6. Drag Force on Coffe Filters

Purpose: To study relationship between air drag forces and the velocity of a falling body.


Equipment: Computer with Logger Pro software, lab pro, motion detector, nine coffee filters, meter stick

Introduction: When an object moves through a fluid, such as air, it experiences a drag force that opposes its motion. This force generally increases with velocity of the object. In this lab we are going to investigate the velocity dependence of the drag force. We still start by assuming the drag force, Fd, has a simple power law dependence on the speed given by Fd= k |v| ^n, where the power n is to be determined by the experiment.
This lab will investigate drag forces acting on a falling coffee filter. Because of the large surface area and low mass of these filters, they reach terminal speed soon after being released.

Procedure:

NOTE: You will be given a packet of nine nested coffee filters. It is important that the shape of this packet stays the same throughout the experiment so do not take the filters apart or otherwise alter the shape of the packet. Why is it important for the shape to stay the same? Explain and use a diagram.

1. Login to your computer with username and password. Start the Logger Pro software, open the

Mechanics folder and the graphlab file. Don’t forget to label the axes of the graph and create an appropriate title for the graph. Set the data collection rate to 30 Hz.

2. Place the motion detector on the floor facing upward and hold the packet of nine filters at a minimum height of 1.5 m directly above the motion detector. (Be aware other of nearby objects which can cause reflections.) Start the computer collecting data, and then release the packet. What should the position vs time graph look like? Explain.

Verify that the data are consistent. If not, repeat the trial. Examine the graph and using the mouse, select (click and drag) a small range of data points near the end of the motion where the packet moved with constant speed. Exclude any early or late points where the motion is not uniform.

3. Use the curve fitting option from the analysis menu to fit a linear curve (y = mx + b) to the selected data. Record the slope (m) of the curve from this fit. What should this slope represent? Explain.

Repeat this measurement at least four more times, and calculate the average velocity. Record all data in an excel data table.

4. Carefully remove one filter from the packet and repeat the procedure in parts 2 and 3 for the remaining packet of eight filters. Keep removing filters one at a time and repeating the above steps until you finish with a single coffee filter. Print a copy of one of your best x vs t graphs that show the motion and the linear curve fit to the data for everyone in your group (Do not include the data table; graph only please).

5. In Graphical Analysis, create a two column data table with packet weight (number of filters) in one column and average terminal speed (|v|) in the other. Make a plot of packet weight (y-axis) vs. terminal speed not velocity (x-axis). Choose appropriate labels and scales for the axes of your graph. Be sure to remove the “connecting lines” from the plot. Perform a power law fit of the data and record the power, n, given by the computer. Obtain a printout of your graph for each member of your group. (Check the % error between your experimentally determined n and the theoretical value before you make a printout – you may need to repeat trials if the error is too large.)

6. Since the drag force is equal to the packet weight, we have found the dependence of drag force on speed. Write equation 1 above with the value of n obtained from your experiment. Put a box around this equation. Look in the section on drag forces in your text and write down the equation given there for the drag force on an object moving through a fluid. How does your value of n compare with the value given in the text? What does the other fit parameter represent? Explain.


Our data:

The slope of Time vs. Position graph represents the terminal velocity (positive direction is upward). So on this graph, the terminal velocity is -1.66m/s. The terminal speed is 1.66m/s.



On this graph, the terminal velocity is -1.99m/s. The terminal speed is 1.99m/s.



The average terminal speed of 1-9 trails

We find that when the number of filter decreases, the average terminal speed also decreases.

We use the power law fit to find the best fitting curve that Y=2.14X^1.95 (X= terminal speed; Y= number of coffee filters). We find n is 1.95, which is almost 2. That means it matches to the value that given in the text that n=2.

Question:

(1)Why is it important for the shape to stay the same?
---Because the drag relates to objects' cross section area. If filter's shape changes, then it's drag will also change.

(2)What should the position vs time graph look like?
---The curve decreases faster and faster at first. Then, at a moment, the curve turns to be linear, decreases with constant slope.

(3) What should this slope represent (y = mx + b)?
---The slope m represents the terminal velocity.

(4)How does your value of n compare with the value given in the text? What does the other fit parameter represent?
---We find n is 1.95, which is almost 2. That means it matches to the value that given in the text that n=2.
Because Mg= (1/4)Av^2 (A= cross section area). We found that Y=2.14X^1.95 (X= terminal speed; Y= number of coffee filters). So. Y× mg= (1/4)A×(X^2) (m= mass of each coffee filter). Also, Y= [A/ (4gm)]× (X^2). So 2.14 represents A/ (4gm): the cross section area of coffee filter divides four times a coffee filter's gravity force.

