Roberto Lam
MAT202.6671 (Final Project)
Gordon Crandall
Fall2005
The Shot Put
The shot put has been a test of strength for centuries. Early versions of the shot
were heavy stones. Today’s athletes use a shot made of metal weighing 7.26 kilograms
(16 pounds). The men’s shot put record by the end of 1960 was little more than 20
meters. By 1990, the record had moved past the 23-meter mark.
The angle of elevation of the shot’s release from the athlete’s hand is very
important. If the angle it too great, too much of the initial velocity imparted to the shot
will be used to send the shot up, and horizontal distance will be sacrificed. On the other
hand, if the angle is too low, the shot will not stay above ground long enough to travel a
significant distance. The purpose of this project is to analyze the men’s shot put and to
determine both experimentally and analytically the optimal angle of release that
maximizes horizontal distance.1
The years in which records were set in the men’s shot put, along with the record distance
are given in the tabulated data below, x=0 correspond to the year 1900.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
1
Year, x
(mdata)
60
60
60
60
60
62
62
64
64
65
67
73
76
76
78
83
85
86
87
87
87
88
90
Distance, y (fdata)
19.38
19.45
19.67
19.99
20.06
20.08
20.10
20.20
20.68
21.52
21.78
21.82
21.85
22.00
22.15
22.22
22.62
22.64
22.79
22.81
22.91
23.06
23.12
Information from project guidelines.
1
Roberto Lam
MAT202.6671 (Final Project)
Gordon Crandall
Fall2005
(Question 1)Creating a scatter plot.2
1. Using Maple, the following scatter plot is created for the data above.
2. As it can be observed, the points plot is approximately linear. So, to prove the
linear approximation the “strength” of the linear relationship between time in
years and distance is obtained by calculating the correlation coefficient using
Maple; and the value is the following:
> linearcorrelation(mdata,fdata);
0.9531344480
The number obtained is 0.95 which is the “strength” of the line. This means that the
closer to 1 the more strength it has.
3. Later on, the equation of the regression line is also found by using Maple with the
following command line:
> regline:=leastsquare[[x,y],y=a*x+b]([mdata,fdata]);
regline := y = 0.1076794795 x + 13.56515106
2
Maple calculations can be found on page 12
2
Roberto Lam
MAT202.6671 (Final Project)
Gordon Crandall
Fall2005
4. With this equation, we can predict the record distance for a shot put in the year
2006 by plugging the value 106 into x (years) and evaluate the equation for
y(distance).
> y:= x->.1076794795*x+13.56515106;
y := x ® 0.1076794795 x + 13.56515106
> y(106);
24.97917589
We observed that the predicted record distance for the year 2006 will be
approximately 24.98 meters.
(Question 2)Obtaining the equations for projectile (shot put) motion.
The position function for a projectile t seconds after release has two natural components,
horizontal X(t) and vertical Y(t). The equations for X(t) and Y(t) are found by solving an
“uncoupled” system of differential equations by simple integration. The two equations
are the following elementary differential equations:
d2 X
dt 2
=0
d2 Y
dt 2
= -g
(-g) is the acceleration due to gravity (9.8 meters/sec2 in the metric system).
Recall from Calculus I that the two above equations are second derivatives of the position
equations X(t) and Y(t) functions, meaning that these equations are the acceleration of the
object.
1. If we solve the pair of equation by integrating both side of each equation we will
get the velocity; then if we integrate again we will get the position functions. But
remember that integration introduces arbitrary constants. To evaluate these
constants, initial conditions on velocity and position are used. The initial
conditions on the velocity for X and Y components at (t=0), are the following:
dX
= V cos(A)
dt
dY
= V sin(A)
dt
3
Roberto Lam
MAT202.6671 (Final Project)
Gordon Crandall
Fall2005
The initial velocity (V) imparted to the shot is supposed to be constant and the angle of
elevation (A) is more than zero and less than ninety degrees (0<A<90)0.
