(*^

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:[font = title; inactive; preserveAspect; startGroup]
Exploring times that a particle projected upward passes through the same point -- once going up and once coming down
:[font = section; inactive; preserveAspect; startGroup]
BRIEF ABSTRACT
:[font = subsection; inactive; preserveAspect; endGroup]
First find out maximum height H a given projectile will go.  Then test
the time T(a) (0 < a < 1) it takes between when the projectile passes
distance aH going up and then coming down. Develop T(a) as a function of a.
:[font = section; inactive; Cclosed; preserveAspect; startGroup]
GENERAL INFORMATION
:[font = subsection; inactive; preserveAspect; endGroup]
FileName:  UPDOWN

Full title:  An exploration of the times at which a particle projected upward passes through the same point - once going up and once coming down.

Developer:   Brian J. Winkel, Department of Mathematical Sciences, United States Military Academy, West Point NY 10996 USA.
Phone: 914-938-3200.  Email: ab3646@usma2.usma.edu.
FAX: 914-938-2409. 

Contact:  
Brian J. Winkel, Department of Mathematical Sciences, 
United States Military Academy, West Point NY 10996 USA.
Phone: 914-938-3200.  Email: ab3646@usma2.usma.edu.
FAX: 914-938-2409. 

Support:  
The production of this material is supported by the National Science Foundation under Division of Undergraduate Education grant DUE-9352849:  Development Site for Complex, Technology-Based Problems in Calculus with Applications in Science and Engineering and the Arvin Foundation of Columbus IN. 
:[font = section; inactive; Cclosed; preserveAspect; startGroup]
STATEMENT OF PROBLEM
:[font = subsection; inactive; preserveAspect]
Consider a particle, projected upwards in a uniform gravitational field with no resistance. 
:[font = subsection; inactive; preserveAspect; startGroup]
(a) Say the initial velocity is 100 m/sec and the gravitational field is due to earth's gravity, i.e. g = - 9.81 m/sec^2.  

We fire a projectile directly up from ground zero.
:[font = subsubsection; inactive; preserveAspect; endGroup]
(i) Determine the maximum height, H,  the particle attains.  

(ii) For heights h = 0, .1 H, .2H, .3 H, . . . , .9 H, H determine the time (from t = 0) it takes to reach height h (on the UPward path) and the time (from t = 0) it takes to reach height h again (on the DOWNward) path.

(iii) Look at the numbers you get and make a conjecture about the relationship between these times and the height h. 
:[font = subsection; inactive; preserveAspect]
(b) Test your conjecture with another initial velocity.  Do you get the same results?  the same relationship?  Does the initial velocity seem to enter your relationship? 
:[font = subsection; inactive; preserveAspect; endGroup]
(c) Verify your conjecture you made in (a) (iii) in general for the general equation for a projectile fired vertically with constant acceleration a, initial velocity v0, and initial vertical displacement x0.
:[font = section; inactive; Cclosed; preserveAspect; startGroup]
KEYWORDS 
:[font = subsection; inactive; preserveAspect; endGroup]
Projectile motion in one dimension.
:[font = section; inactive; Cclosed; preserveAspect; startGroup]
TEACHER NOTES
:[font = subsection; inactive; preserveAspect]
ISSUES RELATED TO THE PROBLEM
:[font = subsection; inactive; preserveAspect; startGroup]
Prerequisites
:[font = subsubsection; inactive; preserveAspect; endGroup]
Students need to know the equation describing projectile motion in a gravitational field in one dimension without resistance.
:[font = subsection; inactive; preserveAspect]
Time allotment - time management
:[font = subsection; inactive; preserveAspect]
Expectations 
:[font = subsection; inactive; preserveAspect]
Future payoffs
:[font = subsection; inactive; preserveAspect; startGroup]
Extensions  
:[font = subsubsection; inactive; preserveAspect]
The students might consider the same problem with resistance.  
:[font = subsubsection; inactive; preserveAspect]
As is stated in the References and Sources section below this product of times is NOT independent of initial velocity.
:[font = subsubsection; inactive; preserveAspect; endGroup]
One might also consider the problem for a bouncing object with 100% coefficient of restitution or less than 100% coefficient of restitution and ask for relationships between later passages through the given height.
:[font = subsection; inactive; preserveAspect; startGroup]
References and Sources
:[font = subsubsection; inactive; preserveAspect; endGroup; endGroup]
From "A Problem Characterization of Uniformly Accelerated Motion" in the Problems and Solutions section of  SIAM Review: June 1995, Volume 37, Number 2, pp. 249-250.  Author of the problem M. S. Klamkin.

The problem posed in the journal was:

"It is a known result that if a particle is projected upwards in a uniform gravitational field with no resistance, then the product of the two times it takes to pass through any point of its path is independent of the initial velocity of the projection. 

Prove or disprove that the result cannot hold if additionally the particle was subject to a resistance as some function of the velocity."

