Hello Eleven Eleven Originally Posted by

**Eleven Eleven** I've spent over an hour on this and I'm getting nowhere.

Here's my question-

In rugby, a player scores a 'try' by getting the ball into the try area (see diagram I uploaded) Once a try has been scored the scoring team is allowed to take a goal-kick from any point on an imaginary line drawn out from the point where the try was scored and parallel to the side of the field. A player scores a try 16 metres to the left of the left-hand goal post.

Using trigonometry find the best position to place the ball for the kick at goal after this try has been scored to maximise the chance of a successful goal.

Secondly - it is harder to kick a goal from near the sideline than from the centre of the field. Use trigonometry to try and measure how much more difficult it is to kick a goal from near the sideline then from near the centre of the field. Quantify the answer (e.g is it twice, three or four times more difficult?)

In the attached diagram:$\displaystyle d$ is the distance to the left of the left-hand post where the try was scored.

$\displaystyle w$ is the distance apart of the posts.

$\displaystyle x$ is the distance back from the goal-line from which the kick is taken.

Then we need to make a couple of assumptions about the nature of the kick at goal:For very small values of $\displaystyle x$ and $\displaystyle d$ it will not be possible for the kicker to achieve the elevation required to clear the bar. So we'll assume that for the values we shall consider, $\displaystyle x$ and/or $\displaystyle d$ are large enough for this not to be a problem.

We'll also assume that as $\displaystyle x$ and $\displaystyle d$ become larger, the directional accuracy of the kick remains constant. (In practice this is not necessarily the case: a kick taken from the half-way line requires considerably more energy than one from the 22 metre line, and therefore may be subject to a greater degree of error in its direction.)

Therefore we shall concern ourselves only with the angle subtended by the posts at the point from which the kick is taken (the angle $\displaystyle \theta$ in the diagram): the greater the angle $\displaystyle \theta$, the greater the chance that the kick is successful.

In the diagram, then:$\displaystyle \tan\phi = \frac{d}{x}$

$\displaystyle \tan(\theta +\phi)=\frac{w+d}{x}$

At this stage, we have a choice about how to find the value of $\displaystyle x$ that gives the maximum value of $\displaystyle \theta$. If you are allowed to use graphical methods, it is very easy to put some values into a spreadsheet to produce a graph. If you have Excel, you can open the attached spreadsheet file. If not, I've extracted the graph.

For $\displaystyle d = 16$ and $\displaystyle w = 5.6$, the optimum value of $\displaystyle x$ is around $\displaystyle 18.6$ metres.

If you're not allowed to use a graphical method, I have a calculus method which gives the value of $\displaystyle x$ as $\displaystyle \sqrt{d(w+d)}$, giving $\displaystyle 18.59$ as the answer.

I'll post this solution for you if you'd like to see it.

As far as the final part of the question is concerned, this is pretty open ended. In the diagram the angle subtended from a point directly in front of the posts is $\displaystyle \psi$, where$\displaystyle \tan\tfrac12\psi= \frac{w}{2x}$

You could work out some values of $\displaystyle \theta$ and $\displaystyle \psi$ for various values of $\displaystyle d$ and $\displaystyle x$. (If the kick is on the touchline, $\displaystyle d$ will be around $\displaystyle 30$ metres, depending on the width of the pitch.) Then if, for certain values, $\displaystyle \psi \approx 2\theta$, for example, you could say that it was twice as difficult from the touchline than from under the posts.

Grandad