📚 Understanding Projectiles | 理解抛体运动
In CIE A Level Physics, projectile motion is the motion of an object launched into the air and moving under the influence of gravity alone, after any initial propulsion has ceased. It combines a constant horizontal velocity with a constant vertical acceleration, which makes it one of the clearest and most examinable applications of kinematics and vector resolution on the syllabus.
在 CIE A Level 物理中，抛体运动是指物体被抛入空中后，仅在重力作用下运动的模型；研究对象一旦离开抛射装置，就不再受推进力影响。它的核心特征是水平方向速度恒定、竖直方向加速度恒定，因此是运动学与向量分解在考纲中最清晰、也最常被考查的应用之一。
1. What is a Projectile? | 什么是抛体？
In CIE A Level Physics, a projectile is defined as an object that moves through the air under the influence of gravity alone after being launched. Once it is in flight, the only force considered in an ideal model is its weight, acting vertically downward. The word "projectile" therefore refers to the motion model, not to a particular type of object.
在 CIE A Level 物理中，抛体被定义为抛出后仅在重力作用下运动的物体。进入飞行阶段后，理想模型中唯一考虑的力是竖直向下的重力。因此 "抛体" 指的是一个运动模型，而不是某一种特定的物体。
Common examples include a kicked football, a bullet after leaving the barrel, a cannonball, and a tennis ball in flight. In each case the launch force has already finished acting, so only the weight remains to change the motion of the object.
常见例子包括被踢出的足球、离开枪管后的子弹、炮弹以及飞行中的网球。在这些情形中，抛射的力都已经作用完毕，因此只剩下重力在改变物体的运动状态。
Projectile motion is usually studied without air resistance, with a uniform gravitational field and no Earth-curvature effects. Under these ideal conditions the gravitational field strength is taken as constant and directed vertically downward, so the acceleration of the projectile is uniform throughout the flight.
研究抛体运动时通常忽略空气阻力，并假设重力场均匀、不考虑地球曲率的影响。在这些理想条件下，重力场强度被视为恒定且竖直向下，因此抛体在整个飞行过程中的加速度都是匀定的。
2. Independence of Horizontal and Vertical Motion | 水平与竖直运动的独立性
The most fundamental principle is that the horizontal and vertical components of motion are independent. This means acceleration in one direction does not change the motion in the perpendicular direction, so the two directions can be analysed separately and then recombined.
最基本的原则是水平与竖直分量相互独立。一个方向的加速度不会改变垂直方向上的运动状态，因此两个方向可以分别分析，最后再用向量合成还原整体的运动。
onstant in magnitude and direction for the whole flight, and the horizontal displacement is simply x = vₓ t.
忽略空气阻力时，水平方向没有力，因此水平加速度为零。水平速度 vₓ 在整个飞行过程中大小和方向都保持不变，水平位移就是 x = vₓ t。
fall: the projectile experiences a constant downward acceleration g = 9.81 m s⁻². Galileo' that both reach the ground at the same time because their vertical motions are identical.
竖直方向则与自由落体完全相同：物体受到恒定向下的加速度 g = 9.81 m s⁻²。伽利略的经典演示——一个小球自由下落，另一个小球同时从同一高度水平抛出——说明两者同时落地，因为它们的竖直运动完全相同。
3. Resolving the Initial Velocity | 初速度的分解
t-angled triangle gives the horizontal component vₓ = v₀ cos θ and the vertical component vy = v₀ sin θ.
以初速率 v₀、与水平方向成 θ 角抛出的物体，其初速度可分为两个分量。用直角三角形分解速度向量，得到水平分量 vₓ = v₀ cos θ 与竖直分量 vy = v₀ sin θ。
vₓ = v₀ cos θ vy = v₀ sin θ
horizontal component vₓ is constant throughout the flight, while the vertical component vy n, but it must be applied consistently: with upward positive, the vertical acceleration is −g, so a = −9.81 m s⁻².
水平分量 vₓ 在整个飞行过程中保持不变；竖直分量 vy 在上升阶段每秒减少 g，在最高点变为零，随后向下增大。取向上为正只是一种约定，但必须全程一致：若取向上为正，则竖直加速度为 −g，即 a = −9.81 m s⁻²。
ase worth remembering is a projectile launched horizontally, where θ = 0, so vₓ = v₀ and vy ject then starts with no vertical velocity and falls under gravity from the first instant.
一个值得记住的特殊情形是水平抛出，此时 θ = 0，因此 vₓ = v₀、vy = 0。物体从抛出瞬间起竖直初速度为零，立即开始自由下落。
4. Key Equations of Projectile Motion | 抛体运动的核心方程
Because the horizontal acceleration is zero and the vertical acceleration is constant, the SUVAT equations can be applied to each direction separately. The table below lists the equations that cover almost every CIE question on projectiles.