NB:  The solution by W. B. Jordan presented proves that the product of the two times is NOT independent of the initial velocity when resistance is present.
;[s]
5:0,0;108,1;120,2;770,3;771,4;795,-1;
5:1,10,8,Times,1,12,0,0,0;1,10,8,Times,3,12,0,0,0;1,10,8,Times,1,12,0,0,0;1,10,8,Times,1,12,0,0,0;1,10,8,Times,1,12,0,0,0;
:[font = section; inactive; Cclosed; preserveAspect; startGroup]
POSSIBLE SOLUTION(S)
:[font = subsection; inactive; Cclosed; preserveAspect; startGroup]
(a) We enter a model equation for vertical projectile motion.
:[font = input; preserveAspect]
g = -9.81;
:[font = input; preserveAspect]
p[t_] = 1/2 g t^2 + 100 t; 
:[font = subsubsection; inactive; preserveAspect; startGroup]
We determine the maximum height and corresponding time. 
:[font = input; Cclosed; preserveAspect; startGroup]
TopData = {p[t],t}/.Solve[p'[t]==0,t][[1]]
:[font = output; output; inactive; preserveAspect; endGroup]
{509.683995922528, 10.19367991845056}
;[o]
{509.684, 10.1937}
:[font = input; preserveAspect; endGroup]
H = TopData[[1]];
:[font = subsubsection; inactive; preserveAspect; startGroup]
We set out a list of the heights we wish to examine.
:[font = input; preserveAspect; endGroup]
data = {0, .1 H, .2 H, .3 H, .4 H, .5 H, .6 H, 
					.7 H, .8 H, .9 H, H};
:[font = subsubsection; inactive; preserveAspect; startGroup]
We determine the times at which these heights are attained.  We list the times in pairs - first element of the pair is the time at which the height is encountered in the UPward flight and the second element of the pair is the time at which the height is encountered in the DOWNward flight.
:[font = input; preserveAspect]
time = {};
:[font = input; preserveAspect]
Do[AppendTo[time, t/.Solve[p[t] == data[[i]],t]],
	{i,1, Length[data]}];
:[font = input; Cclosed; preserveAspect; startGroup]
time
:[font = output; output; inactive; preserveAspect; endGroup; endGroup]
{{0., 20.38735983690112}, 
 
  {0.5231060341435903, 19.86425380275753}, 
 
  {1.076175423038574, 19.31118441386255}, 
 
  {1.665035407399841, 18.72232442950128}, 
 
  {2.29768940630496, 18.08967043059616}, 
 
  {2.985659722869036, 17.40170011403209}, 
 
  {3.746630662245913, 16.64072917465521}, 
 
  {4.610371483127766, 15.77698835377336}, 
 
  {5.634927670744567, 14.75243216615655}, 
 
  {6.970155290348235, 13.41720454655289}, 
 
  {10.19367991845054, 10.19367991845058}}
;[o]
{{0., 20.3874}, {0.523106, 19.8643}, {1.07618, 19.3112}, 
 
  {1.66504, 18.7223}, {2.29769, 18.0897}, {2.98566, 17.4017}, 
 
  {3.74663, 16.6407}, {4.61037, 15.777}, {5.63493, 14.7524}, 
 
  {6.97016, 13.4172}, {10.1937, 10.1937}}
:[font = subsubsection; inactive; Cclosed; preserveAspect; startGroup]
Here are some trial data plots which give possible relationships between the times at which the particle attains a given height is a linear function of the respective height.   For the discussion below we define T1 to be the first time the particle passes a given height and T2 the second time the particle passes the given height.
:[font = input; preserveAspect]
Trial1Data = Table[{data[[i]],time[[i,1]] + time[[i,2]]},
				{i,1,Length[time]}];
:[font = input; Cclosed; preserveAspect; startGroup]
Trial1Plot = ListPlot[Trial1Data,PlotStyle->PointSize[.02],
			AxesLabel->{"height", "T1 + T2"}]
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[(300)] .58438 .104 0 2 Msboxa
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[(500)] .9581 .104 0 2 Msboxa
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Graphics["<<>>"]
;[o]
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This does not appear to demonstrate any relationship. So we try another combination of T1 and T2.
:[font = input; preserveAspect]
Trial2Data = Table[{data[[i]],time[[i,2]] - time[[i,1]]},
				{i,1,Length[time]}];
:[font = input; Cclosed; preserveAspect; startGroup]
Trial2Plot = ListPlot[Trial2Data,PlotStyle->PointSize[.02]]
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This does not appear parabolic, but perhaps rather like a square root function in h.  We could try to fit it and see if when we change v0 the fit changes.
:[font = input; Cclosed; preserveAspect; startGroup]
Trial3Data = Table[{data[[i]],time[[i,2]]/time[[i,1]]},
				{i,1,Length[time]}]
:[font = message; inactive; preserveAspect]
                                 1
Power::infy: Infinite expression -- encountered.
                                 0.
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{{0, DirectedInfinity[]}, 
 
