Figure 1 is a graph showing the trajectory of a rugby ball. The height of the ball above the ground, $H$ metres, has been plotted against the horizontal distance, $x$ metres, measured from the point where the ball was kicked. The ball travels in a vertical plane. The ball reaches a maximum height of 12 metres and hits the ground at a point 40 metres from where it was kicked. (a) Find a quadratic equation linking $H$ with $x$ that models this situation. (b) The ball passes over the horizontal bar of a set of rugby posts that is perpendicular to the path of the ball. The bar is 3 metres above the ground. (c) Use your equation to find the greatest horizontal distance of the bar from $O$. (c) Give one limitation of the model.

A-Level Edexcel Maths Pure Proof: Figure 1 is a graph showing the

Figure 1 is a graph showing the trajectory of a rugby ball - Edexcel - A-Level Maths Pure - Question 9 - 2018 - Paper 2

Question 9

Figure 1 is a graph showing the trajectory of a rugby ball. The height of the ball above the ground, $H$ metres, has been plotted against the horizontal distance, $... show full transcript

Worked Solution & Example Answer:Figure 1 is a graph showing the trajectory of a rugby ball - Edexcel - A-Level Maths Pure - Question 9 - 2018 - Paper 2

Step 1

Find a quadratic equation linking H with x that models this situation.

96%

114 rated

Answer

To find the quadratic equation, we know that the vertex form of a quadratic function is given by:

$H = a(x - h)^2 + k$

where (h, k) is the vertex. Here, the maximum height (12) is at the vertex, so we have:

$H = a(x - 20)^2 + 12$

Since the ball hits the ground at 40 m (where $H = 0$ ), we can substitute these values to find $a$ :

0 = 100a + 12 100a = -12 a = -0.12$$ Thus, the quadratic equation is: $$H = -0.12(x - 20)^2 + 12$$

Step 2

The ball passes over the horizontal bar of a set of rugby posts that is perpendicular to the path of the ball. The bar is 3 metres above the ground.

99%

104 rated

Answer

To find where the ball is at the height of 3 m, we set:

$3 = -0.12(x - 20)^2 + 12$

Solving for $x$ gives:

-0.12(x - 20)^2 = -9 (x - 20)^2 = 75 x - 20 = ext{±}\sqrt{75} x = 20 ext{±} 5 ext{√}3\

This means the ball passes over the bar at two points: $20 + 5 ext{√}3$ and $20 - 5 ext{√}3$ .

Step 3

Use your equation to find the greatest horizontal distance of the bar from O.

96%

101 rated

Answer

The greatest horizontal distance is at the point:

$x = 20 + 5 ext{√}3 \ ext{ or } 20 - 5 ext{√}3$

Calculating this:

\approx 28.66 m\

So, the greatest horizontal distance of the bar from $O$ is approximately 28.66 m.

Step 4

Give one limitation of the model.

98%

120 rated

Answer

One limitation of the model is that it assumes a perfect parabolic trajectory. In reality, air resistance and factors such as wind may alter the path of the ball, making it deviate from this idealized curve.

Figure 1 is a graph showing the trajectory of a rugby ball - Edexcel - A-Level Maths Pure - Question 9 - 2018 - Paper 2

Worked Solution & Example Answer:Figure 1 is a graph showing the trajectory of a rugby ball - Edexcel - A-Level Maths Pure - Question 9 - 2018 - Paper 2

Find a quadratic equation linking H with x that models this situation.

The ball passes over the horizontal bar of a set of rugby posts that is perpendicular to the path of the ball. The bar is 3 metres above the ground.

Use your equation to find the greatest horizontal distance of the bar from O.

Give one limitation of the model.

Join the A-Level students using SimpleStudy...