📚 Vector Methods with Projectiles | 抛体运动的向量方法
Vector methods give projectile motion a single consistent framework: position, velocity and acceleration are all written in terms of i and j components, and calculus links them directly. Instead of juggling two separate lists of SUVAT equations, you work with one vector equation, then read off the horizontal and vertical information only when you actually need it. This is especially valuable in Edexcel A Level Mechanics, where exam questions often ask you to integrate or differentiate vector functions rather than just quote scalar formulae.
向量方法为抛体运动提供了一个统一框架：位置、速度和加速度都写成 i、j 分量形式，并通过微积分直接联系起来。你不必再分两套清单去背 SUVAT 公式，而是只用一个向量方程，等到真正需要时再分别读出水平与竖直方向的信息。这在 Edexcel A Level 力学中特别有用，因为考试题常常要求你对向量函数求导或积分，而不仅仅是套用标量公式。
1. Why Use Vectors for Projectiles? | 为何用向量处理抛体运动？
Projectile motion is two-dimensional, so scalar equations alone can hide the fact that horizontal and vertical motion are independent. Time is the only quantity shared by the two directions, and every other quantity - displacement, velocity, acceleration - has its own horizontal and vertical version. Writing them as one vector keeps that structure visible, so you are far less likely to mix up a horizontal speed with a vertical one under exam pressure.
抛体运动是二维运动，仅使用标量方程容易掩盖水平与竖直运动相互独立这一事实。两个方向唯一共享的量是时间，而位移、速度、加速度等每一个量都有各自的水平与竖直版本。把它们写成一个向量，这个结构就始终清晰可见，考试紧张时也就不容易把水平速度与竖直速度混为一谈。
Vector notation also handles the awkward cases far more naturally: projection from a cliff or a moving platform, an initial velocity already given in component form, or a target whose position is stated as a vector. In each case the horizontal and vertical data slot into the same equation instead of forcing you to restart the algebra.
向量记号在处理那些麻烦情形时也自然得多：从悬崖或运动平台上抛出、初速度已经以分量形式给出、目标位置用向量表示等。每一种情形下，水平与竖直数据都能塞进同一个方程，而不必重新另起一套代数运算。
Finally, vectors are the bridge to calculus. Differentiating a position vector gives the velocity vector; integrating a velocity vector gives the position vector back. That two-way link is exactly what many of the harder A Level questions are testing.
最后，向量是通往微积分的桥梁。对位置向量求导得到速度向量，对速度向量积分又回到位置向量。这种双向联系正是许多较难的 A Level 题目真正考查的内容。
2. Setting Up the Vector Model | 建立向量模型
Take the point of projection as the origin O. Let i be the unit vector in the horizontal direction of motion and j the unit vector vertically upwards. Choosing j upwards means gravity appears with a negative sign, which is the convention used in almost all A Level mark schemes.
以抛出点为原点 O。设 i 为水平运动方向的单位向量，j 为竖直向上的单位向量。取 j 向上意味着重力以负号出现，这也是几乎所有 A Level 评分标准采用的约定。
If a particle is projected with speed U at an angle α to the horizontal, its initial velocity vector is the sum of two perpendicular components:
若粒子以速率 U、与水平方向成 α 角抛出，其初速度向量就是两个互相垂直分量之和：
u = U cos α i + U sin α j
directed downwards at an angle below the horizontal, the j component becomes negative: u = U cos α i - U sin α j.
若粒子从离地高度 h 处出发、而原点设在地面上，则初始位置向量为 r₀ = h j。若抛出方向朝下、与水平方向成一定角度，则 j 分量取负：u = U cos α i - U sin α j。
Two decisions must be made before any working: where the origin sits, and which way is positive. Fix them at the start, state them in your answer, and never change them mid-question.
动笔之前必须做两个决定：原点取在哪里，以及哪个方向为正。一开始就固定好并在解答中写明，中途绝不再改。
3. The Constant Acceleration Vector | 常加速度向量
The key advantage of the vector approach is that the acceleration due to gravity is a constant vector pointing vertically downwards. With j taken upwards, the acceleration vector is
向量方法的关键优势在于：重力加速度是一个方向竖直向下的常向量。取 j 向上时，加速度向量为
a = -g j
The i component of acceleration is exactly zero, which is precisely why the horizontal component of velocity never changes during flight. The j component is a constant -g, which is why the vertical velocity decreases at a steady rate until the particle reaches its highest point, then becomes negative and increases in magnitude as the particle falls.
