Year 10 CIE Advanced Mathematics: Case Study Practical Exercise | 十年级CIE进阶数学：案例分析实战演练

📚 Year 10 CIE Advanced Mathematics: Case Study Practical Exercise | 十年级CIE进阶数学：案例分析实战演练

In CIE IGCSE Additional Mathematics (0606), real-world applications often combine several topics into a single challenging problem. This case study focuses on two vertically thrown balls, requiring you to use quadratic functions, differentiation, solving simultaneous equations, and careful domain checking. Working through this step-by-step analysis will build your confidence in tackling multi-part exam questions efficiently.

在 CIE IGCSE 附加数学 (0606) 中，实际应用题常将多个知识点融合到一个综合性问题里。本案例研究两个垂直上抛的小球，需要你运用二次函数、求导、联立方程求解以及仔细检查定义域。通过逐步分析，你将更有信心高效应对考试中的多小问综合题。

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1. Case Overview | 案例概述

Two balls, P and Q, are thrown vertically upwards at the same instant. Ball P leaves the ground, while ball Q is launched from a platform 5 metres above the ground. Their heights, h₁ and h₂ metres, after t seconds are modelled by:

两个小球 P 和 Q 在同一时刻被竖直上抛。P 球从地面抛出，Q 球从离地 5 米高的平台上抛出。它们的离地高度 h₁ 与 h₂（单位：米）随时间 t（单位：秒）的关系如下：

h₁ = -2t² + 12t   and   h₂ = -t² + 4t + 5

We are to find the maximum height of each ball, the instant they are at the same height, their landing times, and—while both are in the air—the greatest vertical distance between them.

我们需要求出每个球的最大高度、它们处于同一高度的时刻、各自的落地时间，以及当两者都还在空中时，它们之间的最大竖直距离。

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2. Understanding the Domain | 理解定义域

ball hits the ground when its height returns to zero. We will solve h = 0 for each equation later; for now, note that the common time interval—when both are airborne— to the earlier of the two landing times. Every answer must be checked against this domain.

由于高度方程只在球在空中时有效，我们必须先确定有效的时间区间。两球都从 t = 0 时刻开始。当球落回地面时高度为零，之后方程不再适用。稍后我们会分别求解 h = 0；目前只需记住，两者共同在空中的时间区间是从 t = 0 到较早的那个落地时刻。所有求出的解都必须对照这个定义域进行检验。

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3. Maximum Height of Ball P | 球 P 的最大高度

To find the maximum height of ball P, we differentiate h₁ with respect to t and set the derivative to zero:

为求球 P 的最大高度，我们对 h₁ 关于 t 求导，并令导数为零：

h₁'(t) = -4t + 12

Setting h₁'(t) = 0 gives t = 3 seconds. Substitute back into h₁:

令 h₁'(t) = 0 解得 t = 3 秒。代入原高度方程：

h₁(3) = -2(3)² + 12(3) = -18 + 36 = 18 m

ld also complete the square: h₁ = -2(t - 3)² + 18, which clearly gives the vertex (3, 18).

因此，球 P 的最大高度为 18 米，出现在第 3 秒。你也可以用配方法：h₁ = -2(t - 3)² + 18，顶点 (3, 18) 一目了然。

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4. Maximum Height of Ball Q | 球 Q 的最大高度

Similarly, for ball Q we differentiate h₂:

类似地，对球 Q 的高度函数求导：

h₂'(t) = -2t + 4

Set the derivative equal to zero → t = 2 s. The corresponding height is:

令导数为零 → t = 2 s。代入得高度：

h₂(2) = -(2)² + 4(2) + 5 = -4 + 8 + 5 = 9 m

of 9 metres at t = 2 s. The completed-square form h₂ = -(t - 2)² + 9 confirms the vertex.

