To study the relationships of linear fluid drag to zero drag, in two-dimensional projectile behavior 

>	x0:=v0*cos(theta0)*t; y0:=v0*sin(theta0)*t-g*t^2/2;

 

(1)

 

(1)

 

>	tau:=vterm/g;

 

(2)

 

>	xL:=v0*cos(theta0)*tau*(1-exp(-t/tau)); yL:=(v0*sin(theta0)+vterm)*tau*(1-exp(-t/tau))-vterm*t;

 

(3)

 

(3)

 

Now we choose some of these parameters: 

>	g:=9.8; vterm:=2; v0:=1; theta0:=1;

 

(4)

 

(4)

 

(4)

 

(4)

 

>	qu:=[solve(y0,t)];

 

(5)

 

>	t1:=qu[2];

 

(6)

 

>	plot([x0,y0,t=0..t1]);

 

>	plot([xL,yL,t=0..t1]);

 

>	plot({[x0,y0,t=0..t1],[xL,yL,t=0..t1]});

 

>	tL1:=fsolve(yL,t,avoid={t=0});

 

(7)

 

>	down:=evalf(subs(t=tL1,[xL,yL]));

 

(8)

 

Notice that this numerical calculation agrees with the graph above, but gives us more significant figures. 

            NOW, let's change the terminal velocity, to something less than the initial velocity. 

>	vterm:=0.2;

 

(9)

 

>	t2:=fsolve(y0,t,avoid={t=0});

 

(10)

 

Notice, as expected, that t2 is the same as t1, since that involves no drag at all. 

>	plot({[x0,y0,t=0..t1],[xL,yL,t=0..t1]});

 

>	t2L:=fsolve(yL,t,avoid={t=0});

 

(11)

 

>	evalf(subs(t=t2L,[xL,yL]));

 

(12)

 

We see, interestingly enough, that in this case, the time to again reach the ground is about 60% or so of the time  

    for the case without drag; however, the distance traveled is truly much less! 

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