MATLAB ODE function problem: 150lb skydiver steps out of an airplane 2 miles up.
ID: 3759449 • Letter: M
Question
MATLAB ODE function problem: 150lb skydiver steps out of an airplane 2 miles up. Gravity accelerated the diver's fall but is decelerated by the drag on the chute. 1st order differential equation modelling for this problem is m(dv/dt)=-mg+Kv^2 Initial velocity is 0, air resistance K=4/15.
1. Find the numerical solution using MATLAB ode solvers, ode45, and ode23.
2. On same graph, plot velocity vs time graph for both ode solver, label them.
3. What's the terminal velocity in ft/s according to the solvers?
4. What's the terminal velocity in mph?
5. At what time the diver's reach terminal velocity? Output in seconds I would be very grateful if anyone can explain the commands.
Thank You!
Explanation / Answer
MATLAB ODE function problem: 150lb skydiver steps out of an airplane 2 miles up. Gravity accelerated the diver's fall but is decelerated by the drag on the chute. 1st order differential equation modelling for this problem is m(dv/dt)=-mg+Kv^2 Initial velocity is 0, air resistance K=4/15.
1. Find the numerical solution using MATLAB ode solvers, ode45, and ode23.
2. On same graph, plot velocity vs time graph for both ode solver, label them.
3. What's the terminal velocity in ft/s according to the solvers?
4. What's the terminal velocity in mph?
5. At what time the diver's reach terminal velocity? Output in seconds I would be very grateful if anyone can explain the commands.
Thank You!
Answer:
Numerical solution for the equation
m(dv/dt)=-mg+Kv^2
In this
m= mass
g=gravity(ft/s^2)
v=velocity(m/s)
and initial velocity assumed to be zero.
Then the equation becomes
m(dv/dt)=-mg+4/15 v^2
Preparing to solve numerically
first, we put the equation in the form of
dy/dt=f(t,y)
for ,our example the equation becomes
dv/dt=-g+4/15 v^2/m
Solving numerically
we have variety of ode solvers
most commonly used is:ode 45
we have to provide
i. intial value for v
ii. time range for which solution to be sought
The function
functionrk=f(t,y) here y is velocity
mass=150;
g=9.81;
rk=-g+4/15*y^2/mass;
The solution
time range=[0 30]; /* where one is starting time and another one is finishing time. assumed as 30 seconds*/
initial velocity-0;
[t,y]=ode45(@f,time range,initial velocity)
plot(t,y)
ylabel('velocity(m/s)')
xlabel('time(s)')
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