Conclusion:

  According to our graph and data, we notice that when an object begins to fall, its speed gradually increases at first. The net force of object is Mg-Drag, so acceleration= (Mg-Drag)/M. The Drag is increasing, the acceleration is decreasing. When Drag is equal to mass of the object, the acceleration turns to be zero. So the object will keep falling will constant velocity.

 

 

 

10.Motion in One Dimension with Air Drag

Purpose: To analyze how changing force affects motion in one dimension.

Introduction:
So far we have assumed objects move with constant acceleration. This assumes that a constant force is applied to the object. What if a changing force (which causes a changing acceleration) is applied to an object?

An object under the influence of air drag is one example of this. Calculus provides a way to solve this problem where variables are changing with respect time. A spreadsheet can be used to perform this calculus. Two basic relationships can be used to analyze the object:

v_new = v_old + a_avg△t (1)

Where v_old is the previous value of velocity and position, and a_avg is the average value of the acceleration and velocity during the time interval △t.


1. By unit analysis, show that the above equation (1) is valid:

Left side unit = m/s
Right side unit = (m/s) + (m/s^2) × s = (m/s) + (m/s) = m/s

So the units of two sides are identical.


2. Why do we use a_avg in equation (1).

Because the can be changing all the time, so the average data is more accurate.


3. Come up with an analogous equation relating y_new and y_old:

y_old = y_o
y_new = y_o + v_old△t + (1/2)a_avg(△t)^2
or y_new = y_o + v_avg△t


4. What is the benefit of changing a small △t:

Because the acceleration is changing all the time, so in a small △t, the a_avg is closer to real value. So, the v_new will also be more accurate.



Since the acceleration depends on the forces acting on the object, we must specify precisely what these forces are. In our problem the drag force will at first be assumed to depend on the first power of the velocity and can be written as:
F_D= -kv
Where k is a proportionality constant.


1. Draw a detailed motion diagram of the object falling down:

a. Now draw a force diagram for the object falling down. Include vectors for all forces, and write a statement of Newton's Second. Solve for acceleration.

 

a= F / m = D - G = (-kv-mg) / m = - (g + kv / m )


b. Give the condition for the object at terminal velocity (a=0). Using this condition solve for k.

a= 0, So net force is 0, mg = kv_t , k= mg / v_tc. Substitute k into your expression for the acceleration from part a.

a = - (g + kv / m )
k = mg / v_t So, a = -g (1+ v / v_t )


2. Open the spreadsheet in the Physics Apps file called air drag. Describe what the spreadsheet is calculating.

a. What are the assumption? (i.e. initial values? )

t_o = 0s , g = -9.8m/s^2 , initial velocity = 20m/s , terminal velocity = -40m/s , △t = 0.1s

b. What is v-halfstep?

V-halfstep is the average velocity in 0.1s.

c. What is the a-halfstep?

A-halfstep is the average acceleration in 0.1s.


3. Using the graph paper provided, draw scales graphs of position vs. time, velocity vs. time and acceleration vs. time for the object no air drag.

position vs. time graph when there is no drag

p= -4.9t^2 + 20t +1000

velocity vs. time graph when there is no drag

v= -9.8t + 20

Make predictions on another sheet of graph paper about how position and velocity graphs would change if you include air drag (D= -kv).

velocity vs. time graph when drag is equal to -kv

Now look at a drag force that is dependent on the square of the velocity. Assuming a drag force,
F_D= | kv^2 |, find the new formula for the acceleration.

a= F / m = D - G = (kv^2 - mg) / m
mg = k v_t^2 , k = mg / v_t ^2
So, a= [(v^2/ v_t^2) -1] g

position vs. time graph when drag is equal to |kv^2|

velocity vs. time graph when drag is equal to |kv^2|

Conclusion:

When there is no drag, the net force on the ball is gravity force, so the acceleration is equal to -9.8m/s^2. The motion trajectory is parabola. The velocity is changing with constant slope, decreasing to 0 first then increasing in negative position. Changing force affects motion in one dimension. We plotted v-t and p-t graphs in three cases: with no drag; drag is equal to -kv and drag is equal to | kv^2 |.

