📚 Analysis Methods for Projectile Motion at Any Angle | 任意角度抛体运动的分析方法
Projectile motion is a classic topic in A-level Mathematics and Mechanics. When an object is launched at an arbitrary angle, its motion can be analysed by separating the initial velocity into horizontal and vertical components, treating each direction independently. This article provides a structured set of methods for solving any projectile problem, from deriving the trajectory equation to applying symmetry and numerical techniques.
抛体运动是A-level数学与力学中的经典专题。当物体以任意角度抛出时，可将初速度分解为水平与垂直两个分量，并将两个方向上的运动分别独立处理。本文提供一套系统化的分析方法，涵盖从轨迹方程推导到对称性应用与数值解法等内容。
1. Core Assumptions and Coordinate System | 核心假设与坐标系统
Projectile problems in A-level mechanics typically assume that air resistance is negligible, the gravitational field is uniform, and the Earth’s surface is flat over the range of motion. These assumptions make the trajectory a perfect parabola.
A-level力学中的抛体问题通常假定空气阻力可以忽略、重力场均匀、在运动范围内地表近似平坦。在这些假设下，运动轨迹是一条完美的抛物线。
Place the origin at the point of launch. Let the x-axis be horizontal in the direction of launch, and the y-axis be vertical upward. The acceleration components are then ax = 0 and ay = −g.
将原点设在发射点。取x轴为水平方向且指向抛射方向，y轴竖直向上。于是加速度分量为 ax = 0，ay = −g。
The value of g is often taken as 9.8 m/s², but in some exam boards 10 m/s² is used. You should always state your assumption clearly in written solutions.
g的取值通常为9.8 m/s²，但有些考试局采用10 m/s²。在书面解答中应明确说明你采用的取值。
2. Resolving Initial Velocity | 初速度分解
If the object is launched with an initial speed u at an angle θ above the horizontal, the horizontal component is u cos θ and the vertical component is u sin θ. For negative angles (launched downward), the vertical component is negative.
若物体以初速度u、仰角θ抛出，则水平分量为u cos θ，垂直分量为u sin θ。若θ为负（向下抛出），则垂直分量为负。
ux = u cos θ, uy = u sin θ
Angles can be given in degrees or radians, but most A-level mechanics questions use degrees. Make sure your calculator is in the correct mode before evaluating trigonometrical functions.
角度可以以度或弧度给出，但大多数A-level力学题目使用度。在计算三角函数之前，请确保计算器处于正确的角度模式。
3. Principle of Independent Motion | 独立运动原理
The horizontal and vertical motions are completely independent. Gravity affects only the vertical component, while the horizontal component remains constant throughout the flight. This principle allows us to treat the motion as two separate one-dimensional problems connected by time.
水平运动与垂直运动完全独立。重力只影响垂直分量，水平分量在整个飞行过程中保持不变。这一原理使我们能够将运动视为两个独立的一维问题，并通过时间联系起来。
For example, a ball dropped from rest and a ball projected horizontally from the same height hit the ground at the same time, because the vertical motion in both cases is identical.
例如，从同一高度自由释放的小球与水平抛出的小球会同时落地，因为二者在垂直方向的运动完全相同。
4. Kinematic Equations | 运动学方程
Using the constant-acceleration formulae separately in the x and y directions gives the position and velocity at any time t.
在x和y方向分别使用匀加速运动公式，可以得到任意时刻t的位置和速度。
For the horizontal direction:
水平方向：
sx = uxt = (u cos θ)t
For the vertical direction:
垂直方向：
vy = u sin θ − gt
sy = (u sin θ)t − ½gt²
These equations assume upward is positive. If you choose downward as positive, the signs change accordingly. Always draw a diagram and label the positive direction.
这些方程假定向上为正。如果你选择向下为正，则相应符号会改变。务必画出受力示意图并标出正方向。
5. Deriving the Trajectory Equation | 轨迹方程的推导
The trajectory equation expresses y directly in terms of x, eliminating time. From the horizontal equation, t = x/(u cos θ). Substituting this into the vertical displacement equation gives the parabola equation.
