Flyballs with tailwinds

Modified 4PM Friday, Frb 1,2002.

The magnitude of the drag force acting on the ball is F_D = -k v_w².
The velocity relative to the ground is v = v_x i + v_y j, so the velocity relative to the wind is v_w = ( v_x - tailwind) i + v_y j
The speed relative to the wind is then v_w = sqrt( ( v_x - tailwind)² + v_y² ).
The direction, relative to the wind is u_w = [( v_x - tailwind)/v_w] i + [ v_y/v_w] j.
The force vector is parallel to v_w, so F_D = F_D u_w and
F_D =-k v_w² { [( v_x - tailwind)/v_w] i + [ v_y/v_w] j }.
Therefore, F_D = [-k ( v_x - tailwind)v_w] i + [-k v_y v_w] j.

Then the acceleration in the x-direction satisfies dv_x / dt = [(-k/m) ( v_x - tailwind) sqrt( ( v_x - tailwind)² + v_y² ) ]
The acceleration in the y-direction satisfies dv_y / dt = -g - [(k/m) v sqrt( ( v_x - tailwind)² + v_y² ) ]

g = 32 ft/s/s, k/m = 32/140², at sea level. (remember, that k/m will decrease at higher altitudes)

Thus, the Maple code will look something like
v0:= 180; theta:= .9; tailwind:=10; fly:= dsolve( { diff(x(t),t,t) = -32/140^2 * sqrt( (diff(x(t),t)-tailwind)^2 + diff(y(t),t)^2)* (diff(x(t),t)-tailwind) , diff(xyt),t,t) = -32 -32/140^2 * sqrt( (diff(x(t),t)-tailwind)^2 + diff(y(t),t)^2) * diff(y(t),t) , x(0)=0, y(0)=0, D(x)(0)=v0*cos(theta), D(y)(0)=v0*sin(theta) }, {x(t),y(t)}, numeric, output=listprocedure ); xt:= subs( fly, x(t)) ; yt:= subs( fly, y(t)); vxt:= subs( fly, diff(x(t),t)); vyt:= subs( fly, diff(y(t),t));

You may modify the code in flyball2.mws to study these trajectories.