2. Similarly the initial conditions for the position functions X(t) and Y(t) are:
X(0) = 0
Y(0) = H
where 0 and H are the points of the position of the shot (0,H) where it is thrown from; H
is the height of the shot above the ground.
Shot
(0,H)
Ground
(0,0)
3. Now, with the initial conditions set on the velocity and position functions, the
equations of position X(t) and Y(t) can be found by using Maple and introducing
the data(initial conditions) in the command’s line. The results are the following:
>dsolve({diff(X(t),t$2)=0,D(X)(0)=V*cos(A),X(0)=0},X(t));
X(t) = V cos(A) t
And
>dsolve({diff(Y(t),t$2)+g=0,D(Y)(0)=V*sin(A),Y(0)=H},Y(t));
Y(t) = -
1 g t2
+ V sin(A) t + H
2
(Question 3) Length of time.
1. Solving the quadratic equation Y (t )  H  (V sin A)t 
1 2
gt for t, using Maple we
2
get that the length of time t(in seconds that the shot will remain in the air) is given
4
Roberto Lam
MAT202.6671 (Final Project)
Gordon Crandall
Fall2005
by the following formula. Keep in mind that using the quadratic the positive
answer is selected for this case.
> solvefor[t](H+V*sin(A)*t-1/2*g*t^2);
é
ê t = V sin(A) +
ê
ë
2
2
V sin(A) + 2 g H
V sin(A) ,t=
g
ù
2
2
V sin(A) + 2 g H ú
ú
g
û
(Question 4) Horizontal distance.
V sin( A)  V 2 sin 2 ( A)  2 gH
1. Then, knowing that, t 
, the horizontal distance
g
X, traveled by the shot is given by the formula below by solving X(t).
(V 2 cos A) (sin 2 A) 
X 
2 gH
V2
g
(Question 5) Q&A
Taking in consideration that the angle A and g are constants in the above equation; then
we can have the following equation:
X V H
1. What effect does increasing the initial height H have on the horizontal distance X
traveled?
A: If we increase H to a power, let say 2, on the formula above the horizontal
distance X will increase:
X  VH
2. What effect does increasing the initial velocity V have on X?
A: If we increase also V to the power 2, the distance X will also increase:
X V 2 H
3. To maximize the horizontal distance X, should a shot putter try to maximize the
initial height H, the initial velocity V, both or neither?
A: To maximize the horizontal distance X, the shot putter should try to
maximize both H and V, for instance if both increase to a second power we
get the following:
X  V 2H
5
Roberto Lam
MAT202.6671 (Final Project)
Gordon Crandall
Fall2005
(Question 6)Angle of elevation.
1. Using Maple we defined X as a function of A with the previous equation from
2 gH
(V 2 cos A) (sin 2 A)  2
V
Question 4 X 
. And assuming that H=2.1 and
g
V=13.2 we obtained the following formula: (Convert to degrees using Pi radians
=180 degrees).
> X:=A->17.78*cos(A)*sqrt(sin(A)^2+0.24);
X := A ® 17.78 cos(A)
sin(A)2 + 0.24
2. Using an example we plug in:
> X(3.1416/4);
Which gives us
10.81514534
This means that at an angles of 45 degrees (3.1416/4) assuming that H=2.1
and V=13.2, the distance for the shot will be 10.82 meters.
(Question 7) Optimal angle of elevation.
1. It can be shown that the maximal horizontal distance Xmax and the optimal angle
of elevation Aopt are related by the functional equation:
X max  H tan( 2 Aopt )
Then, solving for A in Maple we obtained the following formula:
> solvefor[A](X=H*tan(2*A));
A=
1
æX ö
arctanç ÷
2
è Hø
Now, using H=2.1meters, for the height used in the release of the shot and
1
180
multiply by
to convert radians to degrees, we define A as a function
2

of X in Maple as the following:
> A:=X->28.65*arctan(X/2.1);
æXö
A := X ® 28.65 arctanç
÷
è 2.1ø
6
Roberto Lam
MAT202.6671 (Final Project)
Gordon Crandall
Fall2005
2. Next, using the above formula to find the angle of release for a horizontal distance
X=25 meters, the angle is obtained.
