#$&*
course Phy 231
4-12-11 about 11:00am
Ball and Ramp Projectile BehaviorAbout 1.5 hours
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The basic activity here is to allow a ball to roll off a gently inclined ramp and fall to the floor, observing its horizontal range..
As the ramp gets steeper the landing position of the ball changes.
You are going to see how far the ball travels in the horizontal direction, after leaving the ramp, for three different slopes.
Then you'll use a complicated formula to figure out how fast the ball was traveling as it left each ramp, and you'll use these results to figure out the corresponding accelerations.
The ultimate goal is to figure out how the acceleration of the ball changes with the slope of the ramp.
You will set up the longer of the two grooved ramps in your lab kit as shown in the picture below, on the edge of a table or counter between half a meter and a meter above a level floor.
Two sheets of paper, as shown, will be placed on the floor (perhaps with the rectangular piece of plywood that came with your kit used as a backing to protect your floor) so that when the steel ball rolls down the incline and off the end, it will fall freely to the floor and strike the paper. The small piece of carbon paper that came with your kit will be placed on top of the paper, giving you a clear mark on the paper when the ball strikes. The carbon paper might be stapled inside the pack of copies of paper rulers; if it is not there it should be clearly visible when you unpack your lab kit.
The system should be within reach of your computer, so you can make some timing measurements.
To begin the setup, place a piece of 8.5 x 11 inch paper on a table, desk or counter as shown in the picture below. An 8.5 inch edge of the paper should be perfectly aligned with the edge of the table, and a single domino aligned with that edge, as shown. Place a stack of 2 dominoes, aligned with the other edge of the paper, as shown.
Place the long grooved ramp across the two stacks of dominoes so that the lower end protrudes 2 cm beyond the domino at the edge of the table.
Observe what happens when a ball is released from the top of the ramp:
Release the ball from rest at the far end of the ramp and let it roll off the edge of the ramp and fall freely to the floor (the ball will fall while also traveling in the horizontal direction).
Place a mark or an object at the approximate position where the ball hit the floor.
Notice that the ramp will vibrate a bit as the ball rolls down, and will as a result tend to slide on the dominoes. Use some means to prevent this (tape can be helpful), but be sure that whatever you do it doesn't interfere with the slope of the ramp or the motion of the ball.
Determine where a ball dropped from just below the edge of the ramp lands:
Position the piece of thin plywood that came with your kit so that when dropped from directly below the edge of the ramp the ball strikes near the middle of the plywood.
Place a sheet of paper on top of the plywood.
This paper will move during trials of the experiment, and needs to be in the same position at the beginning of each trial.
Devise the best means you can to ensure that, when the paper moves, you can place it back in exactly the same position it previously occupied.
Place the small piece of carbon paper that came with your kit over the piece of paper, with the carbon side facing the paper so that when the ball is dropped, it will make a mark on the paper.
Drop the ball from directly below the edge of the ramp and observe the landing position.
If the paper has moved, reposition it back to its original position and drop the ball again.
Repeat until you've dropped the ball 5 times, obtaining 5 indications of the landing position.
The landing should be very close together, but nobody is good enough to hit the exact same spot 5 times in a row. However, you should do your best.
If you think you can improve your consistency, give it another try. It doesn't take long to drop the ball 5 times.
Determine where a ball allowed to roll down the ramp and to fall freely to the floor will land:
Now move the board (or book) to the position where the ball previously landed after rolling from rest down the ramp. The board should be positioned so the ball hits close to the center.
Reposition the first piece of paper to its original position, then tape a second piece, and if necessary a third, so you have a continuous sheet reaching to the new position of the board.
Place the carbon paper face-down over the paper at the position where the ball is expected to hit (if you don't have carbon paper, you might place the ruler at this position)
Mark the starting point for the ball on the ramp, so you can be sure you are starting the ball from the same position every time.
Allow the ball to roll from rest down the length of the ramp and fall freely to the floor, and note the point where the resulting mark is made on the paper.
If the ball missed the carbon paper then simply reposition the paper so that the first sheet is again in its original position, change the position of the carbon paper as appropriate, make sure the ramp is in its original position and try again.
Repeat until you have five marks indicating the position where the ball lands. Be sure you reposition the paper for every trial, since the paper is likely to move slightly with every impact. The paper must always be in its original position before the ball strikes it.