 

 

 

9.Human Power

Purpose: To determine the power output of a person

Equipment: Two meter metersticks, stopwatch, kilogram bathroom scale

Introduction: Power is defined to be the rate at which work is done or equivalently. the rate at which energy is converted from one form to another. In this experiment you will do some work by climbing from the first floor of the science building to the second floor. By measuring the vertical height climbed and knowing your mass, the change in your gravitational potential energy can be found:

△ PE = mgh

Where m is the mass, g is the acceleration of gravity, and h is the verticle height gained.

Your power output can be determined by:

Power = △PE / △t Where △t is the time to climb the vertical height h

Data:

m: 61.02kg

mg: 598N

h: 4.29m

△t₁ = 8.66s

△t₂ = 10.06s

Power₁= mgh / △t₁= 309N × 4.29m / 8.66s = 290.21 J/s = 297.21 W = 0.38 Hp

Power₂= mgh / △t₂= 266N × 4.29m / 10.06s = 284.1 J/s = 286.1 W = 0.38 Hp

Power_avg = (Power₁ + Power₂ ) / 2 = (0.39 Hp +0.38 Hp) / 2 = 0.385 Hp

Question:

1. Is it okay to use your hands and arms on the handrailing to assist you in your climb up the stairs? Explain why or why not.

If we use our hands and arms on the handrailing to assist us, our calculated value of power won't change. Because when you put your hands on the handrailing, the handrailing will give you a normal force upward, but your weight doesn't change, which is still equal to mg. So, according to the formula △PE = mgh, the change of the gravitational potential energy will be the same.

2. Discuss some of the problems with the accuracy of this experiment.

The height of the floor, the mass and the time are all measured values that are not really accurate, so our calculated values of power are also have error.

Followup questions:

1. Two people of the same mass climb the same flight of stairs. Hinrik climbs the stairs in 25 seconds. Valdis takes 35 seconds. Which person does the most work? Which person expands the most power? Explain your answers.

Because △PE = mgh, their mass and the height are the same, so they both did the same work.

Hinrik expands more power. Because Power = △PE / △t , △PE are the same, but Hinrik used less time, so he had a higher power.

2. A box that weights 1000 Newtons is lifted a distance of 20.0 meters straight up by a rope and pulley system. The work is done in 10.0 seconds. What is the power developed in watts and kilowatts.

Power = △PE / △t = (1000N× 20m) / 10s= 2000W= 2KW

3. Brynhildur climbs up a ladder to a height of 5.0 meters. If she is 64 kg:

a) What work dose she do?

Work = mgh = 64kg × 9.8m/s^2 × 5m =3136J

b) What is the increase in the gravitational potential energy of the person at this height?

△PE = mgh = 64kg × 9.8m/s^2 × 5m =3136J

Increase 3136J

c) Where does the energy come from to cause this increase in P.E.?

the force of gravity do the negative work on the person, so the gravitational potential energy will increase by the same amount. The energy the person gets to do the work come from this person's chemical energy. The consumption of chemical energy increases this person's gravitational potential energy.

4. Which requires more work: lifting a 50 kg box vertically for distance of 2m , or lifting a 25kg box vertically for a distance of 4 meters?

Lifting a 50 kg box:

Work = mgh = 50kg × 9.8m/s^2 × 2m = 980N

Lifting a 25 kg box:

Work = mgh = 25kg × 9.8m/s^2 × 4m = 980N

So, they require the same work.

Conclusion:

In this lab, we determined the power output when people climb the floors. We also learned the definition of the potential energy and power.

    

 

12.Balanced Torques and Center of Gravity

Purpose: To investigate the conditions for rotational equilibrium of a rigid bar and to determine the center of gravity of a system of masses.

Equipment: Meter stick, meter stick clamps (knife edge clamp), balance support, mass set, weight hangers, unknown masses, balance.

Introduction: The condition for rotational equilibrium is that the net torque on an object about some point in the body, O, is zero. Remember that the torque is defined as the force times the lever arm of the force with respect to the chosen point O. The lever arm is the perpendicular distance from O to the line of action of the force.

Procedure:

Note: In each of the following steps, where appropriate, make a careful sketch showing the meter stick with the applied forces and mark their locations. Also, show the point, O, about which you are calculating torque.

1. Balance the meter stick in the knife edge clamp and record the position of the balance point. What point in the meter stick does this correspond to?