因此，球 Q 在 t = 2 秒时到达最大高度 9 米。配方形式 h₂ = -(t - 2)² + 9 也印证了该顶点。

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5. Time When Both Balls are at the Same Height | 两球等高的时刻

Set h₁ equal to h₂ and simplify:

令 h₁ 等于 h₂，化简：

-2t² + 12t = -t² + 4t + 5

Bring all terms to one side:

将所有项移到一边：

-t² + 8t - 5 = 0 ⇒ t² - 8t + 5 = 0

Apply the quadratic formula with a = 1, b = -8, c = 5:

使用求根公式，其中 a = 1, b = -8, c = 5：

t = [8 ± √64 - 20] / 2 = [8 ± √44] / 2 = [8 ± 2√11] / 2 = 4 ± √11

Approximately, t₁ = 4 - √11 ≈ 0.683 s and t₂ = 4 + √11 balls are airborne. Therefore the only valid instant they share the same height is t = 4 - √11 ≈ 0.683 s. Substitute back to find the height:

近似计算，t₁ = 4 - √11 ≈ 0.683 s，t₂ = 4 + √11 ≈ 7.317 s。但我们需要检查定义域。下面将求得球 Q 在 t = 5 s 时落地，因此 t₂ 超出两者同时在空中时段。故唯一有效的等高点为 t = 4 - √11 ≈ 0.683 s。代入任一高度函数可得等高时的高度。

Height at t = 4 - √11: h₁ = -2(4-√11)² + 12(4-√11) = ... ≈ 7.34 m

(Exact value can be left in surd form when required.)

（考试中可根据需要用根式表示精确值。）

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6. Landing Times of Both Balls | 两球的落地时刻

For ball P, set h₁ = 0:

对球 P，令 h₁ = 0：

-2t² + 12t = 0 ⇒ -2t(t - 6) = 0

Thus t = 0 (launch) and t = 6 s (landing).

得 t = 0（抛出时刻）和 t = 6 s（落地时刻）。

For ball Q, set h₂ = 0:

对球 Q，令 h₂ = 0：

-t² + 4t + 5 = 0 ⇒ t² - 4t - 5 = 0

(t - 5)(t + 1) = 0

The positive root is t = 5 s, so ball Q lands after 5 seconds.

正根为 t = 5 s，因此球 Q 在 5 秒后落地。

Consequently, the common interval when both balls are in the air is 0 ≤ t ≤ 5.

因此，两球同时在空中的时间区间为 0 ≤ t ≤ 5。

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7. Vertical Distance Between the Balls | 两球间的竖直距离

e ball P starts from the ground and reaches a higher peak, we define d(t) = h₁(t) - h₂(t):

在两球都飞行的阶段，定义竖直距离 d(t) 为高度差。由于球 P 从地面出发且最高点更高，我们取 d(t) = h₁(t) - h₂(t)：

d(t) = (-2t² + 12t) - (-t² + 4t + 5) = -t² + 8t - 5

This is a quadratic function opening downwards (coefficient of t² is negative), so it has a maximum. We must maximise d(t) for t in [0, 5].

这是一个开口向下的二次函数（t² 系数为负），因而存在最大值。我们需要在区间 [0, 5] 上最大化 d(t)。

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8. Maximising the Vertical Distance with Calculus | 用微积分求最大竖直距离

Differentiate d(t):

对 d(t) 求导：

d'(t) = -2t + 8

Set the derivative to zero to locate the stationary point:

令导数为零以确定驻点：

-2t + 8 = 0 ⇒ t = 4

To confirm it yields a maximum, evaluate d(t) at the critical point and at the endpoints:

t = 4 落在 [0, 5] 内。为确认它为最大值，我们计算驻点及两端点处的 d 值：

d(0) = -5,   d(4) = -16 + 32 - 5 = 11,   d(5) = -25 + 40 - 5 = 10

. Hence the maximum vertical distance between the balls while both are in the air is 11 m, occurring after 4 seconds.

最大值出现在 t = 4 s，d = 11 米。因此，两球同时在空中的最大竖直距离为 11 米，发生于第 4 秒。

2a). With a = -1, b = 8, we obtain t = -8/(2 × -1) = 4, which matches the calculus result.

另一种方法是使用二次函数顶点公式，顶点横坐标为 -b/(2a)。这里 a = -1, b = 8，得 t = 4，与求导结果一致。

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9. Interpretation and Verification | 解释与验证

At t = 4 s, ball P is at h₁ = -2(16) + 48

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