-------------------------------------------------------------- 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 --------------------------------------------------------------