加速度的 i 分量为零，这正是水平速度在飞行过程中始终保持不变的原因。j 分量为常数 -g，因此竖直速度会稳定减小，直到粒子到达最高点；随后竖直分量变为负值，其大小随下落而增大。
Air resistance is neglected throughout the A Level specification unless the question states otherwise: the particle is modelled as a point mass moving under gravity alone, and g is usually taken as 9.8 m s⁻² or 9.81 m s⁻² depending on what the paper tells you.
除非题目另有说明，A Level 大纲全程忽略空气阻力：粒子被建模为仅在重力作用下的质点，g 通常取 9.8 m s⁻² 或 9.81 m s⁻²，具体以试卷给出的数值为准。
a = 0 i + (-g) j
This single line replaces both of the scalar statements "horizontal acceleration = 0" and "vertical acceleration = -9.8". Everything else in the topic follows from it.
这一行就同时替代了标量表述中的两句话："水平加速度 = 0"与"竖直加速度 = -9.8"。本专题的其他结论都由它推出。
4. Velocity as a Function of Time | 速度随时间的变化
UVAT equation, v = u + a t, substitute the initial velocity and the constant acceleration:
使用第一个 SUVAT 方程的向量形式 v = u + a t，代入初速度与常加速度：
v = (U cos α i + U sin α j) + (-g j) t
Collecting components gives the velocity vector at any time t:
合并分量后即得任意时刻 t 的速度向量：
v = U cos α i + (U sin α - g t) j
The i component is a constant, U cos α. The j component is a linear function of time, U sin α - g t, which starts positive, passes through zero at the highest point and then becomes negative.
i 分量为常数 U cos α。j 分量是时间的一次函数 U sin α - g t，它起初为正，在最高点处取零，随后变为负值。
The speed at any instant is the magnitude of the velocity vector, and the direction of motion is found from the components:
任意时刻的速率为速度向量的大小，运动方向则由分量求出：
|v| = √((U cos α)² + (U sin α - g t)²), tan θ = (U sin α - g t) / (U cos α)
i and the particle is moving horizontally; the time taken to reach it is t = U sin α / g.
其中 θ 为速度与水平方向的夹角。在最高点处 j 分量为零，故 v = U cos α i，粒子沿水平方向运动；到达最高点所用的时间为 t = U sin α / g。
5. Displacement and the Position Vector | 位移与位置向量
r₀ + u t + ½ a t², gives the position of the particle measured from the origin. Starting from the origin with the initial velocity of Section 2:
第二个向量形式的 SUVAT 方程 r = r₀ + u t + ½ a t² 给出粒子相对于原点的位置。从原点出发、初速度如第 2 节所述时：
r = (U cos α · t) i + (U sin α · t - ½ g t²) j
describes constant downward acceleration. That is the whole content of the topic in two lines.
把分量分开读，就回到了熟悉的标量结论：x = U cos α · t 描述水平匀速运动，y = U sin α · t - ½ g t² 描述竖直方向恒定向下加速。整个专题的内容就浓缩在这两行里。
ected from a point at height h above the ground, add the initial position vector r₀ = h j:
若粒子从离地高度 h 处抛出，则需加上初始位置向量 r₀ = h j：
r = (U cos α · t) i + (h + U sin α · t - ½ g t²) j
The i component is unchanged, because a vertical shift of the origin cannot affect horizontal motion. Only the j component carries the extra h.
i 分量不受影响，因为原点的竖直平移不会影响水平运动。只有 j 分量额外带上 h。
6. The Cartesian Equation of the Path | 轨迹的直角坐标方程
From x = U cos α · t we get t = x / (U cos α), and substituting into the j component gives
在两个分量之间消去 t，可看出轨迹是一条抛物线。由 x = U cos α · t 得 t = x / (U cos α)，代入 j 分量得
y = x tan α - g x² / (2 U² cos² α)
sing the identity 1 / cos² α = 1 + tan² α, an equivalent and often more convenient form is
利用恒等式 1 / cos² α = 1 + tan² α，可以得到一个等价且常常更方便的形式
y = x tan α - (g x² / (2 U²)) (1 + tan² α)
Since the coefficient of x² is negative, the parabola opens downwards, as expected. This form is useful when a question gives the gradient of the path at a point, or when you are asked to show that a projectile passes through a given coordinate.
