Figure 1 is a graph showing the trajectory of a rugby ball - Edexcel - A-Level Maths Pure - Question 10 - 2018 - Paper 2

To find the x distance where the ball is at least 3 metres high:

Set H = 3 in the derived equation like so:

$3 = -\frac{3}{25}(x - 20)^2 + 12$

Rearranging gives:

$\frac{3}{25}(x - 20)^2 = 12 - 3$

Simplifying further:

$\frac{3}{25}(x - 20)^2 = 9 \Rightarrow (x - 20)^2 = 75$

Taking the square root yields two potential x-values:

$x - 20 = \pm 5\sqrt{3}\Rightarrow x = 20 \pm 5\sqrt{3}$

The distances are thus approximately 20 - 8.66 m and 20 + 8.66 m.

The greatest horizontal distance of the bar, which is at 3 m above the ground, is found in the previous calculation. Using:

$x = 20 + 5\sqrt{3}\approx 28.66 ext{ metres}$

This signifies that the bar's greatest horizontal distance from point O is about 28.66 metres.

One limitation of the model is that it assumes the ball follows a perfect parabolic trajectory, neglecting factors such as wind resistance and the ball being affected by spin, which could alter its path in real situations.