轨迹方程直接以x表示y，并消去时间t。由水平方程得 t = x/(u cos θ)，代入垂直位移方程，即可得到抛物线方程。
y = x tan θ − (g x²)/(2u² cos² θ)
This form is useful for checking whether a point lies on the trajectory, and for solving problems that specify the path rather than the time.
该形式可用于判断某点是否位于轨迹上，也可用于解决那些直接给出路径而不是时间的题目。
6. Time of Flight | 飞行时间
The time of flight is the total time until the projectile returns to the launch height. Setting sy = 0 in the vertical displacement equation gives the solutions t = 0 and t = 2u sin θ / g.
飞行时间是抛体从发射点到返回同一高度所需的总时间。令垂直位移方程中 sy = 0，可得 t = 0 和 t = 2u sin θ / g 两个解。
T = (2u sin θ)/g
If the projectile lands at a different height, you cannot use this shortcut. Instead, set sy equal to the vertical displacement relative to the launch point and solve the resulting quadratic equation.
如果抛体落点高度与发射点不同，则不能使用该简便公式。此时应令 sy 等于相对于发射点的垂直位移，并解相应的二次方程。
7. Maximum Height | 最大高度
At the highest point, the vertical velocity becomes zero. Setting vy = 0 gives the time to reach maximum height: t = u sin θ / g. Substituting this into the vertical displacement equation yields the maximum height.
在最高点，垂直速度为零。令 vy = 0，可得到达最大高度的时间：t = u sin θ / g。代入垂直位移方程，即可得到最大高度。
H = (u² sin² θ)/(2g)
Notice that the maximum height depends only on the vertical component of the initial velocity, not on the horizontal component.
注意最大高度只取决于初速度的垂直分量，与水平分量无关。
8. Horizontal Range | 水平射程
The horizontal range is the horizontal distance travelled during the total time of flight. Since the horizontal velocity is constant, multiply the horizontal velocity by the time of flight.
水平射程是整个飞行时间内水平方向走过的距离。由于水平速度恒定，只需用水平速度乘以飞行时间即可。
R = (u cos θ) × (2u sin θ)/g = (u² sin 2θ)/g
The range is maximised when sin 2θ = 1, i.e. when θ = 45°. For any other range value below the maximum, there are two complementary launch angles that give the same range.
当 sin 2θ = 1，即 θ = 45° 时，射程最大。对于任何小于最大值的射程，总有两个互补的发射角可以得到相同的射程。
9. Special Cases and Symmetry | 特殊情况与对称性
If the projectile is launched horizontally, θ = 0, the time of flight depends only on the height of launch. If it is launched vertically, θ = 90°, the range is zero and the motion is one-dimensional.
若抛体水平抛出，θ = 0，飞行时间只取决于发射高度。若竖直上抛，θ = 90°，射程为零，运动退化为直线运动。
Another important symmetry is that launch angles θ and 90° − θ produce the same range, because sin(2θ) = sin(180° − 2θ). This is often used in thinking questions to compare ranges quickly.
另一个重要的对称性是：发射角 θ 与 90° − θ 产生相同的射程，因为 sin(2θ) = sin(180° − 2θ)。这常用于思考题中快速比较射程大小。
At the highest point, the vertical velocity is zero but the horizontal velocity remains u cos θ. Therefore the projectile is never momentarily at rest unless it is thrown straight upward.
在最高点，垂直速度为零，但水平速度仍为 u cos θ。因此除非竖直上抛，否则抛体永远不会瞬时静止。
10. Numerical and Vector Methods | 数值与向量方法
For more complex situations, such as launching from an elevated position or when the terrain is not flat, vector notation provides a compact description. The displacement vector is written as:
对于更复杂的情形，例如从高处发射或地形不平坦时，向量记号提供了紧凑的描述。位移向量可写为：
r = u t + ½ g t²
Here u is the initial velocity vector and g is
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