由于 x² 的系数为负，抛物线开口向下，与预期一致。当题目给出轨迹在某点的斜率，或要求证明抛体经过某个已知坐标点时，这个形式特别有用。
t, which in turn gives the direction of motion there: dy/dx = tan α - (g x / (U² cos² α)).
将直角坐标方程对 x 求导，可得轨迹在任意点处的斜率，进而得到该点处的运动方向：dy/dx = tan α - (g x / (U² cos² α))。
7. Time of Flight, Range and Maximum Height | 飞行时间、射程与最大高度
U sin α · t - ½ g t² = t (U sin α - ½ g t), the two roots are the launch instant t = 0 and
水平地面问题只需令 r 的 j 分量等于零。由 y = U sin α · t - ½ g t² = t (U sin α - ½ g t)，两个根分别是抛出时刻 t = 0 与
T = 2 U sin α / g
Substituting this total time of flight into the i component gives the horizontal range:
将这个总飞行时间代入 i 分量，即得水平射程：
R = U cos α × 2 U sin α / g = U² sin 2 α / g
setting the j component of velocity to zero, so t = U sin α / g, and substituting into y:
最大高度则由速度的 j 分量等于零得到，此时 t = U sin α / g，代入 y 得：
H = U² sin² α / (2 g)
Note the symmetry: the time to the highest point is exactly half the total time of flight on level ground, and the speed at the highest point equals the horizontal component U cos α, which is also the minimum speed during the flight.
注意其对称性：在水平地面上，到达最高点的时间恰好是总飞行时间的一半，最高点处的速率等于水平分量 U cos α，这也是整个飞行过程中的最小速率。
8. Projection From a Height or On to a Lower Level | 从高处抛出或落到更低平面
When the particle is projected from a cliff of height h and lands on ground level, the j component must be set to -h rather than 0 relative to the projection point, or the origin must be placed on the ground and the height built into r₀. Both routes give the same quadratic in t:
当粒子从高度 h 的悬崖上抛出并落到地面时，j 分量应相对抛出点设为 -h 而不是 0；或者把原点放在地面上，把高度写进 r₀。两种做法得到同一个关于 t 的二次方程：
h + U sin α · t - ½ g t² = 0
Solve with the quadratic formula and take the positive root. Never discard the negative root without comment: say explicitly that t must be positive, as this is a standard communication mark.
用求根公式求解并取正根。不要不加说明地舍去负根：应明确指出时间 t 必须为正，这是一分标准的表述得分点。
Once the correct landing time is known, substitute it into the horizontal component to obtain the horizontal distance travelled. The horizontal velocity is still constant throughout, since gravity has no horizontal component even when the particle starts above or below the origin in the vertical sense.
确定正确的落地时间后，把它代入水平分量即可得到水平飞行距离。由于重力没有水平分量，即使粒子在竖直方向上从原点上方或下方出发，水平速度在整个过程中仍然保持不变。
If the projection is directed downwards at an angle α below the horizontal, the initial velocity is U cos α i - U sin α j and the j component of displacement becomes -U sin α · t - ½ g t², so the particle reaches the ground sooner than if it were simply dropped.
若抛出方向朝下、与水平方向成 α 角，则初速度为 U cos α i - U sin α j，位移的 j 分量变为 -U sin α · t - ½ g t²，因此粒子比自由下落更快到达地面。
9. When the Initial Velocity Is Given as a Vector | 初速度直接以向量形式给出
ctly, for example u = 7 i + 24 j m s⁻¹. Then U cos α = 7 and U sin α = 24, so the speed is √7² + 24² = 25 m s⁻¹ and tan α = 24 / 7, giving α ≈ 73.7° above the horizontal.
