Computer Algebra Recipes

3.2. SECOND-ORDER MODELS 119
> tf:=solve(x=xf,t);
μ
xb k + V cos(Á) ¡ xf k
ln
V cos(Á)
tf := ¡
k
In the y direction, the gravitational force on the ball must be included as well
as air resistance. Newton's second law yields the ODE given in yeq.
> yeq:=diff(y(t),t,t)=-k*diff(y(t),t)-g;
yeq := d2
μ
d
y(t) =¡k
dt2 dt y(t)
¡ g
Solving yeq with the dsolve command, subject to the initial condition y(0) =
yb, _y(0) = V sin(Á), where yb is the ball's initial y-coordinate, and taking the
rhs, yields y.
> y:=rhs(dsolve(fyeq,y(0)=yb,D(y)(0)=V*sin(phi)g,y(t)));
y := ¡ e(¡kt) (V sin(Á) k + g)
¡ gt
k 2
k + yb k2 + V sin(Á) k + g
k2 Evaluating y at t=tf , and equating to the fence height yf , yields the following
formidable looking transcendental equation eq for the unknown angle Á.
> eq:=eval(y,t=tf)=yf;
(xb k + V cos(Á) ¡ xf k)(V sin(Á) k + g)
eq := ¡
k2 V cos(Á)
μ
xb k + V cos(Á) ¡ xf k
g ln
V cos(Á)
+
+ yb k2 + V sin(Á) k + g
k 2
k 2
= yf
To solve the transcendental equation for Á, thegivenvaluesxb =0m,yb =2m,
xf =20m,yf =3:5m, V =15m/s,andg =9:8m/s 2 , are entered. The drag
coe±cient is taken to be k =0:01 s ¡1 .
> xb:=0: yb:=2: xf:=20: yf:=3.5: V:=15: g:=9.8: k:=0.01:
The transcendental equation must be solved numerically. This is done in the
following two command lines, two di®erent search ranges (in radians) being
speci¯ed for Á to determine the two possible angles, ©1 and ©2, atwhichthe
ball can be thrown to just clear the fence. Making use of the ditto operator, we
convert the ¯rst angle to degrees.
> Phi[1]:=fsolve(eq,phi,0..0.8); evalf(convert(%,degrees));
©1 := 0:6696475750 38:36797980 degrees
> Phi[2]:=fsolve(eq,phi,0.8..Pi/2);
©2 := 0:9693759172
Without air resistance, it was previously found that the smaller of the two
possible angles was 37.47 degrees, about 1 degree less than the value found
above. It is left as an exercise for you to compare the values for the upper
angle, with and without air resistance.
To animate the ball, the ¯rst angle ©1 will be selected.