How to solve a problem in relative motion?

Question

I have a project due tomorrow that I have been struggling with for a month. I am utterly stumped, any help would be appreciated. The problem is:

There is a ferris wheel, 30m in radius, set 80m above water level. There is a boat 150m away from the center of the ferris wheel, and is moving toward the wheel at 10m/s. The wheel is turning at 0.2rad/s. Assuming no initial velocity and no friction, write a Mathematica program that will give the range for a person to successfully jump off the wheel and land in a boat, which is 1m in length.

I solved this analytically, but the assignment requires recursive programming. What I want to do is output the range of X coordinates when the person lands in the water and the corresponding location of the boat, and if they match, output the time value; otherwise go to the next increment. Here's what I have:

Clear[n, h];
Vx[0] = 0; Pyi[0] = 110; h = 300; t[0] = 0; Vyf[0] = 0;
t[n_] := t[n] = t[n - 1] + h
Vx[n_] := Vx[n] = 6 Cos[-0.2 t[n]]
Vy[n_] := Vy[n] = 6 Sin[-0.2 t[n]]
Pxi[n_] := Pxi[n] = 30 Cos[0.5 Pi - 0.2 t[n]]
Pyi[n_] := Pyi[n] = 80 + 30 Sin[0.5 Pi - 0.2 t[n]]
Pyf[n_] := Pyf[n] = Pyi[n - 1] + Vy[n] h - 4.9 h^2
Pxf[n_] := Pxf[n] = Pxi[n - 1] + Vx[n] h
Sxf[n_] := Sxf[n] = 150 - 10 t[n]
Do[If[Pxf[n] == Sxf[n], Print[t[n]*0.2], " "], {n, 0, 10}]

Execution yields no output. What should I do?

For reference:

t - time
Vx - initial x velocity (yielded from Vi Cos[theta])
Vy - initial y velocity
Pxi - location at which the person leaves the wheel (x)
Pyi - the same as above, y direction
Pyf - final location, y direction (should equal 0, don't know if I need to set that)
Pxf - final location, x direction
Sxf - location of boat

I executed t[14] and got 0.0014. I'm not sure why.

EDIT:

I've fixed the Do[If[Pxf[n] == Sxf[n], Print[t[n]*0.2], " "], {n, 0, 10}] to Do[If[Pxf[n] - Sxf[n] < 0.1, Print[t[n]*0.2], " "], {n, 0, 20}]. It produces an output now, but the answer is wrong (does not output the range of 9.6 to 9.91). I think I'm having two problems:

-I don't know when to put [n-1] and when to just use t[n]

-Since the velocity equations for initial velocity and the free-fall velocity are different, I don't get how to tie them in. Here is my new code:

Clear[Vx, Vy, Pxi, Pyi, n, h, Sxf];
Vx[0] = 0; Pyi[0] = 110; h = 0.001; t[0] = 0; Vyf[0] = 0;
t[n_] := t[n] = n h
Vx[n_] := Vx[n] = 6 Cos[-0.2 t[n]]
Vy[n_] := Vy[n] = 6 Sin[-0.2 t[n]]
Pxi[n_] := Pxi[n] = 30 Cos[0.5 Pi - 0.2 t[n]]
Pyi[n_] := Pyi[n] = 80 + 30 Sin[0.5 Pi - 0.2 t[n]]
Pyf[n_] := Pyf[n] = Pyi[n - 1] + Vy[n] h - 4.9 h^2
Pxf[n_] := Pxf[n] = Pxi[n - 1] + Vx[n] h
Sxf[n_] := Sxf[n] = 150 - 10 t[n]

Do[If[Pxf[n] - Sxf[n] < 0.1, Print[t[n]*0.2], " "], {n, 0, 20}]

Answer 1

One of the nice things about Mathematica is that it supports many different styles of programming. I think your code has the aspect of more "traditional" code that one might write in a different programming language. Perhaps your code is correct in spirit, but it seems to me overly complicated and as written obviously is not okay because it throws a recursion error.

You could increase the $RecursionLimit and depending on what exactly is wrong that might help, but I think in this case it is better to rewrite.

This is more along the lines of what you need perhaps,

 Clear[Vx, Vy, Pxi, Pyi];
 Pxi = 30 Cos[0.5 Pi - 0.2 t];
 Pyi = 80 + 30 Sin[0.5 Pi - 0.2 t];
 Vx = D[Pxi, t];
 Vy = D[Pyi, t];
 Sxf = 150 - 10*t;
 Table[({Pxi /. t -> u + Vx*u, Sxf} /. u -> T) /. 
 Solve[Pxi + Vx*T - (9.8/2)*T^2 == 0, T][[2]], {t, 0, 15, 0.1}]

Given expressions for position on the ferris wheel over time, it computes the components of the velocity vector. Then for a range of values of time, it prints out the pairs you requested. Needless to say this code is still not as elegant as it could be, I suspect, and I'm not sure the physics of the situation is correctly represented here, but you should be able to check against your analytical solution (this code produces an intersect of the boat and the jumpers at a position of around 10).

Note that you will run into trouble using == to check for equality since you are using discrete time steps and inexact numbers. You may need to use FindRoot or NSolve.