📚 Year 11 CCEA Further Mathematics: Case Study Practice | CCEA 进阶数学案例分析实战演练
Real-world applications bring Further Mathematics to life, especially when you tackle an engineering challenge that combines kinematics, trigonometry, and calculus. In this case study, we will design a theme park water slide exit, modelling the rider's projectile motion and optimising the launch angle to maximise horizontal range. Each step mirrors the type of reasoning expected in CCEA GCSE Further Mathematics examinations, showing how core techniques from the specification are used in a practical context.
真实世界中的应用能让进阶数学变得生动，尤其是当你要解决一个融合了运动学、三角学和微积分的工程设计挑战时。在本案例分析中，我们将设计一条主题公园水滑道的出口区域，模拟游客的抛体运动，并优化发射角度以使水平射程最大化。每一步都反映了 CCEA GCSE 进阶数学考试中所要求的推理类型，展示课程核心技巧如何用于实际情境中。
1. Introducing the Theme Park Ride Design Challenge | 介绍主题公园滑道设计挑战
Imagine you are an engineer designing a new water slide. The slide finishes at a point above a splash pool, and riders fly through the air before landing in the water. Your goal is to choose the exit angle so that they travel as far as possible horizontally – adding thrill while still landing safely within the pool's length.
想象你是一名工程师，正在设计一条新的水滑道。滑道终点位于溅水池上方，游客会腾空飞出一段距离后才落入水中。你的目标是选择合适的出口角度，使水平飞行距离尽可能远——增加刺激感，同时确保他们安全地落入水池范围内。
This challenge combines the geometry of the slide, the speed acquired during the descent, and the parabolic path after take‑off. We will build a step‑by‑step mathematical model and then use differentiation to find the optimal angle, demonstrating exactly how Further Mathematics supports engineering design.
这一挑战将滑道的几何形状、下滑获得的速度以及腾空后的抛物线路径结合在一起。我们将逐步建立数学模型，然后运用微分法求出最优角度，恰好展示了进阶数学如何支撑工程设计。
2. Defining the Coordinate System and Known Values | 定义坐标系和已知量
We place the exit point of the slide at coordinates (0, h), where the positive x‑axis is horizontal and the positive y‑axis is vertically upward. The take‑off height is set to h = 2.0 m above the water surface, which lies at y = 0.
我们将滑道出口点置于坐标 (0, h)，正 x 轴为水平方向，正 y 轴为竖直向上。起飞高度设为 h = 2.0 m（相对水面），水面位于 y = 0 处。
at an angle θ above the horizontal. The speed depends on the vertical drop from the slide' mes entirely from the loss in gravitational potential energy over a vertical distance of H – h = 6.0 m.
游客以初速度 v₀、与水平方向成 θ 角腾空而起。速度大小取决于从滑道最高点下降的高度，最高点高度为 H = 8.0 m。假设能量守恒且摩擦力可忽略，出口处的动能完全来自重力势能在垂直距离 H – h = 6.0 m 上的减少。
½ m v₀² = mg(H – h) ⇒ v₀ = √[2g(H – h)]
½ m v₀² = mg(H – h) ⇒ v₀ = √[2g(H – h)]
Taking the gravitational field strength g = 9.8 m s⁻², we compute v₀ = √2 × 9.8 × 6.0 unch speed is fixed for all choices of angle θ because the vertical drop remains the same.
取重力场强度 g = 9.8 m s⁻²，计算得 v₀ = √2 × 9.8 × 6.0 = √117.6 ≈ 10.84 m s⁻¹。因为下落高度差不变，所以无论角度 θ 选多少，这个发射速度大小都是固定的。
3. Writing Parametric Equations for Projectile Motion | 写出抛体运动的参数方程
ation. Taking time t = 0 at the instant of launch, we can write separate equations for the horizontal and vertical displacements.
离开滑道后，唯一作用在游客上的力是重力，产生向下的恒定加速度 g。水平方向没有加速度。取起飞瞬间 t = 0，我们可以分别写出水平和竖直位移方程。
x = v₀ cos θ · t
水平位移 x = v₀ cos θ · t
y = h + v₀ sin θ · t – ½ g t²
竖直位移 y = h + v₀ sin θ · t – ½ g t²
Here, v₀ cos θ is the constant horizontal component of velocity, while the vertical component starts at v₀ sin θ and decreases linearly due to gravity. These parametric equations are the foundation for tracing the parabolic trajectory.
这里，v₀ cos θ 是速度的水平分量（恒定），而竖直分量初值为 v₀ sin θ，并因重力而线性减小。这组参数方程是描绘抛物线轨迹的基础。
4. Eliminating Time to Obtain the Trajectory Equation | 消去时间得到轨迹方程
ituting this into the vertical displacement equation eliminates t and gives a direct relationship between y and x – the equation of the path.
由水平方程可解出时间 t = x / (v₀ cos θ)。将其代入竖直位移方程即可消去 t，得到 y 与 x 的直接关系——飞行路径方程。
y = h + x tan θ – (g x²) / (2 v₀² cos² θ)
y = h + x tan θ – (g x²) / (2 v₀² cos² θ)
This is a quadratic function in x, confirming the trajectory is a parabola opening downwards. The coefficient of x² is negative because gravity pulls the rider down. For any given angle θ, this equation can predict the rider's height at any horizontal distance from the exit.
这是关于 x 的二次函数，证实轨迹是一条开口向下的抛物线。x² 项系数为负，因为重力将游客往下拉。给定任意角度 θ，这个方程都能预测离开出口后任意水平距离处游客的高度。
5. Formulating the Range Function R(θ) | 构建射程函数 R(θ)
ectory equation and solve for the positive horizontal distance, which we call the range R.
当竖直坐标变为零，即轨迹与水面 y = 0 相交时，游客落入水中。我们将 y = 0 代入轨迹方程，解出正的水平距离，即射程 R。
0 = h + R tan θ – (g R²) / (2 v₀² cos² θ)
0 = h + R tan θ – (g R²) / (2 v₀² cos² θ)
This is a quadratic equation in R: (g / (2 v₀² cos² θ)) R² – (tan θ) R – rmula gives two roots; the positive root represents the physical range. Using trigonometric identities, the expression can be simplified to a more compact form that shows dependence on θ
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