Your paper will look something like the one pictured below, with 5 marks clustered at the straight-drop point and the other at the projectile-landing point..
Measure the positions of the marks:
Place your papers on a tabletop or countertop.
Draw a straight line from the center of the group of marks resulting from straight drops to the center of the group of landing positions that result from the ball being rolled down the ramp. We will refer to the first as straight-drop marks and the second as landing-position marks.
Extend the line so that it goes beyond the last mark on each end.
This line should be very close to the average line along which the ball moved. We will call this line the axis of motion.
Through the center of each mark draw a line perpendicular to the original line, and extending about 3 cm on either side of that line.
Mark a point on the axis as your origin.
Your markings will look similar to those pictured below, with an axis, an origin and a 10 lines crossing the axis to indicate the distances along the axis at which each the ball landed for each trial. Make your lines must more nearly perpendicular to the axis than some of those depicted below:
Measure the position at which each of the short lines crosses the axis.
Measure also the vertical drop of the ball. Measure the vertical distance from the bottom of the ball as it leaves the ramp to the piece of plywood directly below.
The measurements you report here must be in centimeters. If your measurements need to be converted to standard cm, please do so.
In the space below, in comma-delimited format report in the first line the five positions resulting from straight drops. In the second line report the five landing positions of the ball after rolling down the ramp. In the third line report the vertical drop. Again, be sure all measurements are in cm.
Starting in the fourth line briefly explain how you obtained your measurements.
-------->>>>>>> straight drops, landing positions, vertical drop
Your answer (start in the next line):
5.30, 5.51, 5.70, 5.81, 6.00
14.95, 15.00, 15.20, 15.20, 15.31
75.40
I followed the rules for set-up as followed, did 5 straight line drops, did 5 ramp drops, and recorded the positions of each. I measured each line's distance from the origin to obtain my measurements.
#$&*
Find the mean and standard deviation of your five straight-drop positions and your five landing position:
Using the data analysis program convert a copy of the lines in the above space to columns (just copy the lines and click on Change Rows to Columns).
Click the Mean and Standard Deviation button, and note the mean and standard deviation of the first column.
Then click on the Select Column button and specify the second column. Isolate those values and compute their mean and standard deviation.
Report the mean and standard deviation of the straight drop positions in the first line of the space below, and the mean and standard deviation of the 5 landing positions in the second line, using comma-delimited format in each line.
Starting in the next line briefly indicate how you obtained your results.
-------->>>>>>>>>> mean & sdev straight drop, same for landing positions
Your answer (start in the next line):
5.664, .2701
15.13, .1512
I put the data in the data program as instructed to obtain my results.
#$&*
You will now place your taped-together sheets of paper back in the original position, add first one then two dominoes to the stack, and obtain something that looks very much the pictures shown below. Specific instructions follow the pictures.
Add another domino to the stack (the stack will now contain 3 dominoes) and obtain 5 more marks for the new ramp slope.
You may use the same paper you used previously, being sure to position it exactly as you did before.
Be sure to keep the overhang of the ramp consistent from one setup to another, so the marks made with the straight drop continue to apply to a straight drop.
Be sure the ramp does not slip due to vibrations of the rolling ball, and carefully recheck the ramp position with each trial.
You will obtain 5 new marks on your sheet of paper.
Measure the positions of those marks relative to the origin of your axis.
Report the positions of the five new marks, in comma-delimited format on the first line below. Starting in the second line give a brief explanation of the meaning of your data and how they were obtained.
---------->>>>>>>> 3 dominoes five new positions, explanation
Your answer (start in the next line):
22.50, 22.25, 22.28, 22.35, 22.48
I added one more domino to my set-up, did 5 more drops, and measured the distance of the drops from the origin.
#$&*
Repeat the procedure once more with a fourth domino added to the stack, obtaining 5 new marks.
Report the positions of the five new marks, in comma-delimited format on the first line below:
---------->>>>>>>> 4th domino 5 positions
Your answer (start in the next line):
25.40, 25.70, 25.72, 26.00, 26.15
#$&*
You will now determine for each setup the mean distance traveled by the ball after leaving the ramp:
Report again the mean and standard deviation for the 2-domino setup, using comma-delimited format in the first line below.
Find the mean and standard deviation of the 5-mark trials for the 3-domino stack and report them in the second line.
