📚 Motion in Two Dimensions | 二维运动
Motion in two dimensions is a central topic in Edexcel A-level mechanics, especially in projectile problems. It extends the equations of linear motion by requiring you to handle two perpendicular directions at the same time. This topic brings together vector resolution, the constant-acceleration equations, and the independence of components.
二维运动是 Edexcel A-level 力学中的核心专题，尤其在抛体问题中经常出现。它扩展了直线运动方程，要求你同时处理两个互相垂直的方向。本专题将矢量分解、匀加速运动方程以及分量独立性等内容结合在一起。
Mastering this topic is not just about memorising formulas; it is about understanding why horizontal velocity stays constant and why vertical motion is exactly like free fall. The following sections build the method step by step and connect it to standard exam questions.
掌握这个专题不仅仅是记忆公式，更重要的是理解为什么水平速度保持不变，以及为什么竖直运动与自由落体完全相同。下面各节将逐步构建解题方法，并把它与常见考题联系起来。
1. Scalars, Vectors and Resolution | 标量、矢量与分解
Motion in two dimensions requires a vector description. Scalars such as mass, time and speed have magnitude only. Vectors such as displacement, velocity, acceleration and force have both magnitude and direction, so we cannot simply add their numerical values. The standard method is to resolve each vector into two perpendicular components, usually horizontal and vertical.
二维运动需要矢量描述。质量、时间、速率等标量只有大小；位移、速度、加速度、力等矢量既有大小又有方向，因此不能简单地把数值相加。标准方法是把每个矢量分解为两个互相垂直的分量，通常取水平和竖直方向。
asured from the vertical, the sine and cosine swap. Always check which angle the question gives you before substituting.
若矢量 F 与水平方向成 θ 角，则分量为 Fₓ = F cos θ，Fᵧ = F sin θ。若角度从竖直方向量起，正弦和余弦互换。代入前务必确认题目给出的是哪个角。
Fₓ = F cos θ, Fᵧ = F sin θ
2. Independence of Horizontal and Vertical Motion | 水平与竖直运动的独立性
, so it produces no horizontal acceleration. Therefore the horizontal velocity remains constant throughout the flight, while the vertical motion is exactly the same as free fall with acceleration aᵧ = -g.
抛体运动的关键思想是水平运动和竖直运动相互独立。重力竖直向下，因此不会产生水平加速度。所以水平速度在整个飞行过程中保持不变，而竖直运动与加速度 aᵧ = -g 的自由落体完全相同。
This independence means that a stone dropped from rest and a stone projected horizontally from the same height will hit the ground at the same time, provided air resistance is negligible. The horizontal velocity does not change the time of fall, because the vertical initial velocity is zero in both cases.
这种独立性意味着，从同一高度由静止释放的石子和水平抛出的石子会同时落地，前提是空气阻力可忽略。水平速度不会改变下落时间，因为两种情况下竖直初速度都为零。
In exam questions, this idea allows you to split a complicated two-dimensional problem into two simpler one-dimensional problems. You solve them separately and then reconnect them through time.
在考试题中，这一思想让你把一个复杂的二维问题拆成两个简单的一维问题。分别求解后，再通过时间把它们联系起来。
3. The Projectile Model and Assumptions | 抛体模型与假设
tance is negligible. The Earth is treated as flat over the range of the projectile, and the projectile does not spin. These assumptions are usually stated or implied in Edexcel questions.
标准抛体模型有几个假设：重力加速度恒定，g = 9.81 m s⁻²，方向向下；空气阻力忽略不计；在抛体射程范围内地面视为平坦；抛体不旋转。Edexcel 题目通常会直接说明或默认这些假设。
Under these assumptions, the horizontal acceleration is zero and the vertical acceleration is constant. The equations of motion can then be applied separately in each direction. If air resistance were included, the horizontal velocity would decrease and the path would no longer be a symmetric parabola.
在这些假设下，水平加速度为零，竖直加速度恒定。于是运动方程可以在两个方向上分别使用。如果考虑空气阻力，水平速度会减小，轨迹也不再是对称抛物线。
The model works well for dense, compact objects moving at moderate speeds over short distances, such as a cricket ball or a shot put. It is less accurate for light objects with large surface area, such as a shuttlecock or a sheet of paper.
该模型适用于密度大、体积小、速度适中且距离较短的物体，例如板球或铅球。对于表面积大、质量轻的物体，例如羽毛球或纸片，模型精度会下降。
4. Equations of Motion in Component Form | 分量形式的运动方程
uₓ and x = uₓt. Vertically, aᵧ = -g, so vᵧ = uᵧ - gt, y = uᵧt - ½gt², and vᵧ² = uᵧ² - 2gy.
在匀加速运动中，SUVAT 方程可以在每个方向上分别使用。水平方向 aₓ = 0，所以 vₓ = uₓ，x = uₓt。竖直方向 aᵧ = -g，所以 vᵧ = uᵧ - gt，y = uᵧt - ½gt²，vᵧ² = uᵧ² - 2gy。
| Direction 方向 | Acceleration 加速度 | Velocity 速度 | Displacement 位移 |
|---|---|---|---|
| Horizontal 水平 | aₓ = 0 | vₓ = uₓ | x = uₓt |
| Vertical 竖直 | aᵧ = -g | vᵧ = uᵧ - gt | y = uᵧt - ½gt² |
nsistent throughout the whole solution, and state your positive direction in your working.
