Case Study & Practical Drill for Year 12 OCR Mathematics | Year 12 OCR 数学：案例分析实战演练

📚 Case Study & Practical Drill for Year 12 OCR Mathematics | Year 12 OCR 数学：案例分析实战演练

In Year 12 OCR Mathematics, students are expected to apply pure mathematical techniques, mechanics and statistics to solve real-world problems. This comprehensive case study blends all three strands through a basketball shooting scenario. You will model the trajectory of a shot using kinematics and quadratic functions, use differentiation to find the maximum height, determine launch parameters, and then perform a binomial hypothesis test on a player’s free-throw success rate. This integrated approach mirrors the style of exam questions and deepens your understanding of how mathematics describes the world around us.

在 Year 12 OCR 数学中，学生需要运用纯数、力学和统计来解决现实问题。这个综合性案例通过篮球投篮情景融合了三个分支。你将用运动学和二次函数对投篮轨迹进行建模、使用微分求最大高度、确定出手参数，然后对球员的罚球成功率进行二项分布假设检验。这种整合式方法反映了考试题的风格，并加深你对数学如何描述周围世界的理解。

1. The Case Study Scenario | 案例情景介绍

A student is investigating the physics of a basketball shot. Using a video tracking app, she records the vertical height y (in metres) of the ball at various horizontal distances x (in metres) from the release point. The data collected is shown in the table below. She then fits a quadratic model to the data. Meanwhile, her coach claims that her free-throw success rate is 70%. To check this claim, she records 12 free throws and counts 8 successes. The case study will take you through the modelling, calculus and statistical inference step by step.

一位学生正在研究篮球投篮的物理原理。通过视频跟踪应用，她记录了球在出手点不同水平距离 x（米）处的垂直高度 y（米）。收集的数据如下表所示。她随后对数据拟合了一个二次模型。同时，她的教练声称其罚球命中率为 70%。为了验证这一说法，她记录了 12 次罚球，命中 8 次。本案例将带领你逐步完成建模、微积分和统计推断。

The measured points (x, y): (1.0, 2.5), (2.0, 4.4), (3.0, 5.4), (4.0, 5.6), (5.0, 5.0), (6.0, 3.6).

测量点 (x, y) 为：(1.0, 2.5), (2.0, 4.4), (3.0, 5.4), (4.0, 5.6), (5.0, 5.0), (6.0, 3.6)。

2. Modelling the Basketball Shot – Kinematics | 投篮建模 – 运动学

A projectile launched with initial speed u at an angle θ to the horizontal, ignoring air resistance, follows the parametric equations x = u cosθ t and y = h + u sinθ t − ½gt², where h is the initial height and g = 9.8 m/s². If we set h = 0 for the release point, eliminating t gives the Cartesian trajectory equation: y = x tanθ − (g/(2u² cos²θ)) x².

以初速度 u、与水平夹角 θ 发射的抛体（忽略空气阻力）遵循参数方程 x = u cosθ t 和 y = h + u sinθ t − ½gt²，其中 h 为初始高度，g = 9.8 m/s²。若将出手点设定为 h = 0，消去 t 得到笛卡尔轨迹方程：y = x tanθ − (g/(2u² cos²θ)) x²。

This shows that the path is a quadratic function of the form y = ax² + bx + c, with a = −g/(2u² cos²θ), b = tanθ and c = h. The negative coefficient of x² confirms the downward-opening parabola observed in the data.

这表明轨迹是形如 y = ax² + bx + c 的二次函数，其中 a = −g/(2u² cos²θ)，b = tanθ，c = h。x² 的负系数证实了数据中观察到的开口向下的抛物线。

3. Fitting a Quadratic Model to Data | 数据拟合二次模型

Using a calculator’s quadratic regression capability, the student obtains the least‑squares fit: y = −0.49x² + 3.51x + 0.18. The coefficient of determination R² is 0.998, indicating an excellent fit to a parabolic shape. The small constant term 0.18 reflects the fact that the release point is very close to y = 0, validating the simplification h ≈ 0.

