Cristina-Alex Manolache
Prof. Gordon Crandall
MAT 201 # 6227
Maple Lab Project
Project: Study of Motion of a Falling Ball
Using First and Second Derivatives
Part 1
The position and velocities of a falling ball at time intervals of 0.02 second are given by the data entered below:
> Tdata:=[0.00,0.02,0.04,0.06,0.08,0.10,0.12,0.14,0.16,0.18,0.20,0.22];
> Hdata:=[0.290864,0.284279,0.274400,0.260131,0.241472,0.219520,
0.189885,0.160250,0.126224,0.086711,0.045002,0.000000];
> Vdata:=[-0.16405,-0.32857,-0.49403,-0.71322,-0.93309,-1.09409,-1.47655,-1.47891,-1.69994,-1.96997,-2.07747,-2.25010];
> with(stats[statplots]);
> scatterplot(Tdata,Hdata,color=red,symbol=cross);
> scatterplot(Tdata,Vdata,color=red, symbol=cross);
> with(stats[fit]);
> leastsquare[[x,y],y=a+b*x+c*x^2]([Tdata,Hdata]);
> with(plots);
Warning, the name changecoords has been redefined
> display(scatterplot(Tdata,Hdata,color=blue,symbol=cross),
plot(0.2914702143-0.2374347153*x-4.967839660*x^2,x=0.0..0.22));
> leastsquare[[x,y],y=a*x+b]([Tdata,Vdata]);
> with(plots);
> display(scatterplot(Tdata,Vdata,color=blue,symbol=cross),plot(-0.1469526923-9.785270979*x,x=0.0..0.22));
> plot([-9.935679320*x-0.2374347153,-9.785270979*x-0.1469526923],x=-0.6..0.6,y=-0.6..0.6,linestyle=[1,4],color=[red,blue]);
Question # 1
Use Maple to enter the three columns of data, and then use the scatter plot command to make a graph of Time versus Velocity. Examine the graphs. What model (function) seems to give the best fit for the scatter plot of heights? What type of model (function) seems to give the best fit for the scatter plot of velocities of the falling ball? Describe any relationships that you observe between the two graphs. Use Maple to find the best fit position function determined by the data.
The model (function) which seems to give the best fit for the scatter plot of heights is a parabola. The model (function) which seems to give the best fit for the scatter plot of velocities is a linear function. Analyzing the two graphs sketched above, enable us to observe that the ball is being dropped, thus falling. Considering the graphs and the data provided by the given table, the height is max at time 0. Additionally, we observe that the height is positive and decreasing, while the velocity is negative.
Question # 2
A model for the position function of the falling body (in the absence of air resistance which acts like a friction to retard the fall) is given by:
s (t) = -1/2*g*t2 + v0 + s0,
> S:=t->-0.5*g*t^2+v*t+s;
where g is the acceleration due to gravity and v0 ands0 are the initial position of the ball and the initial velocity imparted to the body respectively. Use the Maple least squares command to find the best quadratic model for the Time (x) versus Height (y) data. Use your result to determine the ball’s initial height and initial velocity, the velocity function and the acceleration function. Record your results below:
Position Function: s (t) = 0.2914702143-0.2374347153 *x-4.967839660*x2
Initial Height: s0 = 0.2914702143
Initial Velocity: v0 = -0.2374347153*x
Velocity Function: v (t) = s' (t) = -9.93567932*x-0.2374347153
Acceleration Function: a (t) = s" (t) = -9.93567932
Question # 3
Use Maple to plot the Time versus Velocity data and the velocity function that you determined in Question # 2 on the same axes. Is there a good fit between the data and the function? Why or why not? Explain any discrepancies.
As shown by the graph obtained, the fit is not precise, but is a great fit. The fit is considered being good as the data is on or very closely situated to the line of the equation.
Question # 4
A model for the velocity function has the form:
v (t) = -g*t + v0,
where again g is the acceleration due to gravity and is v0 the initial velocity of the falling body. Use the least squares method to find the best linear fit for the Time versus Velocity data. Use this velocity function to determine the acceleration function, and record your results below:
Velocity Function: v (t) = -9.785270979*x – 0.1469526923
Acceleration Function: a (t) = v' (t) = -9.785270979
Question # 5
Use Maple to plot the Time versus Velocity data along with the velocity functions from Questions 2 and 4 on a single pair of axes. Which velocity function gives the better fit to the data? Explain why these two velocity functions, which describe the same data, are different.