Report in the third line the same information for the 4-domino stack.
The mean distance traveled by the ball for each setup will be the magnitude of the difference between the mean straight-drop position and the mean projectile-landing position.
In the fourth, fifth and sixth lines below report the respective mean horizontal distances traveled by the the ball, as it falls, from the 2-, 3- and 4-domino setups. Report a single number in each line.
Starting in the seventh line give a brief explanation of the meaning of your numbers and how they were obtained.
---------->>>>>>>> mean std dev landing positions 2 dom, same 3 dom, same 4 dom, mean, mean horiz dist 2 dom, same 3 dom, same 4 dom
Your answer (start in the next line):
15.13, .1512
22.37, .1139
25.79, .2909
15.13
22.37
25.79
These numbers are the mean horizontal distances of the ball drops. The 2nd three numbers in the first three lines are the standard deviations, or the average distance from the mean drop for the 2-, 3-, and 4- domino set-ups.
#$&*
From the angle of the ramp, the vertical drop and the horizontal distance traveled by the ball after leaving the ramp, it is possible to use the equations of uniformly accelerated motion to determine the velocity of the ball as it left the ramp.
First do an approximate calculation, based on the not-quite-accurate assumption that the ball is moving only in the horizontal direction as it leaves the ramp, with 0 velocity in the vertical direction:
When you dropped the ball, it accelerated from rest at 980 cm/s^2 until it hit the floor. Choosing downward as the positive direction, the initial velocity of the ball was v0 = 0, its acceleration was a = 980 cm/s^2, and its displacement `ds was equal to the ramp-to-floor distance you measured (actually since you dropped the ball from below the ram the displacement was a little less, but ignore that difference and calculate based on the distance you measured).
Using the 'fourth' equation vf^2 = v0^2 + 2 a `ds, what therefore was the final velocity of the ball?
Using this information you can easily either use another equation, or use direct reasoning, to find the time required for the ball to reach the floor.
Report that time in the first line of the space below, and starting on the second line explain in detail how you obtained your result.
---------->>>>>>>> time of fall, explanation
Your answer (start in the next line):
0.392
vf^2 = v0^2+2a'ds
vf^2 = 0^2 + 2 (980 cm/s^2) (75.40 cm)
vf^2 = 2 (73892 cm^2/s^2)
vf^2 = 147784 cm^2/s^2
'sqrt(vf) = 'sqrt(147784 cm^2/s^2)
vf = 384.43 cm/s
vAve = (v0 + vf) / 2
vAve = 384.43 cm/s / 2
vAve = 192.22 cm/s
'dt = 'ds / vAve
'dt = 75.40 cm / 192.22 cm/s
'dt = 0.392 s
#$&*
If in every trial the ball required this same time to fall, then based on your mean distances, what was the horizontal velocity of the ball for each of the three setups? Give your answers in comma-delimited format in the first line, in order from the least to the greatest. Starting on the second line explain how you obtained your results, including at least one sample calculation.
---------->>>>>>>> horiz vel each setup, explanation incl sample calc
Your answer (start in the next line):
38.60, 57.07, 65.79
I divided the 'ds by the 'dt for each mean distance: 15.13 cm / 0.392 s = 38.60 cm/s (for the 2-domino set-up)
#$&*
The above analysis was not completely accurate, since the vertical velocity of the ball was not actually zero in any case.
In every case the ramp was sloped, so the ball actually began its fall with a nonzero velocity in the vertical direction.
Would this result in an actual time of fall that is greater or lesser than the one you calculated?
Would this cause the estimates you just made of the horizontal velocity to be greater or less than the actual horizontal velocities?
Would the relative error in your velocity calculation have been greater or less on the steepest ramp, as opposed to the least-steep ramp?
Was the actual speed of the ball as it left the ramp greater or less than its horizontal velocity, and why?
Can you estimate by how much the horizontal velocity would differ from the actual speed of the ball?
Answer the above series of questions in the space below:
---------->>>>>>>> effect of nonzero vertical velocity
Your answer (start in the next line):
The slope of the ramp would decrease the actual time of fall because, although the ball is traveling farther in the horizontal direction, it is coming off of the ramp faster so it is completing the trial at a more rapid pace than it would just being dropped.
This decrease in fall time would cause the estimates I just made of the horizontal velocity to be less than the actual horizontal velocities calculated.