> A(25);
42.60235125
This means that for a distance of 25 meters the angle of release should be approximately
42.60 at a height of 2.1 meters.
3. Using the distance obtained in Question 6, (10.82 meters) and plug it in into
equation on Question 7 we got:
> A(10.81514534);
39.50865653
The angle used for Question 6 was roughly 45 degrees and we obtained 39.5
using the functional equation. The angles differ approximately 5.5 degrees.
(Question 8) Angles in world record distances.
1. Using the functional equation A( X )  28.65 arctan
X
, the angles for eight world
2. 1
record distances is found assuming that the release height
H= 2.1meters. The following table shows them:
Year
1983
Distance(in meters)
22.22
Angle(in degrees)
1985
22.62
42.35110276
1986
22.64
42.35343235
1987
22.79
42.37077511
1987
22.81
42.37307038
1987
22.91
42.38448717
1988
23.06
42.40142823
1990
23.12
42.40814362
42.30363782
7
Roberto Lam
MAT202.6671 (Final Project)
Gordon Crandall
Fall2005
2. Comparing these angles with that on Question 7, the angles used on each record
distances are nearly the same (42 degrees) except for the angle obtained using the
distance from Question 6.
(Question 9) Exponential “fitting”.
1. To fit the data from Question 1 to an exponential curve instead of a linear curve,
we assume an equation of the form y  Ce Kx where C and K are constants to be
found. Taking natural logs of both sides, we obtain:
ln y  ln( Ce Kx )
ln y  ln C  ln C Kx
ln y  Kx  ln C
If this exponential model is a good fit to the data then a plot of X versus lny will
be close to linear and we can find K as the slope of the line and lnC as the vertical
intercept. Then we can recover C by exponentiation.
8
Roberto Lam
MAT202.6671 (Final Project)
Gordon Crandall
Fall2005
2. Taking natural logs of the data on y on Question 1, the following table X vs. lny is
created:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
Year, x
(mdata)
60
60
60
60
60
62
62
64
64
65
67
73
76
76
78
83
85
86
87
87
87
88
90
Distance, y (fdata)
lny
19.38
19.45
19.67
19.99
20.06
20.08
20.10
20.20
20.68
21.52
21.78
21.82
21.85
22.00
22.15
22.22
22.62
22.64
22.79
22.81
22.91
23.06
23.12
2.964241606
2.967847070
2.979094632
2.995232149
2.998727783
2.999724295
3.000719815
3.005682604
3.029167050
3.068982735
3.080992118
3.082826980
3.084200922
3.091042453
3.097837496
3.100992784
3.118834471
3.119718253
3.126321843
3.127199036
3.131573496
3.138099515
3.140698044
9
Roberto Lam
MAT202.6671 (Final Project)
Gordon Crandall
Fall2005
3. Now, with the data in this table, a scatter plot is created:
4. Using Maple, the regression line is found:
> regline:=leastsquare[[x,y],y=a*x+b]([Mdata,Fdata]);
regline := y = 0.005048453605x + 2.694276308
Comparing this equation with the equation ln y  Kx  ln C , it can be say that
K=0.005 and lnC=2.694, therefore C=e2.694 .
5. And, next the correlation coefficient is the following:
> linearcorrelation(Mdata,Fdata);
0.9483814187
The number obtained is close to 1, meaning that is nearly close to a linear.
10
Roberto Lam
MAT202.6671 (Final Project)
Gordon Crandall
Fall2005
(Question 10) Predicting distance and angles.