由于水平加速度为零、竖直加速度恒定，可以对两个方向分别使用 SUVAT 方程。下表列出了几乎覆盖所有 CIE 抛体题目的公式。
| Direction 方向 | Equation 方程 | Note 说明 |
| Horizontal 水平 | x = v₀ cos θ × t | a = 0, velocity constant 速度恒定 |
| Vertical 竖直 | vy = v₀ sin θ − g t | a = −g (upward positive 取向上为正) |
| Vertical 竖直 | y = v₀ sin θ × t − ½ g t² | displacement 位移 |
| Vertical 竖直 | vy² = (v₀ sin θ)² − 2 g y | useful when t is unknown 无 t 时使用 |
The key skill is deciding which quantity is shared by the two directions. Time t is the same for both, so it is almost always the bridge between the vertical and horizontal equations. If a question gives two vertical quantities, find t first, then use it horizontally.
关键技巧是判断两个方向共享哪个物理量。时间 t 对两个方向相同，因此几乎总是竖直与水平方程之间的桥梁。如果题目给出两个竖直方向的量，先求 t，再把它代入水平方向。
5. Time of Flight, Maximum Height and Range | 飞行时间、最大高度与射程
same horizontal level, standard results can be derived from the equations above. Setting vy top, t = v₀ sin θ ÷ g, and doubling it gives the total time of flight T = 2 v₀ sin θ ÷ g.
对于从某一水平面抛出并落回同一水平面的抛体，可以从上面的方程推出标准结论。在最高点令 vy = 0，得到上升时间 t = v₀ sin θ ÷ g，将其加倍即得总飞行时间 T = 2 v₀ sin θ ÷ g。
v₀ sin θ ÷ g H = v₀² sin²θ ÷ (2g) R = v₀² sin 2θ ÷ g
the total time of flight into the horizontal equation gives the range R = v₀² sin 2θ ÷ g.
把上升时间代入竖直位移方程，得到最大高度 H = v₀² sin²θ ÷ (2g)；把总飞行时间代入水平方程，得到射程 R = v₀² sin 2θ ÷ g。
also shows that complementary angles give equal ranges, so 30° and 60° produce the same range for the same speed, though the higher angle gives a greater maximum height and a longer time of flight.
射程表达式表明，在初速率固定时，最大射程出现在 θ = 45°，因为 sin 2θ 在 2θ = 90° 时取最大值 1。它还说明互余角度给出相同射程：同一速率下 30° 与 60° 的射程相同，只是较大角度对应更高的最大高度和更长的飞行时间。
6. Symmetry and the Shape of the Trajectory | 轨迹的对称性与抛物线
constants. This means the trajectory is a parabola, provided air resistance is negligible.
把 x = vₓ t 与 y = v₀ sin θ × t − ½ g t² 联立并消去 t，可得到形如 y = a x − b x² 的方程，其中 a、b 为常数。这意味着在忽略空气阻力的条件下，轨迹是一条抛物线。
The parabola is symmetric about the vertical line through the highest point. The time taken to rise from a given height equals the time taken to fall back to that height, and the speed at a given height on the way up equals the speed at the same height on the way down, although the velocity vectors point in different directions.
该抛物线关于过最高点的竖直线对称。从某一高度上升到最高点所需的时间，等于从最高点落回该高度所需的时间；在同一高度上，上升时的速率等于下降时的速率，只是速度向量的方向不同。
At the highest point the vertical component of velocity is zero, but the horizontal component is unchanged and the acceleration is still g downward, not zero. A very common exam error is to claim that the acceleration of a projectile at the top of its flight is zero, or that the horizontal velocity is zero.
在最高点，竖直分速度为零，但水平分速度不变，加速度仍然是向下的 g，而不是零。考试中非常常见的错误是声称抛体在最高点的加速度为零，或水平速度为零。
7. Horizontal Projection from a Height | 从高处水平抛出
When an object is projected horizontally from a height h with speed u, the initial vertical velocity is zero and the vertical motion is pure free fall. The time of flight therefore depends only on the height, not on the horizontal speed.
当物体从高度 h 处以速率 u 水平抛出时，竖直初速度为零，竖直运动就是纯自由落体。因此飞行时间只取决于高度，与水平速度无关。
t = √2h ÷ g horizontal range = u √2h ÷ g
The horizontal distance travelled before landing is then simply the constant horizontal speed multiplied by this time. Doubling the horizontal launch speed doubles the range but leaves the time of flight unchanged, because the vertical and horizontal motions remain independent.
落地前飞过的水平距离就等于恒定的水平速度乘以这个时间。把水平抛出速率加倍会使射程加倍，但飞行时间不变，因为竖直与水平运动仍然相互独立。
ity on landing, combine the unchanged horizontal component u with the vertical component vy = g t using Pythagoras' theorem, and find the direction using tan θ = vy ÷ u. The resultant speed is √u² + vy² and the velocity is directed below the horizontal.
求落地速度时，用毕达哥拉斯定理把不变的水平分量 u 与竖直分量 vy = g t 合成，并用 tan θ = vy ÷ u 求方向。合速率是 √u² + vy²，速度方向指向水平线下方。
8. Worked Example: Projection at an Angle | 例题：斜抛
and ignore air resistance. Resolving the velocity gives vₓ = 25 cos 40° = 19.2 m s⁻¹ and vy = 25 sin 40° = 16.1 m s⁻¹.
