How can i add a distance to time plot in Matlab

I have a school project to simulate free fall of an object with air resistance. I'm just having trouble figuring out how to plot a distance / time graph. I dont have any distance variable in my code, do i have to add one or is there maybe some kind of work around?

clear all;
V0=0; % initial speed
m=5; % mass in kg
g=9.81; % gravity acceleration kg/m3
rho=1.2; % Air density
A=0.09; % Object area
cw=0.4; % Numerical drag coefficient
k=0.5*cw*rho*A; % Coefficient
N=200; % Time step
V=zeros(1,N); % Speed
V(1)=V0;
deltat=0.2;
for i=1:N-1
V(i+1)=V(i)+deltat*(g-(k/m)*V(i)^2);
end
t=(0:N-1)*deltat;
plot(t,V);
xlabel('time in sec');
ylabel('velocity in m/s');
legend ('Euler Method','location','south');

[Cris LaPierre]

NOTE: User is deleting their comments. You can find the full thread here

Answer 1

[Cris LaPierre]

You have velocity. Has your teacher taught you how to use time and velocity to get distance yet?

[Fares Espiro]

Yes i believe its s = vt + 0.5*at^2
clear all;
V0=0; % initial speed
m=0.62369; % mass in kg
g=9.82; % gravity acceleration kg/m3
rho=1.225; % Air density
A=0.18/4; % Object area
cw=0.6; % Numerical drag coefficient
k=0.5*cw*rho*A; % Coefficient
N=10; % Time step
V=zeros(1,N); % Speed
V(1)=V0;
deltat=0.2;
for i=1:N-1
V(i+1)=V(i)+deltat*(g-(k/m)*V(i)^2);
t=(0:N-1)*deltat;
d=8- V(i+1)*t + 0.5*(g-(k/m)*V(i)^2)*t.*t % 8 is the starting height
end
vterminal=sqrt(g*m/k); % Terminal velocity
plot(t,d);
xlabel('time in sec');
ylabel('velocity in m/s');
legend ('Euler Method','location','south');
I think this is what you mean right?
But I dont understand why my results vary when i change the Time step
I appreciate your help

[Cris LaPierre]

EDIT: OP response:

Yes i believe its s = vt + 0.5*at^2

clear all;
V0=0; % initial speed
m=0.62369; % mass in kg
g=9.82; % gravity acceleration kg/m3
rho=1.225; % Air density
A=0.18/4; % Object area
cw=0.6; % Numerical drag coefficient
k=0.5*cw*rho*A; % Coefficient
N=10; % Time step
V=zeros(1,N); % Speed
V(1)=V0;
deltat=0.2;
for i=1:N-1
    V(i+1)=V(i)+deltat*(g-(k/m)*V(i)^2);
    t=(0:N-1)*deltat;
    d=8- V(i+1)*t + 0.5*(g-(k/m)*V(i)^2)*t.*t % 8 is the starting height
end

d = 1×10

8.0000 7.8036 8.0000 8.5892 9.5712 10.9460 12.7136 14.8740 17.4272 20.3732

d = 1×10

8.0000 7.4128 7.2144 7.4047 7.9836 8.9513 10.3077 12.0528 14.1866 16.7091

d = 1×10

8.0000 7.0302 6.4370 6.2204 6.3804 6.9170 7.8302 9.1200 10.7865 12.8295

d = 1×10

8.0000 6.6633 5.6838 5.0615 4.7965 4.8887 5.3382 6.1448 7.3088 8.8299

d = 1×10

8.0000 6.3187 4.9693 3.9518 3.2662 2.9126 2.8908 3.2010 3.8431 4.8171

d = 1×10

8.0000 6.0016 4.3055 2.9117 1.8202 1.0311 0.5442 0.3597 0.4775 0.8977

d = 1×10

8.0000 5.7153 3.7008 1.9567 0.4828 -0.7208 -1.6541 -2.3172 -2.7100 -2.8324

d = 1×10

8.0000 5.4614 3.1603 1.0967 -0.7293 -2.3178 -3.6688 -4.7822 -5.6581 -6.2965

d = 1×10

8.0000 5.2398 2.6852 0.3362 -1.8074 -3.7453 -5.4777 -7.0046 -8.3259 -9.4416

vterminal=sqrt(g*m/k); % Terminal velocity
plot(t,d);
xlabel('time in sec');
ylabel('velocity in m/s');
legend ('Euler Method','location','south');

I think this is what you mean right?

But I dont understand why my results vary when i change the Time step

I appreciate your help

REPLY

Because you change the accuracy of your numerical approach. Because you have acceleration, your velocity is contantly changing. Larger time steps are less accurate because they assume a constant velocity during each deltat.

[Fares Espiro]

But the graph im getting have a decreasing derivatives, which means that the velocity also decreases. but in reality the velocity increases when you drop an object. What have i done wrong?

[Cris LaPierre]

I don't know what you are modeling, but the velocity profile makes sense for something that is thrown up in the air. Initially the velocity is high, but slows due to gravity until it stops going up (V=0), it then accelerates as it falls back down.

