? A – Because gravity is acting on it. When the projectile reaches a summit point, gravity causes the projectile to slow down and fall.

5.) How are free falling objects and projectiles similar? A – Both free falling objects and projectiles have acceleration due to gravity (-9.8m/s^2) and have a parabolic trajectory.

= = Main Idea: A projectile has a changing vertical velocity and acceleration, while a projectile has no acceleration and a constant horizontal velocity. = = Question: 1.) Is horizontal acceleration zero when a projectile is launched at an upward angle? <span style="background-color: transparent; color: #000000; font-family: Arial; font-size: 14.6667px; text-decoration: none; vertical-align: baseline;">A.) Yes, horizontal velocity is constant when a projectile is launched upward, so acceleration is zero.
 * <span style="background-color: transparent; color: #000000; display: block; font-family: Arial; font-size: 14.6667px; text-align: center; text-decoration: none; vertical-align: baseline;">Projectiles Motion C **

<span style="background-color: transparent; color: #000000; font-family: Arial; font-size: 14.6667px; text-decoration: none; vertical-align: baseline;">2.) What is the basic information needed to calculate vertical displacement for an projectile launched at an angle? <span style="background-color: transparent; color: #000000; font-family: Arial; font-size: 14.6667px; text-decoration: none; vertical-align: baseline;">A.) To calculate angled vertical displacement, you need the initial vertical velocity amount of time, and acceleration.

<span style="background-color: transparent; color: #000000; font-family: Arial; font-size: 14.6667px; text-decoration: none; vertical-align: baseline;">3.) How does a parabolic trajectory relate to projectiles in this section? <span style="background-color: transparent; color: #000000; font-family: Arial; font-size: 14.6667px; text-decoration: none; vertical-align: baseline;">A.) Since projectiles have a parabolic trajectory, we know that the trajectory is symmetrical, and the velocity of an opposite point can be calculated with its opposite.

<span style="background-color: transparent; color: #000000; font-family: Arial; font-size: 14.6667px; text-decoration: none; vertical-align: baseline;">4.) How does vertical velocity affect horizontal velocity? <span style="background-color: transparent; color: #000000; font-family: Arial; font-size: 14.6667px; text-decoration: none; vertical-align: baseline;">A.) Vertical velocity doesn’t affect horizontal velocity as the two are independent of one another.

<span style="background-color: transparent; color: #000000; font-family: Arial; font-size: 14.6667px; text-decoration: none; vertical-align: baseline;">5.) Does vertical acceleration increase in the positive direction? <span style="background-color: transparent; color: #000000; font-family: Arial; font-size: 14.6667px; text-decoration: none; vertical-align: baseline;">A.) No, vertical acceleration is always decreasing.

= Orienteering Activity = Data:
 * Step || Distance(cm) || Direction ||
 * 1 || 613 || N ||
 * 2 || 2690 || W ||
 * 3 || 900 || S ||
 * 4 || 2655 || E ||
 * 5 || 360 || S ||

Graphical Method:

Analytical Method:

Percent Error:

= Ball In A Cup Activity =

Data, Part A

Data, Cup location calculation Experiment Results, Percent Error



Video: media type="file" key="ballcup.mov" width="300" height="300"

Discussion Of Error and Results: In our experimental results, we found a percent error of 1.44% and a distance of 2.78 meters. This percent error was acceptable, and was close to our theoretical. I believe that there are several forms of error that cause these varied results. The first place for error that could've caused a change in our results would be the launcher was moved throughout testing. The grip used to hold on the launcher only held the front of it, while pushing too hard on the launcher allowed for it to move slightly backwards, or it could very easily change in the left and right directions. For example, when we used the plunger to put the ball inside the launcher, the force used to push the ball could have moved the launcher, and caused it to go in a different direction, which would change the distance. Also, when we got the ball in the cup, it would knock the cup over and move it. Although we placed tape on the bottom of the cup to hold it down, this was not effective. We later placed tape on the floor under the cup, but depending on where we measured on the tape, our results could have been slightly off.

= Shoot Your Grade = Joe, Ben, Sarah Purpose and Rationale: In the shoot your grade lab, out purpose is to use our knowledge of projectiles and their trajectories to calculate what we think its trajectory will be. Our group was given an angle of 15 degrees, which we have to launch our projectile at. This question is an "off the cliff" type question. In the case of this experiment, the launcher is at the top of the cliff, and will land on the floor (or hopefully in the cup), which is the ground part of the question. We will use prior knowledge to space out tape rolls, and calculate the where they should be placed based on several factors that we must calculate. This includes initial velocity of the projectile, the range, max time, and the "height of the cliff."Hypothesis: I think that the projectile will travel in a parabolic trajectory, and our calculated values should be very similar to (if not the same as) our experimental values.