48.5cm

 

2. Select two different masses (100 grams or more each) and using the meter stick clamps and weight hangers, suspend one on each side of the meter stick support at different distances from the support. Adjust the positions so the system is balanced. Record the masses and positions. Is it necessary to include the mass of the clamps in your caculation? EXPLAIN! Sum the torques about your pivot point O and compare with the expected value.

 

mass1: 150g position1: 0.1535m
mass2: 50g position2: 0.1098m

mass of meter stick with clamp: 0.1807 kg

We should include the mass of the clamps, because now the clamp with the mass can be seen as a system, so the mass should be all together.

torqueleft = W1 × Ll = [(50+ 20)/1000]kg × 9.8m/s^2 × [(48.5- 24.1)/100]m = 0.500 N·m
torqueright = W2 × L2 = [(150+ 20)/1000]kg × 9.8m/s^2 × [(58.5- 48.5)/100]m = 0.509 N·m
difference of torque = 0.167-0.166= 0.001 N·m

expected value of L2:
W1·Ll = W2·L2
[(50+ 20)/1000]kg × 9.8m/s^2 × [(48.5- 24.1)/100]m= [(150+ 20)/1000]kg × 9.8m/s^2 × [L2/100]m
L2= 10.047cm
experimental value= 58.5cm-48.5cm= 10cm

percent of error= [(.16971651)/.1651]× 100%= 2.79%



3. Place the same two masses used above at different locations on the same side of the support and balance the system with a third mass on the opposite side. Record the masses and positions. Calculate the net torque on this system about the point support and compare with the expected value.

mass1: 150g position1: 68.5cm
mass2: 50g position2: 58.5cm
mass3: 100g position2: 15.3cm
mass of clamp: 20g


torqueright = W1·Ll+ W2·L2
[(50+ 20)/1000]kg × 9.8m/s^2 × [(58.5- 48.5)/100]m+ [(150+ 20)/1000]kg × 9.8m/s^2 × [(68.5-48.5)/100]m = 0.4018 N·m

torqueleft = W3·L3
[(100+ 20)/1000]kg × 9.8m/s^2 × [(48.5-15.3)/100]m = 0.3904 N·m

difference of torque = 0.4018-0.3904= 0.0114 N·m

expected value of L3:
W3·L3= W1·Ll+ W2·L2
[(100+ 20)/1000]kg × 9.8m/s^2 × L3= 0.4018 N·m
L3= 34.17cm

experimental value of L3= 48.5-15.3= 33.2cm


percent of difference= [(34.17-33.2)/34.17] × 100%= 2.84%



4. Replace one of the above masses with an unknown mass. Readjust the positions of the masses until equilibrium is achieved, recording all values. Using the equilibrium condition for rotational motion, calculate the unknown mass. Measure the mass of the unknown on a balance and compare the two masses by finding the percent difference.

mass1: 150g position1: 68.5cm
mass2: 50g position2: 58.5cm
mass3: unknown position2: 4.8cm
mass of clamp: 20g

expected value of mass3(unknown):
W1·Ll+ W2·L2 = W3·L3
[(50+ 20)/1000]kg × 9.8m/s^2 × [(58.5- 48.5)/100]m+ [(150+ 20)/1000]kg × 9.8m/s^2 × [(68.5-48.5)/100]m = [(m3+ 20)/1000]kg× 9.8m/s^2 × [(48.5-4.8)/100]m
m3= 94g

measured value of the unknown mass: 91.5g

percent of error= [(94-91.5)/94] × 100%= 2.66%



5. Place about 200 grams at 90 cm on the meter stick and balance the system by changing the balance point of the meter stick. From this information, calculate the mass of the meter stick. Compare this with the meter stick mass obtained from the balance. Should the clamp holding the meter stick be included as part of the mass of the meter stick? EXPLAIN!

balance point: 77.8cm

the mass of meter stick obtained from the balance: 84.2g

the calculated value of meter stick:
W1·Ll = W2·L2
m × 9.8m/s^2 × [(77.8- 48.5)/100]m = 0.2kg × [(90- 48.5)/100]m
m= 83.28g

percent of error: [(84.2-83.28)/84.2] × 100%= 1.09%

In this case, the mass of clamp is included in 200grams, so it shouldn't be included as part of the mass of the meter stick. Because in part 1, when we got the balance point of the meter stick (48.5cm) , there is no clamp or mass on it. So, here we still use the balance point as 48.5cm, the mass of clamp also shouldn't be included.