许多考题完全跳过角度，直接给出初速度，例如 u = 7 i + 24 j m s⁻¹。此时 U cos α = 7，U sin α = 24，故速率为 √7² + 24² = 25 m s⁻¹，tan α = 24 / 7，即 α 约 73.7°，方向在水平线上方。
With g = 9.8 m s⁻², the velocity and position vectors become
取 g = 9.8 m s⁻²，速度与位置向量分别为
v = 7 i + (24 - 9.8 t) j
r = 7 t i + (24 t - 4.9 t²) j
Typical questions then ask when the particle moves in a given direction. Moving horizontally means the j component of v is zero; moving parallel to a vector p i + q j means the ratio of the components of v equals q / p; moving perpendicular to the initial velocity means the dot product u · v equals zero. Each of these is a one-line equation in t.
常见问题接着会问粒子何时沿某一给定方向运动。沿水平方向运动意味着 v 的 j 分量为零；平行于向量 p i + q j 意味着 v 的分量之比等于 q / p；垂直于初速度意味着点积 u · v 等于零。每一种情形都只是关于 t 的一行方程。
Being able to move fluently between the (U, α) description and the component description is the single most useful skill in this topic.
能在 (U, α) 描述与分量描述之间自如切换，是本专题中最实用的单项技能。
10. Calculus Questions on Vector Functions | 关于向量函数的微积分题
Some questions give the velocity as a function of time and ask for the position, or give the position and ask for the velocity and acceleration. The rule is simple and always the same: deal with the i and j components separately, and remember that integrating a vector function produces a vector constant.
有些题目把速度表示为时间的函数并要求位置，或给出位置并要求速度与加速度。规则很简单而且始终一致：把 i 与 j 分量分开处理，并牢记对向量函数积分会得到一个向量常数。
cause r = 0 when t = 0. If instead the particle starts at r₀ = 2 i + 3 j, that vector is added on.
若 v = (3 t + 1) i - 5 t j 且粒子从原点出发，则对每个分量积分得 r = (1.5 t² + t) i - 2.5 t² j，由于 t = 0 时 r = 0，常数向量为零。若粒子改为从 r₀ = 2 i + 3 j 出发，则需加上该向量。
Going the other way, differentiating r with respect to t once gives v and twice gives a. For projectile motion the acceleration must come out as a constant vector of the form -g j; if it does not, either the model is not projectile motion or an arithmetic slip has occurred.
反过来，把 r 对 t 求一次导得 v，求两次导得 a。对于抛体运动，加速度必须化为形如 -g j 的常向量；若不是，要么这个模型不属于抛体运动，要么中间出现了计算错误。
Remember that speed is the magnitude of v, not a component of it, and that a negative component of velocity simply means motion in the negative coordinate direction.
切记速率是 v 的大小而非其中某个分量，速度分量为负只是表示沿坐标负方向运动。
11. Common Errors and Marking Points | 常见错误与得分点
Certain mistakes appear again and again in scripts, and recognising them in advance is worth several marks.
有几类错误在答卷中反复出现，提前识别它们往往能多拿好几分。
Mixing sign conventions: using j upwards in one line and downwards in the next, so that the acceleration and the displacement contradict each other.
符号约定混用：一行取 j 向上、下一行又取向下，导致加速度与位移互相矛盾。
Applying the scalar SUVAT equations to the resultant motion instead of to each component separately.
把标量 SUVAT 方程用在合运动上，而不是分别用在每个分量上。
Forgetting the ½ in ½ g t², or writing g t² in the position vector.
忘记 ½ g t² 中的 ½，或在位置向量中直接写成 g t²。
Quoting speed when the question asks for velocity, or giving a vector when a magnitude is required.
题目要求速度向量时却给出速率，或要求大小时却给出向量。
Omitting the constant vector when integrating, and failing to justify why it is zero.
积分时漏掉常数向量，或没有说明它为何为零。
Discarding the negative root of a quadratic in t without a written justification.
舍去 t 的二次方程负根时未作书面说明。
Rounding intermediate values too early, which throws off the final answer by more than the accepted tolerance.
中间结果过早取近似，导致最终答案偏差超出允许范围。
A well-presented solution states the origin and positive directions, writes the vector equations, substitutes, and then answers the specific question asked, with units and an appropriate degree of accuracy.
一份规范的解答应当写明原点与正方向、列出向量方程、代入数值，然后针对题目所问作答，并带上单位与合适的小数位数。
12. Worked Example and Quick Summary | 典型例题与快速总结
². Find the velocity after 2 seconds, the maximum height reached and the horizontal range.
一粒子从
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