📚 Projectiles | 抛体运动
In A-Level Mechanics, projectile motion describes the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory. We analyse projectiles by treating the horizontal and vertical components of motion separately.
在A-Level力学中，抛体运动描述的是物体被投掷或抛射到空中后，仅受重力加速度影响的运动。该物体称为抛体，其运动路径称为轨迹。我们通过独立处理水平和竖直分运动来分析抛体运动。
1. Introduction to Projectiles | 抛体运动简介
Projectile motion is a two-dimensional motion under constant gravitational acceleration. We ignore air resistance and treat the projectile as a particle. The only force acting after launch is weight, causing a constant downward acceleration g (9.8 m s⁻²). This simplifies the analysis into two independent linear motions.
抛体运动是在恒定重力加速度下的二维运动。我们忽略空气阻力，将抛体视为质点。发射后唯一作用的力是重力，产生恒定的向下加速度 g（9.8 m s⁻²）。这使得分析可简化为两个独立的直线运动。
Common examples include a golf ball struck from the ground, a cannonball fired from a cliff, or a football kicked into the air. In each case, the horizontal velocity remains constant, while the vertical velocity changes linearly due to gravity.
常见的例子包括从地面击出的高尔夫球、从悬崖发射的炮弹，或者被踢向空中的足球。在每种情形中，水平速度保持不变，而竖直速度因重力而线性变化。
2. Modelling Assumptions | 模型假设
gligible; acceleration due to gravity is constant and acts vertically downwards; the Earth' he range considered. These allow us to use g = 9.8 m s⁻² and ignore any horizontal forces.
为了应用SUVAT方程，我们做了几个关键假设：抛体为质点（无旋转）；空气阻力可忽略；重力加速度恒定且竖直向下；在所考虑的范围内地球表面是平的。这使我们能使用 g = 9.8 m s⁻² 并忽略任何水平力。
ation. This sign convention is crucial for obtaining correct displacements and velocities.
我们通常取向上为正方向，因此当使用矢量符号时加速度 a = -g。这种符号约定对于获得正确的位移和速度至关重要。
3. Resolving Initial Velocity | 初速度分解
For a projectile launched with speed u at an angle θ to the horizontal, we resolve the initial velocity into components: horizontal ux = u cos θ, vertical uy ponents are used throughout the calculations. Example: if u = 25 m s⁻¹ and θ = 40°, then ux = 25 cos 40° ≈ 19.2 m s⁻¹, uy = 25 sin 40° ≈ 16.1 m s⁻¹.
对于以速率 u、与水平方向夹角 θ 发射的抛体，我们将初速度分解为分量：水平分量 ux = u cos θ，竖直分量 uy = u sin θ。这些分量贯穿所有计算。例如：若 u = 25 m s⁻¹，θ = 40°，则 ux = 25 cos 40° ≈ 19.2 m s⁻¹，uy = 25 sin 40° ≈ 16.1 m s⁻¹。
ux = u cos θ, uy = u sin θ
4. Horizontal Motion Analysis | 水平运动分析
Horizontally there is no acceleration (ax nt of velocity remains constant. The horizontal displacement x at time t is given by x = ux e can be expressed as t = x / (u cos θ), which is useful for eliminating t from equations.
水平方向无加速度（ax = 0），因此水平速度分量保持不变。水平位移 x 与时间 t 的关系为 x = ux t。这是匀速直线运动。因此，时间可表示为 t = x / (u cos θ)，这在消去方程中的 t 时很有用。
Because the horizontal velocity never changes, the horizontal range depends solely on the time of flight. Even if the projectile encounters an obstacle, horizontal motion remains uncomplicated.
由于水平速度始终不变，水平射程仅取决于飞行时间。即使抛体遇到障碍物，水平运动依然简单。
x = ux t
5. Vertical Motion Analysis | 垂直运动分析
Vertically, the motion is subject to constant acceleration ay = -g (taking upward as positive). The relevant SUVAT equations become: vy = uy - g t ; y = uy t - ½ g t² ; vy² = uy² - 2g y. Here y is the vertical displacement from the launch point.
竖直方向上，运动受恒定加速度 ay = -g（取向上为正）影响。相应的SUVAT方程为：vy = uy - g t；y = uy t - ½ g t²；vy² = uy² - 2g y。这里 y 是相对于发射点的竖直位移。
The symmetry of projectile motion is evident: time to reach maximum height equals time to descend from it; speed at a given height during ascent equals speed at the same height during descent (ignoring air resistance).
抛体运动的对称性显而易见：到达最大高度的时间等于从最大高度下降的时间；在上升过程中某一高度的速率等于下降过程中同一高度的速率（忽略空气阻力）。
vy = uy - g t
y = uy t - ½ g t²
vy² = uy² - 2 g y
6. Time of Flight | 飞行
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