Hello, Daniel!
The kicker can kick from anywhere along the line TF (see diagram).
Where is the best place to kick from? Code:
T d A 5.6 B
*    *        *
 * *
 * *
x  * *
 * *
β * *
 * θ *
* *
C *

F *
Let TA = d, .TC = x.
Let /ACB = θ, ./TCA = β
We want to maximize θ.
. . Note that: .tanβ = d/x .[1]
. . . . . . . . . . . . . . . . . . d + 5.6
We have: .tan(θ + β) .= .
. . . . . . . . . . . . . . . . . . . . x
. . . . . . .tanθ + tanβ . . . . d + 5.6
Then: .  .= . 
. . . . . .1 + tanθ·tanβ . . . . . .x
. . . . . . . . . . . . .tanθ + (d/x) . . . d + 5.6
Substitute [1]: .  .= .
. . . . . . . . . . . .1 + (d/x)·tanθ . . . . .x
. . . . . . . . . . . . . . . . . . . . . . .5.6x
This simplifies to: .tanθ .= .
. . . . . . . . . . . . . . . . . . . .x² + d² + 5.6d
. . . . . . . . . . . . . . . . . . . . . . . . . . . .d² + 5.6d  x²
Differentiate: .sec²θ·(dθ/dx) .= .5.6 []
. . . . . . . . . . . . . . . . . . . . . . . . . . .(x² + d² + 5.6d)²
The derivative equals zero when that numerator equals zero.
Hence: . d² + 5.6d  x² .= .0 . . → . . x² .= .d² + 5.6d
. . . . . . . . . . . . . ________
Therefore: .x .= .√d² + 5.6d