Materials and Methods - In our experiment, we used five tape rolls to get the ball through and a .115m high cup to get the ball into. We also used thread to suspend the tape from the ceiling, and duct tape to not only hold the thread to the ceiling, but also to the table that was adjacent to our hoops. One strategy that helped us was to use binder clips attached to the framework of the ceiling, and route the thread through them. This allowed us to easily adjust the thread, as well as quickly secure it. Also used was a laptop to record the video, the projectile launcher and ball to go with it, and several types of measuring devices. We used a standard CM/IN ruler for short distances, a meter stick to measure from the counter top to the floor, and a measuring tape to do all other measurements. To load the launcher, we used a device that would push the spring into place, and to fire, we would pull the string straight up from the launcher. The launcher itself, which contained the ball, was rotated at 30 degrees. We would continually launch the ball throughout tens of runs to make the adjustments needed to get the ball through the hoops. To find the range, we used carbon paper taped to standard printer paper and the ground, which would be hit with the ball in clusters, allowing us to average the distance. Lastly, we used our notebooks and calculators to do the calculations to determine the theoretical trajectory. Observations and Data From Initial Velocity: This chart is the distances from the launcher that were found from firing the ball and having it land in clusters on the carbon copy paper. For the five runs, the ball had scattered landing points, but it averaged to 4.7m, which is our range.
 * Run || Distance(m) ||
 * 1 || 4.76 ||
 * 2 || 4.65 ||
 * 3 || 4.71 ||
 * 4 || 4.68 ||
 * 5 || 4.72 ||
 * Avg: || 4.7 ||

Observations and Data from Performance: This video shows our ball being launched and traveling through all five of the hoops. It represents our best run. media type="file" key="5 hoops.mov" width="300" height="300"

In this chart are our experimental results on the X and Y axes.
 * Ring: || Horizontal Distance (x) (m) || Vertical Distance (y) (m) ||
 * Ring 1: || 1.10 || 1.3323 ||
 * Ring 2: || 1.76 || 1.3126 ||
 * Ring 3: || 2.29 || 1.244 ||
 * Ring 4: || 2.71 || 1.1313 ||
 * Ring 5: || 3.22 || .8886 ||
 * Cup: || NEVER ACHIEVED || NEVER ACHIEVED ||

Physics Calculations: This chart shows our theoretical values for the horizontal and vertical axes, as well as time
 * || Time (s) || Horizontal Distance away; //x values// (m) || Height of the rings; //y values// (m) ||
 * Ring 1: || .165 || 1.10 || 1.331 ||
 * Ring 2: || .26 || 1.73 || 1.302 ||
 * Ring 3: || .34 || 2.26 || 1.21 ||
 * Ring 4: || .407 || 2.71 || 1.084 ||
 * Ring 5: || .484 || 3.22 || .885 ||
 * The Cup: || .706 || 4.7 || .115 ||

Error Analysis

This chart shows our theoretical and experimental values side by side: This chart shows our percent error:
 * || Horizontal Distance away; //x values// (m) || Height Vert. (y) Theor. (m) || Height Vert (y) Exp. (m) ||
 * Ring 1: || 1.10 || 1.331 || 1.3323 ||
 * Ring 2: || 1.73 || 1.302 || 1.3126 ||
 * Ring 3: || 2.26 || 1.21 || 1.244 ||
 * Ring 4: || 2.71 || 1.084 || 1.1313 ||
 * Ring 5: || 3.22 || .885 || .8886 ||
 * The Cup: || 4.7 || .115 || n/a ||
 * Ring: || Horizontal: || Vertical: ||
 * Ring 1: || 0% || 0% ||
 * Ring 2: || 1.73% || .814% ||
 * Ring 3: || 1.33% || 2.81% ||
 * Ring 4: || 0% || 4.36% ||
 * Ring 5: || 0% || .407% ||
 * Cup: || n/a || n/a ||
 * Avg: || .612% || 1.68% ||

Conclusion: In conclusion, I feel that my hypothesis has been proved correct. After our tens of trials, we were finally able to get the ball through the five hoops, but weren't able to attempt the cup, as the launcher was inconsistent. My hypothesis stated that it is possible to calculate the trajectory of a projectile, and position the hoops accordingly so that the ball passes through them. I also stated that our theoretical values would undoubtedly be very close to our experimental values. After collecting our data from the experiment, I have also found this to be true. Of our 10 different percent errors (five for each hoop positioning, and a vertical and horizontal percent error for each), the highest was 4.36 percent, which would be considered acceptable, if not good. After that was 2.81 percent, which is also fantastic. Of the 10 different percent errors, four of them were zero, and two were under one percent, which is extremely good. These percent errors show that the difference between our calculated values and experimental values was very minor, which means that we did a great job at predicting the trajectory of the projectile, which proves both parts of my hypothesis.