1. Using the exponential modeling equation y  Ce Kx and the values for
the constants K and C from the previous question and also using the
functional equation from Question 7 X max  H tan( 2 Aopt ) , the
following predictive table can be fill out:
y:=x->14.79072063*exp(0.00504*x);
y := x ® 14.79072063 e
(0.00504 x)
And
> A := proc (X) options operator, arrow;
28.65*arctan(X/2.1) end proc;
æXö
A := X ® 28.65 arctanç
÷
è 2.1ø
Year
2006
Distance(in meters)
Angle( in degrees)
25.23520390
42.62462617
2016
26.53965420
42.74104324
2026
27.91153376
42.85180784
2036
29.35432807
42.95718807
2046
30.87170287
43.05744029
11
Roberto Lam
MAT202.6671 (Final Project)
Gordon Crandall
Fall2005
Maple Calculations for Final Project
>
mdata:=[60,60,60,60,60,62,62,64,64,65,67,73,76,76,78,83,85,
86,87,87,87,88,90];
mdata := [60, 60, 60, 60, 60, 62, 62, 64, 64, 65, 67, 73, 76, 76, 78, 83, 85, 86, 87, 87, 87, 88, 90]
>
fdata:=[19.38,19.45,19.67,19.99,20.06,20.08,20.10,20.20,20.
68,21.52,21.78,21.82,21.85,22.00,22.15,22.22,22.62,22.64,22
.79,22.81,22.91,23.06,23.12];
fdata := [19.38, 19.45, 19.67, 19.99, 20.06, 20.08, 20.10, 20.20, 20.68, 21.52, 21.78, 21.82, 21.85, 22.00, 22.15,
22.22, 22.62, 22.64, 22.79, 22.81, 22.91, 23.06, 23.12]
Question 1
> with(plots):with(stats[statplots]);
[boxplot, histogram, scatterplot, xscale, xshif t, xyexchange, xzexchange, yscale, yshif t, yzexchange, zscale, zshift]
> with(stats[fit]);
[leastmediansquare, leastsquare]
> regline:=leastsquare[[x,y],y=a*x+b]([mdata,fdata]);
regline := y = 0.1076794795 x + 13.56515106
> pic1:=scatterplot(mdata,fdata,symbolsize=50):
> pic2:=plot(rhs(regline),x=50..110,y=-15..30):
> display({pic1,pic2});
12
Roberto Lam
MAT202.6671 (Final Project)
Gordon Crandall
Fall2005
> with(stats[describe]);
[coeff icientof variation
, count, countmissing, covariance, decile, geometricmean, harmonicmean, kurtosis,
linearcorrelation, mean, meandeviation, median, mode, moment, percentile, quadraticmean, quantile,
quartile, range, sk ewness, standarddeviation, sumdata, variance]
> linearcorrelation(mdata,fdata);
linearcorrelationmdata,
(
fdata)
Question 2
> dsolve({diff(x(t),t$2)=0,D(x)(0)=V*cos(A),x(0)=0},x(t));
x(t) = V cos(A) t
>
dsolve({diff(y(t),t$2)+g=0,D(y)(0)=V*sin(A),y(0)=H},y(t));
13
Roberto Lam
MAT202.6671 (Final Project)
Gordon Crandall
Fall2005
y(t) = H + V sin(A) t -
1 g t2
2
Question 3
> solvefor[t](H+V*sin(A)*t-1/2*g*t^2);
é
ê t = V sin(A) +
ê
ë
2
2
V sin(A) + 2 g H
V sin(A) ,t=
g
ù
2
2
V sin(A) + 2 g H ú
ú
g
û
Question 4
> X:=t->VcosA*t;
X := t ® VcosA t
> X((V*sin(A)+(V^2*sin(A)^2+2*g*H)^(1/2))/g);
VcosA (V sin(A) +
V 2 sin(A)2 + 2 g H )
g
Question 5
Look for Question 5 in the project.
