Changing the launch angle requires only a few small changes.

Input := 

Clear[V0]

Setting theta at 30 degrees

Input := 

theta = 30 Pi/180;

And setting up the same equations as before

Input := 

eq1 = horiz[t]==d[xfar];

Input := 

eq2 = vert[t]==10;

Input := 

s3 = Solve[{eq1,eq2},{V0,t}]

Output =

{{V0 -> -133.038, t -> -4.04939}, 
 
  {V0 -> 133.038, t -> 4.04939}}

Clearly the second solution is the reasonable one

Input := 

V0 = s3[[2,1,2]];

When does the ball get down the first base line?

Input := 

s4 = Solve[horiz[t]==425,t]

Output =

{{t -> 3.68878}}

And how high must the wall be at that point?

Input := 

vert[s4[[1,1,2]]] //N

Output =

30.6601

Let's get the height of the wall as a function of the distance to the plate

Input := 

ht2[dist_] := vert[Solve[horiz[t]==dist,t][[1,1,2]]]

Input := 

b=Plot[ht2[dist],{dist,325,467},
	PlotStyle -> AbsoluteThickness[4]];

Input := 

Show[a,b]

Output =

-Graphics-

So we see that the steeper the launch angle, the higher the wall must be. (The thicker curve corresponds to the 30 degree launch angle.)