Theory If we neglect the air resistance, the flight path of a projectile launche

Question

Theory If we neglect the air resistance, the flight path of a projectile launched at an initial speed vo and an angle of departure relative to the horizontal is a parabola (see Fig. 1) Figure 1: Flightpath of a projectile If we assume that the projectile is launched from the surface of the Earth, i.e., at yo-0, then the altitude of the projectile as a function of time is , where g-9.81 m/sis the gravitational constant, is time (in seconds), and V-Vo sin (in meters per second). The distance of the projectile from the launching position as a function of time is x(t)-Ut, where U-vo cos The velocity of the projectile in the horizontal direction, i.e., parallel to the x-axis, IS u(t)U, and the velocity in the vertical direction, i.e., parallel to the y-axis, is If we include the effects of air resistance, then the altitude of the projectile as a function of time can be approximated as where k is the coefficient of resistance (with the unit 1/s). The distance of the projectile relative to the initial position as a function of time is kt The velocity of the projectile in the horizontal direction is u(t)Uekt, and in the vertical direction

Explanation / Answer

Code

flightpath.m

function [y,x,u,v]=flightpath(v0,theta,k,t)

g = 9.81;

V=v0*sind(theta);

U=v0*cosd(theta);

if (k == 0)

    y=((-1/2)*g*(t.^2))+(V.*t);
    y=y./1000;
    x=U.*t;
    x=x./1000;
    u=U;
    v=(-g.*t)+V;  
else
    y=(((-g.*t)./k)+(((k*V+g)/k^2).*(1-exp(-k.*t))));
    y=y./1000;  
    u=U*exp(-k.*t);
    x=(u/k).*(1-exp(-k.*t));
    x=x./1000;  
    v=(V*exp(k.*t))+((g/k).*(exp(-k.*t)-1));
end
end

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