[John D'Errico]

Hmm. If I just copy and paste in your code, executing it, what is the velocity vector that I see?

V
V =
Columns 1 through 10
0        1.962       3.9207       5.8694       7.8016        9.711       11.592       13.437       15.243       17.005
Columns 11 through 20
18.717       20.376       21.979       23.524       25.008        26.43       27.788       29.083       30.314       31.482
Columns 21 through 30
32.588       33.632       34.617       35.544       36.414        37.23       37.995        38.71       39.377       39.999
Columns 31 through 40
40.579       41.118       41.619       42.085       42.517       42.917       43.287        43.63       43.948       44.241
Columns 41 through 50
44.512       44.762       44.993       45.206       45.402       45.583        45.75       45.904       46.045       46.175
Columns 51 through 60
46.295       46.405       46.507         46.6       46.686       46.765       46.837       46.904       46.965       47.021
Columns 61 through 70
47.073        47.12       47.164       47.204       47.241       47.275       47.306       47.334        47.36       47.385
Columns 71 through 80
47.407       47.427       47.445       47.463       47.478       47.493       47.506       47.518       47.529       47.539
Columns 81 through 90
47.549       47.557       47.565       47.572       47.579       47.585       47.591       47.596       47.601       47.605
Columns 91 through 100
47.609       47.613       47.616       47.619       47.622       47.624       47.627       47.629       47.631       47.633

Do you see a velocoty that decreases or one that increses? Strangely, it looks like V is increasing to me, but maybe this is that new math stuff. :) :)

Eventually, it seems to reach an asymptote, roughly at V=47.653.

Perhaps you are missing that you were computing VELOCITY. And that the derivative of velocity is acceleration, which is indeed decreasing with time. Of course, due to air resistance, the velocity will eventually reach a maximum (that is, terminal velocity) so the acceleration will go to zero. And that just means that eventually, the drag due to air resistance will exactly balance the acceleration due to gravity, once the faling object reaches that terminal velocity.

With some effort, one could even compute the terminal velocity directly from those formulas, finding the velocity where the two accelerations must exactly balance out.

[Fares Espiro]

My project was to calculate how long it would take a falling abject(Basketball) to reach the ground from a specefic height in this case 8m.
I think I found why I had decreasing derivatives it was because i had a minus after the 8 in :
d=8- V(i+1)*t + 0.5*(g-(k/m)*V(i)^2)*t.*t
I dont know if im thinking right but this is what I have now

clear all;
V0=0; % initial speed
m=0.62369; % mass in kg
g=9.82; % gravity acceleration kg/m3
rho=1.225; % Air density
A=0.18/4; % Object area
cw=0.6; % Numerical drag coefficient
k=0.5*cw*rho*A; % Coefficient
N=8; % Time step
V=zeros(1,N); % Speed
V(1)=V0;
deltat=0.15;
for i=1:N-1
V(i+1)=V(i)+deltat*(g-(k/m)*V(i)^2); % velocity
t=(0:N-1)*deltat; %Time
d=V(i+1)*t + 0.5*(g-(k/m)*V(i)^2)*t.*t % Distance that the Basketball have fallen
end
plot(t,d);
xlabel('time in sec');
ylabel('Distande in m');
legend ('Euler Method','location','south');
Now im getting that it takes the Ball around 0.7s secound to reach the ground from a height of 8m --> x = 0.7 when Y = 8
when we experimented in school the time was around 0.9s
Have I done anything wrong in my calculations?
I really appreciate your help

[Cris LaPierre]

EDIT: OP's response:

My project was to calculate how long it would take a falling abject(Basketball) to reach the ground from a specefic height in this case 8m.

I think I found why I had decreasing derivatives it was because i had a minus after the 8 in :

d=8- V(i+1)*t + 0.5*(g-(k/m)*V(i)^2)*t.*t

I dont know if im thinking right but this is what I have now

clear all;
V0=0; % initial speed
m=0.62369; % mass in kg
g=9.82; % gravity acceleration kg/m3
rho=1.225; % Air density
A=0.18/4; % Object area
cw=0.6; % Numerical drag coefficient
k=0.5*cw*rho*A; % Coefficient
N=8; % Time step
V=zeros(1,N); % Speed
V(1)=V0;
deltat=0.15;
for i=1:N-1
V(i+1)=V(i)+deltat*(g-(k/m)*V(i)^2); % velocity
t=(0:N-1)*deltat; %Time
d=V(i+1)*t + 0.5*(g-(k/m)*V(i)^2)*t.*t % Distance that the Basketball have fallen
end
plot(t,d);
xlabel('time in sec');
ylabel('Distande in m');
legend ('Euler Method','location','south');

Now im getting that it takes the Ball around 0.7s secound to reach the ground from a height of 8m --> x = 0.7 when Y = 8

when we experimented in school the time was around 0.9s

Have I done anything wrong in my calculations?

I really appreciate your help

REPLY:

It depends what you want d to be. If it is the current height, then the proper equation is d=8-...

If you just want to know the distance the ball has traveled, then d=V... will give you that.