6. With the 200 grams still at 90 cm mark, imagine that you now position an additional 100 grams mass at the 30 cm mark on the meter stick. Calculate the position of the center of gravity of this combination (two masses and meter stick). Where should the point of support on the meter stick be to balance this system? Check your result by actually placing the 100 g at the 30 cm mark and balancing this system. Compare the calculated and experimental result.

Let the unknown point equal to x cm.

experimental value: x=65.1cm

calculated value:
W1·Ll+ W2·L2 = W3·L3
[100/1000]kg × 9.8m/s^2 × [(x- 30)/100]m+ [(84.2)/1000]kg × 9.8m/s^2 × [(x-48.5)/100]m
= [200/1000]kg× 9.8m/s^2 × [(90-x)/100]m

x=65.29cm

percent of error: [(65.29-65.1)/65.1] × 100%= 0.29%



Conclusion:

In this lab,we learn how  to determine the center of gravity of a system of masses and  how to investigate the conditions for rotational equilibrium of a rigid bar.
 The magnitude of torque depends on three quantities: the force applied, the length of the lever arm connecting the axis to the point of force application, and the angle between the force vector and the lever arm.

 

8. Motion in One Dimension with Air Drag

The purpose of this lab is to analyze how changing force affects motion in one dimension.

Introduction: If a changing force (which causes a changing acceleration) is applied to an object then the object Fnet will be different at every time.

Ø  An object under the influence of Drag is one example. Two basic relationships can be used to analyze the object.              Vnew = Vbold + aavg

1.       By unit analysis we can show that the equation above is valid.

2.       We use aavg  in equation 1 because a is changing therefore taking the average is gives a more accurate value.

3.       We come up with an analogous equation relating Ynew to Yold.

                                       Vnew=VoldΔt+ aavg Δt

                                                                               2

4.       The benefit of choosing a small Δt is that the average of acceleration is closer to the actual acceleration problem.

 

Ø  Since the acceleration depedends on the forces acting on the oblject, we must specify precisely what these forces are. I our problem the drag force will be assumed to depened on the first power of the velocity, and can be written as  F_D = - Kv

Where K is proportionally constant.

 

1.       Motion diagram of the object falling down.

 

a= F / m = D - G = (-kv-mg) / m = - (g + kv / m )


b. Give the condition for the object at terminal velocity (a=0). Using this condition solve for k.

a= 0, So net force is 0, mg = kv_t , k= mg / v_t

c. Substitute k into your expression for the acceleration from part a.

a = - (g + kv / m )
k = mg / v_t So, a = -g (1+ v / v_t )


2. Open the spreadsheet in the Physics Apps file called air drag. Describe what the spreadsheet is calculating.

a. What are the assumption? (i.e. initial values? )

t_o = 0s , g = -9.8m/s^2 , initial velocity = 20m/s , terminal velocity = -40m/s , △t = 0.1s

b. What is v-halfstep?

V-halfstep is the average velocity in 0.1s.

c. What is the a-halfstep?

A-halfstep is the average acceleration in 0.1s.


3. Using the graph paper provided, draw scales graphs of position vs. time, velocity vs. time and acceleration vs. time for the object no air drag.

position vs. time graph when there is no drag

p= -4.9t^2 + 20t +1000

velocity vs. time graph when there is no drag

v= -9.8t + 20

Make predictions on another sheet of graph paper about how position and velocity graphs would change if you include air drag (D= -kv).

position vs. time graph when drag is equal to -kv

velocity vs. time graph when drag is equal to -kv

Now look at a drag force that is dependent on the square of the velocity. Assuming a drag force,
F_D= | kv^2 |, find the new formula for the acceleration.

a= F / m = D - G = (kv^2 - mg) / m
mg = k v_t^2 , k = mg / v_t ^2
So, a= [(v^2/ v_t^2) -1] g

position vs. time graph when drag is equal to |kv^2|

velocity vs. time graph when drag is equal to |kv^2|

Conclusion:

In this lab, we analyzed how changing force affects motion in one dimension. We plotted v-t and p-t graphs in three cases: with no drag; drag is equal to -kv and drag is equal to | kv^2 |.

    

7. Centripetal Force

Purpose: To verify Newton's second law of motion for the case of uniform circular motion.

Equipment: Centripetal force apparatus, metric scale, vernier caliper, stop watch, slotted weight set, weight hanger, triple beam balance.