The relative error of my velocity calculation would be greater on the steepest ramp as opposed to the least-steep ramp because it is the most different from the straight drop in terms of speed and horizontal distance.
The actual speed of the ball as it left the ramp is greater than the velocity because the velocity is affected by the displacement. If I were to let the ball keep rolling on the table after the ramp, the velocity may be higher because the displacement would be much larger in that situation. The drop from the table lessens the velocity, but not the speed.
I'm honestly not very sure how to even guess how much the velocity would differ from the speed.
@& The ball is moving mostly in the horizontal direction, but also, since the ramp is sloped, it moves in the vertical direction as well. Its speed is related to its horizontal and vertical speed by the Pythagorean Theorem.*@
@& It would be possible to sketch the relevant right triangle and estimate the ratio between the hypotenuse and the horizontal leg. The two lengths would be very close, but the horizontal leg will be slightly shorter.*@
#$&*
The complete analysis of the motion of the ball follows.
This analysis neglects the effects of air resistance, which at the level of precision attained in this experiment are insignificant. It also neglects edge effects at the end of the ramp (the ball does remain in contact with the ramp for a short time after it begins its fall; at the speeds encountered here that time will be very short and the effects are likely negligible).
General College Physics and University Physics students will be expected to understand the analysis; Principles of Physics students should understand the general scheme and are encouraged, but not required to understand all the details of this analysis.
You can skim (skim, not skip) this analysis now and come back to it after completing the experiment.
The ball leaves the ramp with an unknown velocity, which we will call v, at an angle theta below horizontal. We can easily determine theta from the slope of the ramp (tan(theta) = ramp slope).
The ball therefore has an initial downward vertical velocity v0_y = v sin(theta) and initial horizontal velocity v0_x = v cos(theta).
Let clock time be t = 0 at the instant the ball leaves the ramp. The horizontal velocity is constant so at a later clock time t the x position of the ball will be
x = v0_x * t.
Let the downward direction be positive. Then the initial vertical velocity is in the positive direction, as is the acceleration of gravity, so the vertical position at clock time t will be
y = v0_y * t + 1/2 g t^2.
v0_x and v0_y can both be expressed in terms of the unknown initial velocity v and the known angle theta.
At the instant of impact, x will be equal to the horizontal range of the projectile, and y will be equal to its distance of fall. Using x_range and y_fall for these distances we have the two equations
x_range = v0_x * t and
y_fall = v0_y * t + 1/2 g t^2.
Writing v cos(theta) and v sin(theta) for v0_x and v0_y the equations become
x_range = v cos(theta) * t and
y_fall = v sin(theta) * t + 1/2 g t^2.
All quantities in these equations are known, except t and v. So we have a system of two simultaneous equations which can be solved for v and t. (note that though we haven't yet solved for theta, we could do so at any time, given the information we have for the slope)
The solution is fairly straightforward. We solve the first equation for t, obtaining t = x_range / (v cos(theta) ). Then we plug this into the second equation to obtain
y_fall = v sin(theta) * (x_range / (v cos(theta)) + 1/2 g ( x_range / (v cos(theta))^2
which simplifies to
y_fall = x_range * tan(theta) + 1/2 g x_range^2 / (v^2 cos^2(theta))
You should be able to write this equation in standard mathematical notation, and when working through this analysis for yourself you should do so. For easy reference the equation looks like this:
It is straightforward to solve this equation for v. We obtain
v = +- sqrt( 1/2 g x_range^2 / (y_fall - x_range * tan(theta) ) / cos(theta).
Since we are only interested in the speed of the ball, we use the positive solution.
In standard simplified form the positive solution would be represented as follows::
We still haven't figured out theta.
We could figure out theta (theta = arcTan(ramp slope); use the calculator to get theta and then cos(theta) and sin(theta)) but we probably won't.
We can more easily use the sides of our ramp-slope triangle to find sin(theta) and cos(theta)
Alternatively we can use the fact that tan(theta) = ramp slope to figure out that
sin(theta) = ramp slope / sqrt(1 + ramp slope^2) and
cos(theta) = 1 / (1 + ramp slope)^2.
Once we have this information we can plug g, x_range, y_fall and our theta-related information into the equation to get v.
