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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
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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
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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
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