  {50.9683995922528, 37.97366596101027}, 
 
  {101.9367991845056, 17.94427190999916}, 
 
  {152.9051987767584, 11.24440017689384}, 
 
  {203.8735983690112, 7.872983346207417}, 
 
  {254.841997961264, 5.828427124746193}, 
 
  {305.8103975535168, 4.441518440112254}, 
 
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This appears hyperbolic and indeed we get a good fit.  We could then see if when we change v0 the fit changes.
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Here is the final data set which gives the relationship that the product of the times at which the particle attains a given height is a linear function of the respective height. 
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{{0, 0.}, {50.9683995922528, 10.39111102798222}, 
 
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And here we see that the product of the times is a linear function of h, indeed, the function appears to go through the origin.
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The plot of the data and the linear function confirm the model and affirm the conjecture that the product of the two times at which the particle reaches height h
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MathPictureEnd

:[font = output; output; inactive; preserveAspect; endGroup]
Graphics["<<>>"]
;[o]
-Graphics-
:[font = text; inactive; preserveAspect]
The small constant term is an indication that the function does go through the origin.
:[font = text; inactive; preserveAspect]
And the slope appears to be - 2/g.
:[font = input; Cclosed; preserveAspect; startGroup]
-2/g
:[font = output; output; inactive; preserveAspect; endGroup; endGroup; endGroup]
0.2038735983690112
;[o]
0.203874
:[font = subsection; inactive; Cclosed; preserveAspect; startGroup]
(c) We offer up a general solution that "if a particle is projected upwards in a uniform gravitational field with no resistance, then the product of the two times it takes to pass through any point of its path is independent of the initial velocity of the projection." 
:[font = subsubsection; inactive; preserveAspect; startGroup]
First we offer the general equation for a projectile fired vertically with constant acceleration a, initial velocity v0, and initial vertical displacement x0.
:[font = input; Cclosed; preserveAspect; startGroup]
q[t_] = -1/2 a t^2 + v0 t + x0
:[font = output; output; inactive; preserveAspect; endGroup; endGroup]
-(a*t^2)/2 + t*v0 + x0
;[o]
     2
-(a t )
------- + t v0 + x0
   2
:[font = subsubsection; inactive; preserveAspect; startGroup]
We determine the two times at which the particle passes through any point (height), say h.
:[font = input; Cclosed; preserveAspect; startGroup]
sol = Solve[q[t]==h,t]
:[font = output; output; inactive; preserveAspect; endGroup]
{{t -> (2*v0 - (4*v0^2 - 4*a*(2*h - 2*x0))^(1/2))/(2*a)}, 
 
  {t -> (2*v0 + (4*v0^2 - 4*a*(2*h - 2*x0))^(1/2))/(2*a)}}
;[o]
                       2
       2 v0 - Sqrt[4 v0  - 4 a (2 h - 2 x0)]
{{t -> -------------------------------------}, 
                        2 a
 
                        2
        2 v0 + Sqrt[4 v0  - 4 a (2 h - 2 x0)]
  {t -> -------------------------------------}}
                         2 a
:[font = input; preserveAspect; endGroup]
t1 = t/.sol[[1]]; t2 = t/.sol[[2]];
:[font = subsubsection; inactive; preserveAspect; startGroup]
And we see that the product of the two times it takes to pass through any point, say h, of its path is independent of the initial velocity.  The product of the times is, however, not constant but rather a linear function of  h.
:[font = input; Cclosed; preserveAspect; startGroup]
Simplify[t1  t2]
:[font = output; output; inactive; preserveAspect; endGroup]
(2*(h - x0))/a
;[o]
2 (h - x0)
----------
    a
:[font = input; Cclosed; preserveAspect; startGroup]
Simplify[t2 - t1]
:[font = output; output; inactive; preserveAspect; endGroup]
(2*(-2*a*h + v0^2 + 2*a*x0)^(1/2))/a
;[o]
                  2
2 Sqrt[-2 a h + v0  + 2 a x0]
-----------------------------
              a
:[font = input; Cclosed; preserveAspect; startGroup]
Simplify[t1/t2]
:[font = output; output; inactive; preserveAspect; endGroup; endGroup; endGroup; endGroup]
(v0 - (-2*a*h + v0^2 + 2*a*x0)^(1/2))/
 
  (v0 + (-2*a*h + v0^2 + 2*a*x0)^(1/2))
;[o]
                     2
v0 - Sqrt[-2 a h + v0  + 2 a x0]
--------------------------------
                     2
v0 + Sqrt[-2 a h + v0  + 2 a x0]
:[font = section; inactive; Cclosed; preserveAspect; startGroup]
ISSUES IN SOLUTION
:[font = subsection; inactive; preserveAspect; endGroup; endGroup]
The difficulty in assigning this problem appears to be in how much of a hint to give the students in manipulating the data to obtain possible relationships.  Students who are familiar with "fooling: with data to obtain relationships and competent data manipulators in Mathematica, other CAS's, or a spreadsheet will be able to massage the data more easily. 
;[s]
3:0,0;268,1;279,2;362,-1;
3:1,12,9,Times,1,14,0,0,0;1,13,10,Times,3,14,0,0,0;1,12,9,Times,1,14,0,0,0;
^*)