This formula will probably fairly meaningless to most students at this point, though most should have a fair idea of how it was obtained. The calculation has been built into the data analysis program. Just click on the Experiment-Specific Calculations button and choose the Ball and Ramp Projectile Experiment. Enter the following upon request:
the change in the ball's vertical position from the end of the ramp to the piece of plywood on the floor as the height of the drop (i.e., y_fall),
the difference between your mean straight-drop and projectile-landing positions as the horizontal range (i.e., x_range), and
the number of dominoes in the stack (from which the program will calculate slope, theta, etc).
Your information and the velocity of the ball, in cm/sec, will appear in the data window.
&#&# The program works, if you enter the data as indicated. If you get an error message, then give it another try, being careful to enter the data using the required syntax. If it doesn't work, copy the contents of each text box used at each step, as it appears just before you click the button, and also specify which button you clicked. Use one line for each textbox entry.
If it worked, you can leave the space below blank.&#&#
-------->>>>>>>> details, in case the program doesn't appear to have worked
Your answer (start in the next line):
#$&*
Use this feature of the program to determine ball velocity information for the first ramp, with two dominoes in the stack.
First sketch your horizontal range information as two intervals on a number line.
The first interval will run from mean - std dev to mean + std dev for the straight-drop positions, with the mean in the middle.
The second interval will run from mean - std dev to mean + std dev for the landing positions, with the mean in the middle.
When prompted enter the information. You measured y_fall; use as x_range the difference between the mean straight-drop position and the mean landing position (in your picture this would be the distance between the centers of your intervals).
Write down the velocity indicated by the program.
Click on the same button again and enter your data once more, but this time using as x_range the maximum distance between points in the two intervals (from the left end of the first interval to the right end of the second). Write down the velocity indicated by the program.
Repeat the calculation once more, but this time use as x_range the minimum distance between points in the two intervals (from the right end of the first interval to the left end of the second). Write down the indicated velocity.
The three velocities you have written down will be the velocity corresponding to the mean of your observed straight-drop and projectile-landing positions, and the lower and upper limits that would result from likely errors in measurement.
Report these three velocities in the space below, in comma-delimited format on the first line. Starting in the second line explain how your velocities were obtained, including a reasonably detailed sample calculation for at least one trial.
Reasonable detail would include the specific information put into the data analysis program, what the data analysis program told you, and how you used these results to obtain a velocity range.
-------->>>>>>>> three velocities 2 dom (mean - std, mean, mean + sdev), explanation & sample calc
Your answer (start in the next line):
24.22, 25.31, 23.16
I followed the instructions as given. For the first velocity listed above I clicked on experiment specific calculation in the data program, entered 1 for ball and ramp projectile experiment, entered in 75.40 for y_fall, and 9.466 for x_range (15.13 - 5.664= 9.466; the mean of the landing position minus the mean of the straight drop landing), and then entered in 2 for the amount of dominoes. The data program then provided the following: """"for drop height 75.4 cm, horizontal range 9.466 cm and 2 dominoes in the stack, the velocity of the ball should be 24.22 cm/s""""
#$&*
Now repeat this series of three calculations for the 3-domino trials, and report your three velocities in the space below, in comma-delimited format on the first line. You need not include the detailed calculation, provided it was done in the same manner as the one you reported in the preceding space .
-------->>>>>>>> same 3-dom
Your answer (start in the next line):
43.20, 44.19, 43.62
#$&*
Repeat once more for the 4-domino stack, and report your three velocities in the space below, in comma-delimited format on the first line. (Again you need not include the detailed calculation, provided it was done in the same manner as the one you reported previously).
-------->>>>>>>> same, 4-dom
Your answer (start in the next line):
52.74, 54.24, 51.25
#$&*
Based on the velocity results for the 2-domino stack, if we assume that the acceleration of the ball was uniform, what are the predicted acceleration and the range of accelerations on the 2-domino ramp? Report in comma-delimited format on the first line, followed by an explanation starting on the second of how you obtained your results, including one sample calculation.
-------->>>>>>>> accel, range of accel, 2-dom, expl with sample calc
Your answer (start in the next line):
2.14, 2.12, 2.16
The first number is the average acceleration; I took the lower end of the interval and the upper end of the interval, added them together, and divided by two. The first half of the interval the acceleration of the ball would be 2.12 cm/s^2, and for the second half of the interval it would be 2.16 cm/s^2, assuming uniform acceleration. For the first interval I took the mean velocity and subtracted the lowest velocity from it (9.466 cm/s - 9.05 cm/s), and divided this answer by half of the time interval (0.39 s / 2 = 0.196 s). This gave me an acceleration of 2.12 cm/s^2.
