📚 A-Level Maths: Projectile Motion Formula Summary | A-Level 数学：抛体运动公式总结
Projectile motion is a core topic in A-Level Mechanics. It describes the motion of an object launched into the air, moving under the influence of gravity alone while air resistance is neglected. This article compiles the essential formulas, derivations and exam strategies you need to tackle projectile motion questions confidently.
抛体运动是 A-Level 力学中的核心专题，描述一个被抛入空中的物体仅在重力作用下（忽略空气阻力）的运动。本文将汇总解题所需的核心公式、推导过程与考试策略，帮助你从容应对抛体运动题目。
1. Resolving Initial Velocity | 初速度的分解
For a particle projected with initial speed u at an angle θ above the horizontal, the initial velocity must be resolved into two perpendicular components. This is the most important first step in any projectile problem.
对于以初速度 u、与水平方向成夹角 θ 抛出的物体，必须先将初速度分解为两个相互垂直的分量。这是解决一切抛体问题最关键的第一步。
Horizontal component: uₓ = u cos θ
Vertical component: uᵧ = u sin θ
The horizontal component remains constant throughout the motion because no horizontal force acts on the projectile. The vertical component changes because gravity acts vertically downwards.
在整个运动过程中，水平分量保持不变，因为抛体在水平方向不受力；竖直分量则因为重力竖直向下作用而发生变化。
2. Equations of Motion for Each Direction | 两个方向上的运动方程
Projectile motion is two-dimensional, but we treat it as two independent one-dimensional motions using the standard SUVAT equations. Time t is the common variable linking both directions.
抛体运动是二维运动，但我们将其视为两个相互独立的一维运动，使用标准 SUVAT 运动学方程。时间 t 是联系两个方向的公共变量。
Horizontal direction: acceleration a = 0, so the horizontal displacement at time t is simply:
水平方向： 加速度 a = 0，因此在时间 t 的水平位移为：
x = (u cos θ) t
Vertical direction: taking upward as positive, acceleration a = -g, so the vertical displacement and velocity are:
竖直方向： 取向上为正，加速度 a = -g，因此竖直位移和竖直速度为：
vᵧ = u sin θ – g t
y = (u sin θ) t – ½ g t²
vᵧ² = (u sin θ)² – 2 g y
Here g is taken as 9.8 m/s² unless the question states otherwise. Always keep the sign convention consistent: if upward is positive, then displacement above the launch point is positive and below is negative.
此处 g 通常取 9.8 m/s²，除非题目另有说明。务必保持正方向约定一致：若向上为正，则发射点以上的位移为正，以下为负。
3. Time of Flight | 飞行时间
The time of flight is the total time the projectile stays in the air. For a particle projected from ground level and landing at the same height, the vertical displacement at landing is y = 0.
飞行时间是抛体在空中停留的总时间。对于从地面抛出且落回同一高度的物体，落地时竖直位移 y = 0。
Substitute y = 0 into the vertical displacement equation:
将 y = 0 代入竖直位移方程：
(u sin θ) t – ½ g t² = 0
t (u sin θ – ½ g t) = 0
The solution t = 0 corresponds to the launch instant. The non-zero solution gives the time of flight:
解 t = 0 对应发射瞬间，非零解给出飞行时间：
T = (2 u sin θ) / g
Notice that the time of flight depends only on the vertical component of velocity and g. It is independent of the horizontal component.
注意飞行时间仅取决于竖直速度分量和 g，与水平分量无关。
4. Maximum Height | 最大高度
At the highest point of the trajectory, the vertical velocity becomes zero. Using vᵧ = 0 in the vertical velocity equation:
在轨迹最高点，竖直速度为零。将 vᵧ = 0 代入竖直速度方程：
0 = u sin θ – g tₕ
tₕ = (u sin θ) / g
This is the time taken to reach the highest point. Substituting tₕ into the vertical displacement equation gives the maximum height H:
这是到达最高点所用时间。将 tₕ 代入竖直位移方程，得到最大高度 H：
H = (u² sin² θ) / (2 g)
Alternatively, use the SUVAT equation vᵧ² = (u sin θ)² – 2 g y with vᵧ = 0:
或者利用 SUVAT 方程 vᵧ² = (u sin θ)² – 2 g y，令 vᵧ = 0：
0 = (u sin θ)² – 2 g H ⇒ H = (u² sin² θ) / (2 g)
For symmetric flight from level ground, the time to reach maximum height is exactly half of the total time of flight. This symmetry is often used to simplify calculations.