一个足球以 25 m s⁻¹ 的速率从水平地面踢出，方向与水平成 40°。取 g = 9.81 m s⁻²，忽略空气阻力。分解速度得 vₓ = 25 cos 40° = 19.2 m s⁻¹，vy = 25 sin 40° = 16.1 m s⁻¹。
The time to reach the highest point is vy ÷ g = 16.1 ÷ 9.81 = 1.64 s, so the total time of flight is 3.28 s. The maximum height is vy = 19.2 × 3.28 = 63 m, which agrees with R = v₀² sin 2θ ÷ g = 625 × sin 80° ÷ 9.81 = 63 m.
到达最高点的时间为 vy ÷ g = 16.1 ÷ 9.81 = 1.64 s，因此总飞行时间为 3.28 s。最大高度为 vy² ÷ (2g) = 16.1² ÷ 19.62 = 13.2 m，射程为 vₓ × T = 19.2 × 3.28 = 63 m，与 R = v₀² sin 2θ ÷ g = 625 × sin 80° ÷ 9.81 = 63 m 一致。
At t = 2.00 s the vertical velocity is vy × 2.00 = −3.55 m s⁻¹, showing the ball is already moving downward. The resultant speed is √19.2² + 3.55² = 19.5 m s⁻¹, directed at tan⁻¹(3.55 ÷ 19.2) = 10.5° below the horizontal.
在 t = 2.00 s 时，竖直速度为 vy = 16.1 − 9.81 × 2.00 = −3.55 m s⁻¹，说明球已经在向下运动。合速率为 √19.2² + 3.55² = 19.5 m s⁻¹，方向为水平线下方 tan⁻¹(3.55 ÷ 19.2) = 10.5°。
9. Worked Example: Horizontal Projection | 例题：水平抛出
y at 8.0 m s⁻¹ from the top of a cliff 45 m high. The vertical motion is free fall, so t = √2h ÷ g = √2 × 45 ÷ 9.81 = √9.17 = 3.03 s. The horizontal distance from the base of the cliff is 8.0 × 3.03 = 24 m.
一块石头从 45 m 高的悬崖顶以 8.0 m s⁻¹ 水平抛出。竖直方向是自由落体，故 t = √2h ÷ g = √2 × 45 ÷ 9.81 = √9.17 = 3.03 s。落地点距崖底的水平距离为 8.0 × 3.03 = 24 m。
The vertical velocity on landing is vy 9.7 m s⁻¹, while the horizontal velocity remains 8.0 m s⁻¹. The resultant landing speed is √8.0² + 29.7² = 30.8 m s⁻¹ at tan⁻¹(29.7 ÷ 8.0) = 75° below the horizontal.
落地时的竖直速度为 vy = g t = 9.81 × 3.03 = 29.7 m s⁻¹，水平速度仍为 8.0 m s⁻¹。落地合速率为 √8.0² + 29.7² = 30.8 m s⁻¹，方向为水平线下方 tan⁻¹(29.7 ÷ 8.0) = 75°。
Notice that the answer to "how long does it take to land?" contains no horizontal information at all, and the answer to "how far does it travel horizontally?" uses that time. This separation of the two directions is the single most useful habit in projectile questions.
注意：求 "多久落地" 完全不需要水平方向的信息，而求 "水平飞多远" 则要用到前面求出的时间。把两个方向分开处理，是抛体题目中最有用的解题习惯。
10. Common Exam Mistakes | 常见失分点
Most lost marks in projectile questions come from a small number of repeated errors. Checking your work against the list below before moving on can protect several marks in a single question.
抛体题目中失掉的分大多来自少数几个反复出现的错误。做完题后对照下面的清单检查，往往能在一道题中挽回好几分。
- Forgetting to resolve the initial velocity, and using v₀ directly in both directions. 忘记分解初速度，在两个方向都直接使用 v₀。
- Using g = 9.81 m s⁻² with the wrong sign. 使用 g = 9.81 m s⁻² 时符号取错。
- Claiming the acceleration is zero at the highest point. 误认为最高点加速度为零。
- Assuming the vertical velocity is zero at the end of a horizontal projection when the object is still above the ground. 物体尚未落地时，误以为竖直速度为零。
- Mixing horizontal and vertical quantities in the same SUVAT equation. 把水平与竖直方向的量混入同一个 SUVAT 方程。
- Rounding intermediate values too early, which shifts the final answer. 中间步骤过早取近似，导致最终答案偏移。
11. Air Resistance and Real Projectiles | 空气阻力与真实的抛体
The ideal model ignores air resistance, but in real situations a drag force acts opposite to the velocity and grows with speed. This reduces both the maximum height and the range compared with the ideal parabola, and the trajectory becomes asymmetric, falling more steeply than it rises.
理想模型忽略空气阻力，但在实际情况中，阻力与速度方向相反，并随速率增大而增大。与理想抛物线相比，最大高度和射程都会减小，轨迹也不再对称，下降段比上升段更陡。
er applies, and the optimum launch angle for maximum range becomes slightly less than 45°. Questions that include air resistance are usually qualitative and ask you to describe or compare, not to calculate.
有空气阻力时，水平速度不再恒定，射程公式 R = v₀² sin 2θ
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