--------------------------------------------------------------
The first part is easy and well known as 45 degrees.
For the second part I got 0.985515 radians or 56.4658 degrees, or
specifically the solution of sin(theta)log((1+sin(theta))/cos(theta))
= 1
For details, see:
http://members.bellatlantic.net/~devjoe/thrown-distance.html
Joseph DeVincentis
--------------------------------------------------------------
After basic calc I get the angle is about 56.47 degrees from ground. I
cannot find a simple way to get a closed form, but it is a root (in
radians) of

Cos[x]^2 ( 2ArcTanh[Tan[x/2]] + Sec[x]Tan[x] )

Chris Lomont
--------------------------------------------------------------
Ed:

Interesting problem!  The solution basically requires setting up an equation 
which expresses the ARC length as a function of an initial fixed velocity [V] 
and the angle [theta] thrown.  Next, take the derivative of that _expression 
with respect to the primary variable theta, set equal to zero and solve for 
value of theta.

ARC = SQRT (4 Ym^2 + (Xm^2)/4)
          + ((Xm^2)/8Ym) ln ((4Ym + 2 SQRT (4 Ym^2 + (Xm^2)/4))/Xm)  

Where Ym = (V cos (90-theta))^2 / 64
          Xm = (V cos (theta))(V cos (90-theta)) / 16

R. Wainwright
--------------------------------------------------------------