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The problem shows a graph of the trajectory of a rugby ball. The height \( H \) of the ball above the ground (in meters) is plotted against the horizontal distance \( x \) (in meters) from the point where the ball was kicked. The equation governing the height of the ball is given as:

\[
H = \frac{3}{100}x(40 - x)
\]

### Key information:
- The ball reaches a maximum height of 12 meters.
- The ball hits the ground at 40 meters horizontally from where it was kicked.
- The ball passes over a horizontal bar 3 meters above the ground.
- We are asked to find the greatest horizontal distance of the bar from \( O \) (the point of kick-off).

### Step-by-step Solution:

1. **Given Equation**:
   \[
   H = \frac{3}{100}x(40 - x)
   \]
   where \( H \) is the height, and \( x \) is the horizontal distance.

2. **Setting Height to 3 Meters**:  
   We need to find the values of \( x \) when the ball passes over the bar, i.e., when \( H = 3 \). Substituting this into the equation:
   \[
   3 = \frac{3}{100}x(40 - x)
   \]
   Multiply both sides by 100:
   \[
   300 = 3x(40 - x)
   \]
   Divide both sides by 3:
   \[
   100 = x(40 - x)
   \]
   Expand the right-hand side:
   \[
   100 = 40x - x^2
   \]
   Rearrange to form a quadratic equation:
   \[
   x^2 - 40x + 100 = 0
   \]

3. **Solving the Quadratic Equation**:  
   Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1 \), \( b = -40 \), and \( c = 100 \).
   \[
   x = \frac{-(-40) \pm \sqrt{(-40)^2 - 4(1)(100)}}{2(1)}
   \]
   \[
   x = \frac{40 \pm \sqrt{1600 - 400}}{2}
   \]
   \[
   x = \frac{40 \pm \sqrt{1200}}{2}
   \]
   \[
   x = \frac{40 \pm 34.64}{2}
   \]
   So, the two possible values of \( x \) are:
   \[
   x_1 = \frac{40 + 34.64}{2} = \frac{74.64}{2} = 37.32
   \]
   \[
   x_2 = \frac{40 - 34.64}{2} = \frac{5.36}{2} = 2.68
   \]

4. **Conclusion**:
   The greatest horizontal distance where the ball passes over the bar is approximately **37.32 meters**.

Would you like a more detailed explanation or have any questions?

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Here are 5 related questions to expand your understanding:
1. What is the maximum height reached by the ball, and how is it calculated from the equation?
2. How would the equation change if the maximum height was 15 meters instead of 12 meters?
3. What role does the quadratic formula play in solving projectile motion problems like this one?
4. Can you derive the time of flight for the rugby ball if you know the equation for \( H \)?
5. How can the range of the projectile be calculated using the given equation?

**Tip**: Always double-check your solutions to quadratic equations by substituting the results back into the original equation!

Use the equation to find the greatest horizontal distance of the bar from O. Given the height equation H = (3/100) x (40 - x), where the ball passes over the bar at a height of 3 meters.

This problem explores how to use a quadratic equation to find the greatest horizontal distance from the origin where a rugby ball passes over a bar 3 meters above the ground. Learn key algebraic concepts like solving quadratic equations and understanding projectile motion. Suitable for Grades 9-11.

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