Introduction: Then centripetal force apparatus is designed to rotate a known mass through a circular path of known radius. By timing the motion for a definite number of revolutions and knowing the total distance that the mass has traveled, the velocity can be calculated. Thus the centripetal force, F, necessary to cause the mass to follow its circular path can be determined from Newton's second law.
F=mv^2/r
Where m is the mass, v is the velocity, and r is the radius of the circular path.
Here we have used the fact that for uniform circular motion, the acceleration, a, is given by:
a=v^2/r


Procedure:
1. For each trail the position of the horizontal crossarm and the verticle indicator post must be such that the mass hangs freely over the post when the spring is detached. After making this adjustment, connect the spring to the mass and practice aligning the bottom of the hanging mass with the indicator post while rotating the assembly.

2.Measure the time for 50 revolutions of the apparatus. Keep the velocity as constant as possible by keeping the pointer on the bottom of the mass aligned with the indicator post. A while sheet of paper placed as a background behind the apparatus can be helpful in getting the alignment as close as possible. Using the same mass and radius, measure the time for three different trials. Record all data in a neat excel table.

3.Using the average time obtained above, calculate the velocity of the mass. From this calculation the centripetal force exerted on the mass during its motion.

4.Independently determine the centripetal force by attaching a hanging a weight to the mass until it once again is positioned over the indicator post (this time at rest). Since the spring is being stretched by the same amount as when the apparatus was rotating, the force stretching the spring should be the same in each case.


Data:
mass= 475grams
radius= 16.5cm
v= 2πr·f
f=50/t
a=v^2/r
F=mv^2/r
(r= radius, f= frequency, t= time of 50 revolutions, m= mass, a= acceleration, F=calculated centripetal force)

The force diagram for the hanging weight:
The average calculated force: 7.36N

The average measured force: 7.066N

Percent difference: 4.44%

5.Add 100g to the mass and repeat steps 2,3,4 above.

Data:

mass= 473grams
radius= 16.5cm

 

The average calculated force: 7.53N

The average measured force: 7.252N

Percent difference: 3.83%

Conclusion:

The sum total of all interactions is directly proportional to the acceleration of the object.

5.Working With Spreadsheets

Purpose: To get familiar with electronic spreadsheets by using them in some simple applications.

Equipment: Computer with EXCEL software.

Part 1:

Create a simple spreadsheet that calculates the values of the following function:

f(x)=Asin(Bx+C)

Initially choose value for of A= 5, B= 3 and C= π/3 (1.047).

Create a column for values of x that run from zero to 10 radians in steps of 0.1 radians. Similarly, create in the next column the corresponding values of f(x) by copying the formula shown above down through the same number of rows (100 in roll).

Then copy and paste our data into the graphing program. Put appropriate labels on the horizontal and vertical axes of the graph. Use Curve Fit to find a function that best fit the data.

The best fit function: y= 5Sin(3x +1.05)-1.49×10^(-10)

Part 2:

Repeat the above process for a spreadsheet that calculates the position of a freely falling particle as a function of time. Start off with g= 9.8m/s^2, vo= 50m/s, xo= 1000m and △t= 0.2s.

The formula for free fall: f(t)= ro+vo△t+(1/2)a(△t)^2

We assuming the direction of vo positive.

(i)When g is positive:

f(t)=1000+ 50t+ (1/2)× 9.8t^2

Use Curve Fit to find a function that best fit the data:


The best fit function: f(t)= 4.9t^2+ 50t+ 1000

(ii)When g is positive:

f(t)=1000+ 50t+ (1/2)× (-9.8)t^2

Use Curve Fit to find a function that best fit the data:

 

The best fit function: f(t)= (-4.9)t^2+ 50t+ 1000

Question: How do data from part 1 and part 2 compare to the values we start with in our spreadsheet?

In part 1, we compare the data(A=5, B=3, C=1.05) from Curve Fit to the values(A=5, B=3, C=1.047) that we start with in our spreadsheet. In part 2, we do the same thing as initial values are "g= 9.8m/s^2, vo= 50m/s, xo= 1000m and △t= 0.2s". We find that the data from Curve fit are almost the same to the initial values.

Conclusion:
In this experiment we use excel spreadsheet to solve the problem.

We find that the data from Curve fit are almost the same to the initial values. This experiment helps us to have many data and efficiently do the formulas.