> plot([-9.935679320*x-0.2374347153,-9.785270979*x-0.1469526923],x=-0.6..0.6,y=-0.6..0.6,linestyle=[1,4],color=[red,blue]);
The Velocity Function: v (t) = -9.785270979*x – 0.1469526923 gives the better fit to the data. Although, I used the same Maple least squares command to find the best models, the two velocity functions are different because we obtained the first for the Time (x) versus Height (y) data, and the second for the Time (x-axis) versus Velocity (y-axis) data.
Question # 6
Of the two acceleration functions that you found in Questions 2 and 4, which give the closer estimate to the actual value of the acceleration due to gravity, -9.8 meters/sec2? Calculate the percent error in the two cases using:
Percent error = Abs (Expected value – Observed value) / Expected value * 100%
Percent error = [(-9.785270979) – (-9.8)] / (-9.785270979) * 100% =
= 0.014729021 / (-9.785270979) * 100% =
= -0.0015052236 * 100% =
~ -0.15 %
Percent error = [(-9.93567932) – (-9.8)] / (-9.93567932) * 100% =
= -0.13567932 / (-9.93567932) * 100% =
= 0.013655766 * 100% =
~ 1.4 %
Yes, the estimated values of g are reasonable. Of the two acceleration functions that I found in Questions 2 and 4, Acceleration Function: a (t) = v' (t) = -9.785270979 m/sec2 is the one which gives the closer estimate to the actual value of the acceleration due to gravity, -9.8 meters/sec2 .
An educational guess would be that the importance of the -0.15 % error depends on the eyes of the beholder. If the beholder would be an engineer, -0.15 % error could not mean too much; however, for a chemist or an astronaut, per se, an error of -0.15 % could be of high importance.
Question # 7
Use your position function from Question 2, and the Maple solve command to find the time when the ball hits the ground. How does this value compare to the observed value of 0.22 seconds. Calculate the percent error as in Question 6.
> solve(0.2914702143-0.2374347153 *x-4.967839660*x^2);
Through my calculation, the expected value of the time when the ball hits the ground is = 0.2195006358, which is approximately equal with 0.22. We will only utilize the positive value/root obtained through solving the equation above. This value is very close to the observed value of 0.22 seconds. I would say that the value of the error percent is non-significant. However, I will show this by calculating it using the formula provided by Professor Crandall.
Percent error = Abs (Expected value – Observed value) / Expected value * 100%
Percent error = [(0.2195006358) – (0.22)] / (0.2195006358) * 100% =
= (-0.0004993642) / (0.2195006358)* 100% =
= (-0.002275001) * 100% =
~ -0.2 %
Therefore, as previously stated, it is noticeable that the percent error is pretty insignificant being that is of only -0.2 %.
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Part 2
A second ball falls according to the data below:
> tdata:=[0.00,0.02,0.04,0.06,0.08,0.10,0.12,0.14,0.16,0.18,0.20,0.22,
0.24,0.26,0.28,0.30,0.32,0.34,0.36,0.38,0.40,0.42,0.44,0.46,0.48,0.50,0.52,0.54,0.56,0.58,0.60,0.62];
> hdata:=[0.806736,0.857225,0.904422,0.946131,0.985644,1.020760,
1.052590,1.080030,1.103080,1.122840,1.137110,1.149180,1.156870,1.160160,1.161260,1.156870,1.149180,1.141500,1.126130,1.105280,1.082230,1.056980,1.026250,0.992230,0.954912,0.913203,0.868201,0.819907,0.767222,0.711244,0.651974,0.589411];
> with(stats[statplots]);
Warning, these names have been redefined: boxplot, histogram, scatterplot, xscale, xshift, xyexchange, xzexchange, yscale, yshift, yzexchange, zscale, zshift
> scatterplot(tdata,hdata,color=red,symbol=cross)
> with(stats[fit]);
Warning, the names leastmediansquare and leastsquare have been redefined
> leastsquare[[x,y],y=a+b*x+c*x^2]([tdata,hdata]);
> with(plots);
> display(scatterplot(tdata,hdata,color=red,symbol=cross),
plot(0.8087991314+2.589810287*x-4.756079168*x^2,x=0.0..0.62));
Question # 8
Use Maple to find the best fit position function determined by the data. Use this position function to find the initial height and initial velocity. Do you think that the ball was dropped or thrown? Calculate the velocity function. Find the time when the ball will hit the ground and the velocity of the ball at that time, often referred as terminalvelocity.
