Model a ski jump curve The ski jump starts off horizontally

Model a ski jump curve: The ski jump starts off horizontally 80 feet above the bottommost part of the jump. It then slopes downward until it bottoms out at a horizontal distance of 120 feet from the starting position. The jump then heads back up and continues an additional 20 feet horizontally and five feet higher than the bottom of the ski jump. Develop your own MATLAB program to determine precisely the ski jumping trajectory and generate a plot for the trajectory curve.

Solution

m = 70; % ski jumper mass (kg)
psi = 0; % Ski angle to horizontal (deg)
rho = 1.03; % air density kg/m^3
vx0 = 28.7; % initial horizontal velocity m/s
vy0 = -1.2; % initial vertical velocity m/s
g = 9.81; % gravitational accel m/s^2
initial_w = [0;0;vx0;vy0]; % initial solution vector
options = odeset(\'Events\',@event);
[times,sols] = ode45(@eom,[0,8],initial_w,options);
figure1 = figure; % Plot the trajectory
plot(sols(:,1),sols(:,2),\'LineWidth\',2,\'Color\',[1 0 0]);
figure2 = figure; % Plot the velocity
vel = sqrt(sols(:,3).^2+sols(:,4).^2);
plot(times,vel)
function dwdt = eom(t,w)
% The equations for the rates of change of x,y,vx,vy
x = w(1); y = w(2); vx = w(3); vy = w(4);
phi = atan(-vy/vx)*180/pi;
alpha = psi + phi; % Angle of attack (degrees)
Fl = 0.5*rho*(-0.8468 + 0.08963*alpha - 0.001292*alpha^2);
Fd = 0.5*rho*(-0.5072 + 0.04398*alpha - 2.861e-4*alpha^2);
dxdt = vx; dydt = vy;

V = sqrt(vx^2+vy^2); % Magnitude of velocity
dvxdt = -Fd*vx*V/m - Fl*vy*V/m;
dvydt = -Fd*vy*V/m + Fl*vx*V/m - g;
dwdt = [dxdt;dydt;dvxdt;dvydt];
end
function [ev,s,dir] = event(t,w)
% Function to detect when skier lands
x = w(1); y = w(2);
q = atan(-y/x)*180/pi; % Angle of skier below start
ev = q-32; % Hits slope when angle-32 degrees = 0
s = 1; % Stop when the event occurs
dir = 0; % Either sign of zero crossing is OK.
end
end