Comp Sci C++ Free Fall Velocity Homework

[RJLiberator]

Homework Statement

I'm using ROOT (c++) to graph a numerical representation of the equation of motion.
For simplicity, I am using initial velocity = 10 and initial height = 10.
change in time = 0.05
so we find velocity by V_n = V_(n-1) -9.81*0.05

For some reason, this code is not giving me a correct reading.

I graph this against the computational method, and the computational method works perfectly.

Homework Equations

The Attempt at a Solution

  float x[100], y[100];
  int n = 100;
  float timestep = 9.81*0.05;
  float v = 10;
  for (int i=0;i<n;i++) {
  x[i] = i*timestep;
  y[i] = v-timestep;
  v = v-timestep;
  }
TGraph *gr = new TGraph(n,x,y);
  gr->SetMarkerColor(kBlue);
  gr->SetMarkerStyle(29);
  gr->Draw();

 

[Dr. Courtney]

I don't see how your code represents the kinematic equations:

x(t) = x_0 + v_x*t
y(t) = y_0 + v_y0*t - 0.5*g*t^2

v_x(t) = v_x0
v_y(t) = v_y0 - g*t

 

[RJLiberator]

So what you are saying is that I am only observing the v_y(t) element? As my code should represent that part of it.

The x-axis should be time in seconds.

This is an example graph I did earlier based on my computational graph (which works out perfectly).

 

[Dr. Courtney]

You don't seem to have given proper consideration to the horizontal component of velocity, nor to how x(t) depends on it.

 

[Mark44]

Homework Statement

I'm using ROOT (c++) to graph a numerical representation of the equation of motion.
For simplicity, I am using initial velocity = 10 and initial height = 10.
change in time = 0.05
so we find velocity by V_n = V_(n-1) -9.81*0.05

Since the object is in freefall (which your title indicates), why is the velocity decreasing?

For some reason, this code is not giving me a correct reading.

I graph this against the computational method, and the computational method works perfectly.

Homework Equations

The Attempt at a Solution

  float x[100], y[100];
  int n = 100;
  float timestep = 9.81*0.05;
  float v = 10;
  for (int i=0;i<n;i++)
  {
     x[i] = i*timestep;
     y[i] = v-timestep;
     v = v-timestep;
  }
  TGraph *gr = new TGraph(n,x,y);
  gr->SetMarkerColor(kBlue);
  gr->SetMarkerStyle(29);
  gr->Draw();

Use

 at the top of your code. This highlights C/C++ reserved words.
Also, indenting loops and other control structures (such as if and switch) makes your code much easier to read.
I would do it like this:
[code=c]
float x[100], y[100];
  int n = 100;
  float timestep = 9.81*0.05;
  float v = 10;
  for (int i=0;i<n;i++) {
  x[i] = i*timestep;
  y[i] = v-timestep;
  v = v-timestep;
  }
TGraph *gr = new TGraph(n,x,y);
  gr->SetMarkerColor(kBlue);
  gr->SetMarkerStyle(29);
  gr->Draw();

One other thing -- declaring timestep like this is not a good idea:
float timestep = 9.81*0.05;
Keep the 9.81 constant out -- just make the time steps actual time intervals, and use the constant when you calculate the values in your x and v arrays. And for vertical distances, I would definitely use y, not x. That confused me for a bit.

 

[gneill]

You've incorporated the acceleration into your timestep variable, no doubt as a nod to efficiency so as not to perform the multiplication g*Δt on each iteration. But you've used timestep to keep track of the time, too.

 

[RJLiberator]

Excellent guys!

That does seem to be the issue! Time for me to get back to work on this, I'll report back if that was the problem!

@Mark44 thank you for the remarks there. I will apply them to my code once I have a working product as right now, things are a bit of a mess :). Coding the night away.

 

[RJLiberator]

Looking good guys! That was the problem!

Still lots of touching up to do, but it is pretty awesome that I got the numerical and computational approaches working here.

Since the object is in freefall (which your title indicates), why is the velocity decreasing?

Good point, thank you for this observation.

Looking pretty solid.

 

[gneill]

Since the object is in freefall (which your title indicates), why is the velocity decreasing?

If the chosen axes have "up" as positive then I don't see a problem with an initial positive upward velocity decreasing and then going negative. The old, "rock is tossed vertically upwards at the edge of a tall cliff" scenario comes to mind.

 

[Mark44]

If the chosen axes have "up" as positive then I don't see a problem with an initial positive upward velocity decreasing and then going negative. The old, "rock is tossed vertically upwards at the edge of a tall cliff" scenario comes to mind.

I agree, but we shouldn't have to guess at what the problem is about.