Please Help  Rugby Goal Kicking Problem
Hi, can someone please help me with this problem? It states:
Suppose a try is scored at T, then the kicker can kick to the posts from anywhere along the line TF perpendicular to the try line PQ. (Refer to Figure 1) Where is the best place to kick from?
Many kickers seem to think that the further they go back (within their kicking limit) the better. But this is not necessarily the case. The best position for the kick to be taken is clearly where angle ACB subtended by the posts along TF is a maximum.
Here is the diagram:
http://img.photobucket.com/albums/v3...ce/figure1.jpg
Basically, I think I have to use trig, and a circle subtending the goal posts, but I'm stuck at that, I have hardly any measurements and angles (the distance between the goal posts is 5.6m.
Thanks for any help,
Daniel
Calculus unnecessary  easy geometric solution
Quote:
Originally Posted by
Soroban Hello, Daniel! 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
Or ....... you could recognise that the spot you want will be the point where the circle passing through A and B is >tangent< to the line TF. This follows from a couple of simple circle geometry theorems. Then some trivial algebra to get the distance from the tryline.
By the way ... just passing by and couldn't resist posting the >simple< approach.
Ciao.