Of our percent errors, the five vertical ones had a higher average than the horizontal percentages. In fact, the vertical percent error average was around 300% larger than the horizontal percentages. Throughout the experiment, I was the official launcher of the projectile. Throughout my many launches, I found that the launcher was inconsistent on an extreme magnitude. For example, after we got the ball through all five hoops, we relaunched and weren't even able to get past the first. I found that the adjustable part of the launcher, that changes the angle, was not only loose, but would loosen extremely quickly after being tightened. This piece would change the angle of the launch by a fraction of a degree whenever I pulled the string to launch it, which was directly upward, most of the time. What would happen is that the front screw would loosen, and by pulling the launch string up, it would move the entire apparatus, changing the angle a different amount on every launch. When I tried holding this piece during launch, I found that the entire launcher would move left or right, depending on where it was held. Because of these two significant issues, I had to reposition the launcher almost every trial, as it never stayed constant. As a result, we were only able to replicate our five hoop pass through once or twice, and it was practically luck that determined whether or not it would go though. If we would attempt to compensate for the change in angle (even by a small amount, it wouldn't stay at 15%, but rather fluctuate around by several fractions of a degree), we would attempt another launch and find that the compensation was useless. This relates to the severe difference between the horizontal and vertical percent error averages because the vertical values had more severe changes than the horizontal values. If one were to look at our hoops, they would see that the the hoops weren't in a straight line, but curved to the left severely. This is another fault in the experiment; the launcher didn't launch the ball straight, it launched it at a curve on the vertical axis, which accounts for the increased percent error on the vertical axis. Lastly, one issue that we encountered was that the string would come loose from the binder clips, and change either slightly, or at a high amount. So, even if we taped the string on the clip, sometimes we would hit a ring directly, and it would fall off, or would fall on one side.

To fix the many fault points in this experiment, the main area of inconsistency. the launcher, should be changed. The launcher was so inconsistent and unreliable that it made the experiment extremely difficult, in that repeating a trial would not yield the same results, or results that were even remotely similar. If we had launchers that were more reliable, it would make the experiment much easier and more consistent. For example, if the launcher was pneumatic and launched the ball at a certain pressure, it would remove one area of fault - that the spring had too many of factors associated with it, and changing one (such as amount of trials run, which would increase its temperature and impact the projectile's launch velocity or angle) would result in a severe change in the experimental results. Another way to help make the lab more reliable would be using a different way to mount the string to the ceiling. For example, if we used something that had a stronger adhesive, or tied the string around the ceiling structure, it could be rid of the unreliable taping method used. A real life application of this could be the mounting of a light fixture in a gym. The light is bound to get hit repeatedly since it's in a gym, and depending on the way the light is held in place, being hit repeatedly could cause it to fall.

= Gourd-O-Rama = Images:

Distance: 11.6m Time: 6.92s

Conclusion: Our project did very well. It traveled 11.6m, which is significantly further than most of vehicles of my peers. In our four launches, the vehicle also proved to be fairly consistent. It traveled within 8m to 8.75m for our first three trials, and then pulled 11.6m on our final run. After trying different pumpkin placements in the last two runs, we found that the placement in the fourth run clearly had a strong impact on the vehicle's distance. In this run, the pumpkin was taped to the far back of the vehicle, which severely increased it's range. I think this is so because having the heaviest part of the vehicle relocated to the back gave it more inertia, allowing it to go further. If I was to redo this project, I would strive to make it lighter. The vehicle had some unnecessary added weights, such as hearty amounts of tape and a little more wood than necessary, but we tried to cut weight by using a light wood called lawon board. I would also use a metal axle instead of a wooden dowel, because I think that although the metal axle could potentially weigh more, it would allow the vehicle to travel further.

D=.5(Vf+Vit)t 11.6=.5(6.92)(vi) 11.6=3.41vi vi=3.35m/s

Vf=Vi+at 0=3.35+a(6.92) -3.35=6.92a a=.494m/s^2