📚 AQA A-Level Maths Question 10: Projectile Motion Against a Wall | AQA A-Level 数学第10题：斜抛运动撞墙问题
Projectile motion is one of the most dependable question types in AQA A-Level Mathematics Paper 3 (Mechanics). A particle is projected at a given speed, and you are told that it must pass through, or strike, a specific point. Because the projection angle is usually unknown, the algebra almost always collapses into a quadratic in tan θ, and the two roots correspond to the two genuine physical trajectories that reach the target. In this worked example we solve Question 10 of a typical AQA mechanics paper: a ball is projected from level ground with speed 25 m s⁻¹ towards a vertical wall 40 m away, and it hits the wall at a height of 12 m. We are asked for the two possible projection angles, correct to the nearest 0.1°.
抛体运动是 AQA A-Level 数学 Paper 3（力学）中最稳定出现的题型之一。题目给定抛射速率，并要求粒子经过或击中某个特定点。由于抛射角通常未知，代数推导几乎总会归结为关于 tan θ 的二次方程，而它的两个根正对应两条真实存在的、能够到达目标的物理轨迹。本文求解一道典型的 AQA 力学第10题：小球从水平地面以 25 m s⁻¹ 的速率斜向上抛出，前方 40 m 处有一堵竖直墙，小球击中墙上离地 12 m 高的位置。题目要求求出两个可能的抛射角，答案精确到 0.1°。
1. The Question | 题目呈现
strikes the wall at a point 12 m above the ground. Find the two possible values of θ, giving your answers to the nearest 0.1°. Take g = 9.8 m s⁻² and ignore air resistance.
一个小球从水平地面上的点 O 以 25 m s⁻¹ 的速率、与水平方向成 θ 角斜向上抛出。距离 O 点 40 m 处立有一堵竖直墙。小球击中墙上离地面 12 m 高的点。求 θ 的两个可能值，答案精确到 0.1°。取 g = 9.8 m s⁻²，忽略空气阻力。
The figures are chosen so that the working stays manageable: the horizontal distance is a whole number of metres, the speed is a perfect square (625), and the resulting quadratic has a clean discriminant once the algebra is set out properly. Do not be tempted to guess angles - the whole mark allocation lies in the modelling and the algebra.
题目数据经过设计，使运算过程可控：水平距离是整数米，速率是完全平方数（625），只要代数整理得当，所得二次方程的判别式就很干净。不要试图靠猜角度作答——本题的得分点几乎全部落在建模与代数推导上。
2. What Is Being Tested? | 考点分析
This question tests several skills that AQA repeatedly examines in the mechanics section. First, you must resolve the initial velocity into horizontal and vertical components. Second, you must apply the SUVAT equations separately in each direction, remembering that the horizontal acceleration is zero. Third, you must eliminate the time t between the two equations, because the time of flight is not given. Fourth, you must recognise the trigonometric identity that converts sec² θ into 1 + tan² θ, producing a quadratic in tan θ. Finally, you must interpret the two roots physically and state both angles to the required precision.
本题综合考查 AQA 力学部分反复出现的几项技能。第一，必须把初速度分解为水平分量与竖直分量。第二，必须在两个方向上分别应用 SUVAT 方程，并牢记水平方向加速度为零。第三，必须在两个方程之间消去时间 t，因为题目没有给出飞行时间。第四，必须识别出把 sec² θ 转化为 1 + tan² θ 的三角恒等式，从而得到关于 tan θ 的二次方程。最后，必须对两个根给出物理解释，并按要求的精度写出两个角度。
- Resolving the initial velocity into perpendicular components | 将初速度分解为相互垂直的两个分量
- Applying SUVAT equations to horizontal and vertical motion | 对水平运动与竖直运动分别应用 SUVAT 方程
- Eliminating time between the two motion equations | 在两个运动方程之间消去时间
- Using the identity sec² θ = 1 + tan² θ correctly | 正确使用恒等式 sec² θ = 1 + tan² θ
- Solving a quadratic and interpreting both roots | 解二次方程并解释两个根
- Rounding angles to the nearest 0.1° and stating both answers | 将角度四舍五入到 0.1° 并写出两个答案
3. Modelling: Resolving the Initial Velocity | 建模：分解初速度
Set up a coordinate system with O at the origin, the x-axis horizontal and pointing towards the wall, and the y-axis vertically upwards. The ball is then a projectile moving freely under gravity, so the horizontal acceleration is zero and the vertical acceleration is −g.
建立坐标系：取 O 为原点，x 轴水平并指向墙的方向，y 轴竖直向上。此后小球即为只受重力作用的抛体，因此水平加速度为零，竖直加速度为 −g。
uₓ = 25 cos θ uy = 25 sin θ
aₓ = 0 ay = −9.8 m s⁻²
The key modelling decision is that the wall is vertical, so the impact point is described by
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