@& Your data indicate that the ball's final velocity on the incline was about 54 cm/s.
The incline is about 30 cm long, and the ball started from rest.
So you know v0, vf and `ds for the motion on the incline.
What therefore is the acceleration on the incline?*@
&&&&vf^2= v0^2 +2a(ds) solving for a gives the equation a= (vf^2 - v0^2)/2ds, substituting gives you a= (25.31cm/s)^2/2(30cm)= 10.7cm/s^2, 9.8cm/s^2, 8.9cm/s^2 for 2 dominoes.&&&&
#$&*
Give the same results for the 3-domino stack, in the same format. No explanation is necessary this time.
-------->>>>>>>> accel, range of accel, 3-dom
Your answer (start in the next line):
2.53, 2.14, 2.91
&&&&32.5cm/s^2, 31.1cm/s^2, 31.7cm/s^2&&&&
#$&*
Give the same results for the 4-domino stack, in the same format. No explanation is necessary this time.
-------->>>>>>>> accel, range of accel, 4-dom
Your answer (start in the next line):
7.63, 7.60, 7.65
&&&&48.6cm/s^2, 46.4cm/s^2, 43.8cm/s^2&&&&
#$&*
Using your results, plot a graph of predicted acceleration vs. number of dominoes. Then sketch and estimate the slope of your best-fit line.
In the space below give the slope and y-intercept of your best-fit line, using comma-delimited format in the first line of the space .
In the second line report in comma-delimited format the horizontal and vertical coordinates of the first point you used to obtain your slope, and in the third line do the same for the second point.
Starting in the third line give a brief explanation of the meaning of your numbers and how they were obtained.
------>>>>>>>> accel vs. # dom: slope and y int best fit, coord of 1st point, coord of 2d point used
Your answer (start in the next line):
2.75, -3.36
2, 2.14
4, 7.63
&&&&19.4, -29
2, 9.8
4, 48.6&&&&
I calculated the rise over run to find the slope: (7.63 - 2.14) / (4 - 2) = 2.75. I plugged back in 2 and 2.14 for x and y to find the y-intercept, which is also very close to where my line intercepts on my illustration.
#$&*
Now extend the points of your graph to form error bars, using for each number of dominoes the maximum and minimum values you obtained for accelerations on those ramps.
Find the slope of the steepest possible straight line that passes through all three error bars, and also the slope of the least steep line.
Report these slopes in the space below, in the first line in comma-delimited format:
Starting in the second line give a brief explanation of the meaning of your numbers and how they were obtained.
------>>>>>>>> steepest slope, least steep with error bars, meaning and how obtained
Your answer (start in the next line):
2.77, 2.72
I used the points (2, 2.12) and (4, 7.65) for the most steep, and (2, 2.16) and (4, 7.60) for the least steep. I found the rise over run, for example: (7.65 - 2.12) / (4 - 2) = 2.77
&&&&19.85, 16.55.
Using the points (2, 8.9) and (4, 48.6) for the steepest slope. For the least steep I used the points (2, 10.7) and (4, 43.8)&&&&
#$&*
NOTE that the next experiment continues using this setup. In that experiment you will simply roll the ball from rest for distances of 10 cm, then 20 cm down each incline, with 5 rolls at each distance for a total of 10 rolls; this will be done for a series of 4 ramp setups. You will also time 10 rolls for each of three different setups. It should take you no more than 20 minutes to obtain the necessary data. You might wish to go ahead and at least get that data now while you have everything set up, using the form Uniformity_of_Acceleration_for_Ball_on_Ramp. However if you don't have time right now, the setup is simple enough and it shouldn't take you long to set this up again.
@& Until the point where you calculated the ball's accelerations, your analysis was fine.
See my note. It shouldn't take you long to correct the acceleration calculations and modify the subsequent results.
&#Please see my notes and, unless my notes indicate that revision is optional, submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.
If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you.
&#
*@"
@& Good.
Your results indicate a ramp slope between about 16 and 20 cm/s^2 per domino.
This result is very consistent with the expected slope.
*@