Question 6
> X:=A->17.78*cos(A)*sqrt(sin(A)^2+0.24);
X := A ® 17.78 cos(A)
sin(A)2 + 0.24
Examples
> X(3.1416/4);
10.81514534
> plot(X(A),A=0..Pi/2);
14
Roberto Lam
MAT202.6671 (Final Project)
Gordon Crandall
Fall2005
Question 7
> A:=X->28.65*arctan(X/2.1);
æXö
A := X ® 28.65 arctanç
÷
è 2.1ø
> A(25);
42.60235125
Question 8
> A(22.22);
42.30363782
> A(22.62);
42.35110276
> A(22.64);
42.35343235
> A(22.79);
15
Roberto Lam
MAT202.6671 (Final Project)
Gordon Crandall
Fall2005
42.37077511
> A(22.81);
42.37307038
> A(22.91);
42.38448717
> A(23.06);
42.40142823
> A(23.12);
42.40814362
Question 9
> Mdata := [60, 60, 60, 60, 60, 62, 62, 64, 64, 65, 67, 73,
76, 76, 78, 83, 85, 86, 87, 87, 87, 88, 90];
Mdata := [60, 60, 60, 60, 60, 62, 62, 64, 64, 65, 67, 73, 76, 76, 78, 83, 85, 86, 87, 87, 87, 88, 90]
> Fdata := [2.964241606, 2.967847070, 2.979094632,
2.995232149, 2.998727783, 2.999724295, 3.000719815,
3.005682604, 3.029167050, 3.068982735, 3.080992118,
3.082826980, 3.084200922, 3.091042453, 3.097837496,
3.100992784, 3.118834471, 3.119718253, 3.126321843,
3.127199036, 3.131573496, 3.138099515, 3.140698044];
Fdata := [2.964241606, 2.967847070, 2.979094632, 2.995232149, 2.998727783, 2.999724295, 3.000719815,
3.005682604, 3.029167050, 3.068982735, 3.080992118, 3.082826980, 3.084200922, 3.091042453,
3.097837496, 3.100992784, 3.118834471, 3.119718253, 3.126321843, 3.127199036, 3.131573496,
3.138099515, 3.140698044]
> with(stats[statplots]);
[boxplot, histogram, scatterplot, xscale, xshif t, xyexchange, xzexchange, yscale, yshif t, yzexchange, zscale, zshift]
> with(stats[fit]);
[leastmediansquare, leastsquare]
> regline:=leastsquare[[x,y],y=a*x+b]([Mdata,Fdata]);
regline := y = 0.005048453605x + 2.694276308
> with(plots):with(stats[statplots]);
Warning, the name changecoords has been redefined
Warning, these names have been redefined: boxplot,
histogram, scatterplot, xscale, xshift, xyexchange,
xzexchange, yscale, yshift, yzexchange, zscale, zshift
[boxplot, histogram, scatterplot, xscale, xshif t, xyexchange, xzexchange, yscale, yshif t, yzexchange, zscale, zshift]
16
Roberto Lam
MAT202.6671 (Final Project)
Gordon Crandall
Fall2005
> pic1:=scatterplot(Mdata,Fdata,symbolsize=40,color=blue):
> pic2:=plot(rhs(regline),x=50..110,y=3..3.5):
> display({pic1,pic2});
> with(stats[describe]);
[coeff icientof variation
, count, countmissing, covariance, decile, geometricmean, harmonicmean, kurtosis,
linearcorrelation, mean, meandeviation, median, mode, moment, percentile, quadraticmean, quantile,
quartile, range, sk ewness, standarddeviation, sumdata, variance]
> linearcorrelation(Mdata,Fdata);
0.9483814187
> Y:=x->0.5048453605e-2*x+2.694276308;
Y := x ® 0.005048453605 x + 2.694276308
K=0.00504, C=e^2.694=14.79072063
17
Roberto Lam
MAT202.6671 (Final Project)
Gordon Crandall
Fall2005
Question 10
I
> exp(2.694);
14.79072063
> y:=x->14.79072063*exp(0.00504*x);
y := x ® 14.79072063 e
(0.00504 x)
> y(106);
25.23520390
> y(116);
26.53965420
> y(126);
27.91153376
> y(136);
29.35432807
> y(146);
30.87170287
II
> A := proc (X) options operator, arrow;
28.65*arctan(X/2.1) end proc;
æXö
A := X ® 28.65 arctanç
÷
è 2.1ø
> A(25.23520390);
42.62462617
> A(26.53965420);
42.74104324
> A(27.91153376);
42.85180784
> A(29.35432807);
42.95718807
> A(30.87170287);
43.05744029
End.
18