Submit Al Question 9. A batsman hits a cricket ball high into the air. It flies into the crowd following a parabolic curve. A mathematician in the crowd observes that the ball was 1 m above the ground as it was hit, 10 m above the ground as it passed over the boundary fence 80 m away and that it was caught 1 m above the ground 10 m into the crowd. She uses this information to calculate that the highest point, in metres, the ball reached as it flew through the air was closest to A. 070

To find the highest point the ball reached, we can model the trajectory of the ball as a parabolic function. We will use the given points to determine the equation of the parabola. ### Known Conditions: 1. The ball was 1 m above the ground when hit (point A). 2. The ball was 10 m above the ground when it passed over the boundary fence 80 m away (point B). 3. The ball was caught 1 m above the ground 10 m into the crowd (point C). ### Points: - Point A: \((0, 1)\) - Point B: \((80, 10)\) - Point C: \((90, 1)\) ### Step 1: Formulate the Parabola The general form of a parabolic equation is: \[ y = ax^2 + bx + c \] We will use the three points to create a system of equations. ### Step 2: Set Up the Equations Using the points: 1. For point A \((0, 1)\): \[ c = 1 \] 2. For point B \((80, 10)\): \[ 10 = 80^2 a + 80b + 1 \] Simplifying gives: \[ 10 = 6400a + 80b + 1 \implies 6400a + 80b = 9 \quad \text{(Equation 1)} \] 3. For point C \((90, 1)\): \[ 1 = 90^2 a + 90b + 1 \] Simplifying gives: \[ 1 = 8100a + 90b + 1 \implies 8100a + 90b = 0 \quad \text{(Equation 2)} \] ### Step 3: Solve the System of Equations We will solve the system of equations formed by Equation 1 and Equation 2: 1. \(6400a + 80b = 9\) 2. \(8100a + 90b = 0\) Let's solve these equations together. I'll proceed to solve the system of equations. Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}6400a+80b=9\\8100a+90b=0\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}a=\frac{9-80b}{6400}\\8100a+90b=0\end{array}\right.\) - step2: Substitute the value of \(a:\) \(8100\times \frac{9-80b}{6400}+90b=0\) - step3: Multiply the terms: \(\frac{81\left(9-80b\right)}{64}+90b=0\) - step4: Multiply both sides of the equation by LCD: \(\left(\frac{81\left(9-80b\right)}{64}+90b\right)\times 64=0\times 64\) - step5: Simplify the equation: \(729-720b=0\) - step6: Move the constant to the right side: \(-720b=0-729\) - step7: Remove 0: \(-720b=-729\) - step8: Change the signs: \(720b=729\) - step9: Divide both sides: \(\frac{720b}{720}=\frac{729}{720}\) - step10: Divide the numbers: \(b=\frac{81}{80}\) - step11: Substitute the value of \(b:\) \(a=\frac{9-80\times \frac{81}{80}}{6400}\) - step12: Calculate: \(a=-\frac{9}{800}\) - step13: Calculate: \(\left\{ \begin{array}{l}a=-\frac{9}{800}\\b=\frac{81}{80}\end{array}\right.\) - step14: Check the solution: \(\left\{ \begin{array}{l}a=-\frac{9}{800}\\b=\frac{81}{80}\end{array}\right.\) - step15: Rewrite: \(\left(a,b\right) = \left(-\frac{9}{800},\frac{81}{80}\right)\) The solution to the system of equations gives us the values of \(a\) and \(b\): 1. \(a = -\frac{9}{800}\) 2. \(b = \frac{81}{80}\) ### Step 4: Find the Vertex of the Parabola The vertex of a parabola given by the equation \(y = ax^2 + bx + c\) can be found using the formula: \[ x = -\frac{b}{2a} \] Substituting the values of \(a\) and \(b\): \[ x = -\frac{\frac{81}{80}}{2 \cdot -\frac{9}{800}} = -\frac{\frac{81}{80}}{-\frac{18}{800}} = \frac{81 \cdot 800}{80 \cdot 18} \] Now, let's calculate this value. Calculate the value by following steps: - step0: Calculate: \(\frac{81\times 800}{\left(80\times 18\right)}\) - step1: Remove the parentheses: \(\frac{81\times 800}{80\times 18}\) - step2: Multiply the numbers: \(\frac{64800}{80\times 18}\) - step3: Multiply the numbers: \(\frac{64800}{1440}\) - step4: Reduce the fraction: \(45\) The x-coordinate of the vertex is \(x = 45\). ### Step 5: Calculate the Maximum Height Now, we will substitute \(x = 45\) back into the parabolic equation to find the maximum height \(y\). Using the equation: \[ y = ax^2 + bx + c \] Substituting \(a\), \(b\), and \(c\): \[ y = -\frac{9}{800}(45^2) + \frac{81}{80}(45) + 1 \] Let's calculate this value. Calculate the value by following steps: - step0: Calculate: \(\frac{-9}{800}\times 45^{2}+\frac{81}{80}\times 45+1\) - step1: Rewrite the fraction: \(-\frac{9}{800}\times 45^{2}+\frac{81}{80}\times 45+1\) - step2: Multiply the numbers: \(-\frac{9^{3}}{32}+\frac{81}{80}\times 45+1\) - step3: Multiply the numbers: \(-\frac{9^{3}}{32}+\frac{729}{16}+1\) - step4: Evaluate the power: \(-\frac{729}{32}+\frac{729}{16}+1\) - step5: Reduce fractions to a common denominator: \(-\frac{729}{32}+\frac{729\times 2}{16\times 2}+\frac{16\times 2}{16\times 2}\) - step6: Multiply the numbers: \(-\frac{729}{32}+\frac{729\times 2}{32}+\frac{16\times 2}{16\times 2}\) - step7: Multiply the numbers: \(-\frac{729}{32}+\frac{729\times 2}{32}+\frac{16\times 2}{32}\) - step8: Transform the expression: \(\frac{-729+729\times 2+16\times 2}{32}\) - step9: Multiply the numbers: \(\frac{-729+1458+16\times 2}{32}\) - step10: Multiply the numbers: \(\frac{-729+1458+32}{32}\) - step11: Add the numbers: \(\frac{761}{32}\) The maximum height \(y\) that the ball reached is approximately \(23.78\) meters. ### Conclusion The highest point the ball reached as it flew through the air was closest to **23.78 meters**. Thus, the answer is not among the options provided (like 070), but the calculated height is approximately **23.78 meters**.