正负号取决于你选择的正方向。若取向上为正，则 aᵧ = -g；若取向下为正，则 aᵧ = +g。整个解题过程必须保持一致，并在步骤中说明正方向。
5. Time as the Connecting Variable | 时间作为连接变量
substitute that time into the horizontal equation. For an object projected horizontally from height h, the vertical motion starts from rest, so h = ½gt² and therefore t = √2h/g.
时间是连接水平运动和竖直运动的变量。在大多数抛体问题中，先由竖直运动求出时间，再把时间代入水平方程。对于从高度 h 水平抛出的物体，竖直运动从静止开始，所以 h = ½gt²，因此 t = √2h/g。
t = √2h/g
at the horizontal distance depends on the initial horizontal velocity, but the time of flight does not, under the standard assumptions. This is a favourite exam point.
一旦知道时间，水平距离就是 x = uₓt。注意，在标准假设下，水平距离取决于初始水平速度，但飞行时间与水平速度无关。这是考试中常考的一点。
and landing heights are the same, the total time of flight is twice this value, T = 2uᵧ/g.
对于斜抛，时间通常由竖直速度方程求出。在最高点 vᵧ = 0，所以 t = uᵧ/g。若发射点和落地点高度相同，总飞行时间是该值的两倍，T = 2uᵧ/g。
6. Horizontal Projection from a Height | 从高处水平抛出
= ½gt² downward; horizontally, x = ut. The time of flight is independent of u, so changing the horizontal speed changes the range but not the time taken to fall.
考虑一个物体从高度 h 水平抛出，初始水平速度为 u。它的初始竖直速度为零。竖直方向 y = ½gt² 向下；水平方向 x = ut。飞行时间与 u 无关，所以改变水平速度会改变射程，但不改变下落时间。
component is still u, while the vertical component is vᵧ = gt. The resultant speed is v = √u² + vᵧ². The direction of the velocity makes an angle θ below the horizontal, where tan θ = vᵧ/u.
落地瞬间，水平速度分量仍为 u，竖直分量为 vᵧ = gt。合速率为 v = √u² + vᵧ²。速度方向与水平方向成 θ 角向下，其中 tan θ = vᵧ/u。
v = √u² + vᵧ², tan θ = vᵧ / u
A common exam task is to find the angle at which the object hits the ground, or to find the height required for a given range. Both are solved by finding t first, then using the horizontal equation or the vertical velocity equation.
常见考题是求物体落地时的角度，或求给定射程所需的高度。两者都先求 t，再用水平方程或竖直速度方程求解。
7. Oblique Projection: Launch at an Angle | 斜抛：以角度发射
izontal component remains constant, while the vertical component decreases at a constant rate until it reaches zero at the highest point.
物体以速度 U、与水平方向成 θ 角发射时，把初速度分解为分量：uₓ = U cos θ，uᵧ = U sin θ。水平分量保持不变，竖直分量以恒定速率减小，直到最高点变为零。
The maximum height is H = U² sin² θ / (2g), and the horizontal range is R = U² sin 2θ / g.
到达最高点的时间由 vᵧ = 0 求出：t = U sin θ / g。若地面水平，总飞行时间 T = 2U sin θ / g。最大高度 H = U² sin² θ / (2g)，水平射程 R = U² sin 2θ / g。
T = 2U sin θ / g, H = U² sin² θ / (2g), R = U² sin 2θ / g
These formulas are convenient, but you should also be able to derive them from the component equations. Examiners often award method marks for a correct derivation even if the final value is numerical.
这些公式很方便，但你也应该能从分量方程推导出来。即使最终答案是数值，考官也常为正确的推导过程给分。
8. Maximum Height, Range and Apex | 最大高度、射程与顶点
At the apex of the trajectory, the vertical velocity is zero, but the horizontal velocity is unchanged. The path is symmetric in the absence of air resistance: the time going up equals the time coming down, the speed at a given height is the same on the way up and on the way down, and the angle above the horizontal on the way up equals the angle below the horizontal on the way down at the same height.
在轨迹顶点，竖直速度为零，但水平速度不变。无空气阻力时轨迹是对称的：上升时间等于下降时间，同一高度上升和下降时的速率相同，上升时与水平方向的夹角等于下降时同一高度与水平方向的夹角。
mplementary angles such as 30° and 60° give the same range, because sin 2θ has the same value for θ and 90° - θ. Air resistance reduces the range and destroys this symmetry, making the descent steeper than the ascent.
对于水平地面且无空气阻力，当 θ = 45° 时射程最大。互余角度如 30° 和 60° 给出相同射程，因为 sin 2θ 在 θ 和 90° - θ 时取值相同。空气阻力会减小射程并破坏这种对称性，使下降轨迹比上升轨迹更陡。
s only true for level ground. Always use the vertical velocity condition vᵧ = 0 to find the apex time.
常见错误是：当发射点和落地点高度不同时，仍假设最大高度出现在总飞行时间的一半处。这只在水平地面时成立。求顶点时间应始终使用竖直速度条件 vᵧ = 0。
9. Velocity, Speed and Direction at a Point | 某点的速度、
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