对于从水平地面抛出的对称飞行，到达最大高度的时间恰为总飞行时间的一半。这一对称性常用于简化计算。
5. Range on Horizontal Ground | 水平地面上的射程
The horizontal range is the horizontal distance travelled during the whole flight. Using the time of flight T and the horizontal velocity u cos θ:
水平射程是飞行全程中水平方向走过的距离。利用飞行时间 T 和水平速度 u cos θ：
R = (u cos θ) × T = (u cos θ) × (2 u sin θ) / g
Using the double-angle identity 2 sin θ cos θ = sin 2θ:
利用二倍角公式 2 sin θ cos θ = sin 2θ：
R = (u² sin 2θ) / g
For a fixed launch speed u, the range is maximum when sin 2θ = 1, that is when 2θ = 90°, so θ = 45°. The maximum possible range is u² / g.
在初速度 u 固定时，当 sin 2θ = 1，即 2θ = 90°，也就是 θ = 45° 时射程最大，最大射程为 u² / g。
Note also that sin 2θ = sin(180° – 2θ), so two different angles, θ and 90° – θ, give the same range. These are called complementary angles.
另注意 sin 2θ = sin(180° – 2θ)，因此两个不同的角度 θ 与 90° – θ 会给出相同的射程，这两个角称为互余角。
6. Equation of the Trajectory | 轨迹方程
To find the path of the projectile, eliminate time t from the horizontal and vertical displacement equations. From the horizontal equation, t = x / (u cos θ). Substituting this into the vertical equation:
为了求抛体的运动路径，从水平位移方程和竖直位移方程中消去时间 t。由水平方程得 t = x / (u cos θ)，代入竖直方程：
y = x tan θ – (g x²) / (2 u² cos² θ)
This is the trajectory equation. It is a quadratic expression in x, so the path is a parabola. The equation is useful when you are given a point on the path and asked to find the launch angle or speed.
这就是轨迹方程。它是关于 x 的二次表达式，因此路径为抛物线。当题目给出路径上的某一点并要求求解发射角或初速度时，此方程非常有用。
Using the identity 1 + tan² θ = sec² θ = 1 / cos² θ, the trajectory equation can also be written as:
利用恒等式 1 + tan² θ = sec² θ = 1 / cos² θ，轨迹方程也可写成：
y = x tan θ – (g x² / (2 u²)) (1 + tan² θ)
If you know the coordinates of one point on the parabola, you can solve for tan θ or u using this form.
若已知抛物线上一点的坐标，可利用此形式求解 tan θ 或 u。
7. Horizontal Projection from a Height | 从高处水平抛出
A special case occurs when a particle is projected horizontally, so θ = 0, from a height h above the ground. The initial vertical velocity is zero.
一种特殊情况是物体从高出地面 h 处水平抛出，此时 θ = 0，竖直方向初速度为零。
Vertical motion starts from rest under gravity. Taking downward as positive or using y = -h with upward positive, the time to hit the ground is found from:
竖直方向从静止开始仅在重力作用下运动。取向下为正，或取向上为正但令 y = -h，落地时间由下式求得：
h = ½ g t²
t = √(2h / g)
The horizontal distance travelled before landing is:
落地前水平方向走过的距离为：
d = u t = u √(2h / g)
Importantly, the time of flight depends only on the height h and g, not on the horizontal speed u. A faster horizontal launch will land further away but not sooner.
重要结论：飞行时间只取决于高度 h 和 g，与水平速度 u 无关。水平抛出速度越大，落点越远，但不会更快落地。
8. Projection from a Height at an Angle | 从高处斜抛
When a projectile is launched from a height h above the ground at an angle θ above the horizontal, the landing point is below the launch point. Using upward as positive, the net vertical displacement at landing is y = -h.
当物体从高出地面 h 处以与水平成角 θ 斜抛时，落点低于发射点。取向上为正，落地时净竖直位移为 y = -h。
Substitute y = -h into the vertical displacement equation:
将 y = -h 代入竖直位移方程：
-h = (u sin θ) t – ½ g t²
Rearranging gives a quadratic in t:
整理后得到关于 t 的二次方程：
½ g t² – (u sin θ) t – h = 0
Solve this quadratic using the quadratic formula. Take only the positive root for time. Then substitute t into x = (u cos θ) t to find the horizontal range from the base of the launch point.
用求根公式解此二次方程，取正的时间根，再将 t 代入 x = (u cos θ) t，即可求出从发射点正下方算起的水平射程。
If the projectile instead lands on a platform of height H above the ground, the net vertical displacement is y = H – h. Always set y equal to the final height minus the initial height.
若抛体落在高于地面 H 的平台上，则净竖直位移为 y = H – h。始终将 y 设为终点高度减去起点高度。
9. Velocity, Speed and Direction at Any Time | 任意时刻的速度、速率与方向
At time t, the velocity components are already known:
在时间 t，速度分量已知：
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