学生使用计算器的二次回归功能，得到了最小二乘拟合：y = −0.49x² + 3.51x + 0.18。决定系数 R² 为 0.998，表明与抛物线形状极为贴合。微小的常数项 0.18 反映了出手点非常接近 y = 0，从而验证了 h ≈ 0 的简化。

Therefore, we adopt the model y = −0.49x² + 3.51x for further analysis, neglecting the y‑intercept as it merely shifts the curve slightly and will not affect the shape or vertex location significantly.

因此，我们采用模型 y = −0.49x² + 3.51x 进行后续分析，忽略 y 截距，因为它只是略微平移曲线，不会显著影响形状或顶点位置。

4. Finding the Maximum Height using Differentiation | 使用微分求最大高度

To locate the vertex of the parabola, differentiate y with respect to x: dy/dx = −0.98x + 3.51. Setting dy/dx = 0 gives −0.98x + 3.51 = 0 → x = 3.51 / 0.98 ≈ 3.58 m.

为求抛物线顶点，对 y 关于 x 求导：dy/dx = −0.98x + 3.51。令 dy/dx = 0 得 −0.98x + 3.51 = 0 → x = 3.51 / 0.98 ≈ 3.58 m。

Substitute this x‑value back into the model: yₘₐₓ = −0.49(3.58)² + 3.51(3.58). Computing gives yₘₐₓ ≈ −0.49 × 12.8164 + 12.5658 ≈ −6.28 + 12.57 ≈ 6.29 m. Thus the maximum height of the basketball is approximately 6.29 metres above the release point.

将此 x 值代回模型：yₘₐₓ = −0.49(3.58)² + 3.51(3.58)。计算得 yₘₐₓ ≈ −0.49 × 12.8164 + 12.5658 ≈ −6.28 + 12.57 ≈ 6.29 m。因此篮球的最高点大约在出手点上方 6.29 米处。

The second derivative d²y/dx² = −0.98 < 0 confirms this is a maximum.

二阶导数 d²y/dx² = −0.98 < 0，确认这是一个极大值。

5. Determining Initial Velocity and Angle | 确定初速度和角度

Comparing the fitted quadratic y = −0.49x² + 3.51x with the theoretical trajectory y = x tanθ − (g/(2u² cos²θ))x² yields two relationships: tanθ = 3.51 and g/(2u² cos²θ) = 0.49.

将拟合的二次函数 y = −0.49x² + 3.51x 与理论轨迹 y = x tanθ − (g/(2u² cos²θ))x² 对比，得到两个关系式：tanθ = 3.51 和 g/(2u² cos²θ) = 0.49。

From tanθ = 3.51, we find θ = tan⁻¹(3.51) ≈ 74.1° (to 1 d.p.). Then, using the identity sec²θ = 1 + tan²θ, we get cos²θ = 1/(1 + 3.51²) = 1/(1 + 12.3201) = 1/13.3201 ≈ 0.07507.

由 tanθ = 3.51，求得 θ = tan⁻¹(3.51) ≈ 74.1°（保留一位小数）。随后，利用恒等式 sec²θ = 1 + tan²θ，得到 cos²θ = 1/(1 + 3.51²) = 1/(1 + 12.3201) = 1/13.3201 ≈ 0.07507。

Now substitute g = 9.8 into the second equation: 9.8/(2u² × 0.07507) = 0.49 → 2u² × 0.07507 = 9.8/0.49 = 20 → u² = 20 / (2 × 0.07507) = 20 / 0.15014 ≈ 133.2. Hence u ≈ √133.2 ≈ 11.5 m/s.

现在将 g = 9.8 代入第二个方程：9.8/(2u² × 0.07507) = 0.49 → 2u² × 0.07507 =

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