Position Function: s (t) = 0.8087991314 + 2.589810287*x – 4.756079168*x2
Initial Height: s0 = 0.8087991314
Initial Velocity: v0 = 2.589810287
Velocity Function: v (t) = s' (t) = -9.512158336*x-2.589810287
Question # 9
For which values of t is s’ (t) positive? For which values of t is s’ (t) negative?
From the graph obtained s’ (t) is positive for t = [0.00, 0.30], and s’ (t) is negative for t = [0.32, 0.62].
Explain what the graph of v (t) = s’ (t) tells you about the graph of s (t).
The graph of v (t) = s’ (t) tells me that s (t) must be a quadratic function; therefore, its graph is a parabola.
What does the graph of a (t) = s” (t) tell you about the graph of s’ (t)?
The graph of a (t) = s” (t) tell me that s’ (t) is a linear function; thus, its graph is represented by a line.
What does the graph of a (t) tell you about the graph of s (t)?
The graph of a (t) tell me that s’ (t) is a linear function; consequently, s (t) must be a quadratic function, and its graph is represented by a parabola.
Question # 10
Choose four distinct points (reasonably separated) in the domain of s (t), and use Maple to plot a graph showing the tangent lines to the graph of s (t) at the chosen points. For what value of t is the tangent line horizontal? What is the value of v (t) at this time?
I picked five (5) distinct points (t) in the domain of s (t), which is between the points [0.00, 0.60]. The values chosen for t1, t2 ,t3, t4 and t5 are:
t = [0.06, 0.24, 0.30, 0.36, 0.60].
Using Maple, I calculated the values for s (t) for each t chosen. These values are to be found below.
s(t)=
> S:=X->0.8087991314+2.589810287*X-4.756079168*X^2;
s(t1)=
> S(0.06);
s(t2)=
> S(0.24);
s(t3)=
> S(0.30);
s(t4)=
> S(0.36);
s(t5)=
> S(0.60);
The slopes, m for each tangent line equation are calculated below, also using Maple. The slope is represented by s’(t), the second derivative of the position function.
Slope m = s’ (t) =
> M:=X->-9.512158336*X+2.589810287;
Slope m for t1=
> M(0.06);
Slope m for t2=
> M(0.24);
Slope m for t3=
> M(0.30);
Slope m for t4=
> M(0.36);
Slope m for t5=
> M(0.60);
Please find below the plotted graphs showing the tangent lines to the graph of s (t) at the chosen points.
The equation of the first tangent line is: Y = 2.019080787*x+0.8259210164
> display(plot(0.8087991314+2.589810287*x-4.756079168*x^2,x=0.0..0.62), plot(2.019080787*x+0.8259210164,x=0.0..0.62));
The equation of the second tangent line is: Y = 0.306892286*x+1.082749291
> display(plot(0.8087991314+2.589810287*x-4.756079168*x^2,x=0.0..0.62), plot(0.306892286*x+1.082749291,x=0.0..0.62));
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The equation of the third tangent line is: Y = -0.263837214*x+1.236846257
> display(plot(0.8087991314+2.589810287*x-4.756079168*x^2,x=0.0..0.62), plot(-0.263837214*x+1.236846257,x=0.0..0.62));
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The equation of the forth tangent line is: Y = -0.834566714*x+1.425186992
> display(plot(0.8087991314+2.589810287*x-4.756079168*x^2,x=0.0..0.62), plot(-0.834566714*x+1.425186992,x=0.0..0.62));
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The equation of the fifth tangent line is: Y = -3.117484715*x+2.520987632
> display(plot(0.8087991314+2.589810287*x-4.756079168*x^2,x=0.0..0.62), plot(-3.117484715*x+2.520987632,x=0.0..0.62));
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