📚 Projectiles (Edexcel A Level Maths) | 抛体运动（Edexcel A Level 数学）
In Edexcel A Level Mathematics, a projectile is modelled as a particle moving freely under gravity with no air resistance. The key idea is to resolve the initial velocity into horizontal and vertical components, then treat the two directions as two separate and completely independent problems linked only by time.
在 Edexcel A Level 数学中，抛体被建模为仅在重力作用下自由运动、不受空气阻力的质点。核心思想是把初速度分解为水平分量与竖直分量，然后把两个方向当作两个彼此独立、仅由时间 t 相联系的问题来处理。
This topic sits in the Year 2 Mechanics content and is examined in Paper 3. It draws together SUVAT equations, vectors, trigonometry and quadratic solving, so a clear, disciplined method matters far more than memorising formulas.
该知识点属于 Year 2 力学部分，在 Paper 3 中考查。它综合了 SUVAT 公式、向量、三角比与二次方程求解，因此清晰、规范的解题流程远比死记公式重要。
1. Modelling Assumptions | 模型假设
Projectile questions in Edexcel usually assume a point mass moving near the Earth's surface. Air resistance is ignored, so the only acceleration is due to gravity, and that acceleration acts vertically downwards throughout the flight.
Edexcel 的抛体问题通常假设物体为质点，在地球表面附近运动。忽略空气阻力，因此唯一的加速度来自重力，并且在整个飞行过程中该加速度竖直向下。
e question states otherwise. Vertical acceleration is written as −g or +g depending on whether upward or downward is taken as positive.
标准重力加速度取 g = 9.8 m s⁻²，除非题目另给数值。竖直加速度写成 −g 还是 +g，取决于你选择向上为正还是向下为正。
You should always state your sign convention explicitly at the start and keep it unchanged through the whole calculation. The model is valid for small vertical displacements, where g is approximately constant and the Earth's surface can be treated as flat and horizontal.
解题时必须先明确写出正方向约定，并在整个计算过程中保持一致。该模型适用于竖直位移较小的情形，此时 g 近似恒定，地球表面可视为水平平面。
A further assumption is that the projectile is unaffected by wind, spin or air density, and that it does not bounce or deform on landing. Any real experiment will differ slightly from the model, which is why exam answers are accepted within a small rounding tolerance.
另一个假设是抛体不受风、旋转或空气密度的影响，且落地时不反弹、不发生形变。真实实验一定与模型有偏差，这也是考试答案允许一定舍入误差的原因。
2. Horizontal Motion | 水平运动
After projection there is no horizontal force acting on the particle, so the horizontal acceleration is zero. This means the horizontal component of velocity is constant for the whole flight.
抛出之后物体在水平方向不受力，因此水平加速度为零。这意味着速度的水平分量在整个飞行过程中保持恒定。
vₓ = u cos θ
The horizontal displacement after time t is simply the horizontal velocity multiplied by time, since there is no acceleration to consider.
时间 t 后的水平位移就是水平速度乘以时间，因为该方向没有加速度需要处理。
x = u cos θ × t
Because horizontal velocity is constant, the graph of x against t is a straight line through the origin, and the particle covers equal horizontal distances in equal time intervals.
由于水平速度恒定，x 对 t 的图像是一条过原点的直线，物体在相等时间内通过相等的水平距离。
eed acts in the horizontal direction. Watch for this special case: it appears frequently in Edexcel questions about objects rolling off a table.
当物体被水平抛出时，抛射角 θ = 0°，于是 u cos θ = u，全部初速度都在水平方向上。注意这个特殊情况：它在 Edexcel 关于物体从桌面滚落的题目中经常出现。
3. Vertical Motion | 竖直运动
ive, the vertical acceleration is a = −g, and the standard SUVAT equations apply directly.
竖直方向上物体具有大小为 g 的恒定向下加速度。若取向上为正，则竖直加速度 a = −g，标准 SUVAT 公式可直接使用。
vy = u sin θ − g t
y = u sin θ × t − ½ g t²
vy² = (u sin θ)² − 2 g y
the highest point of the path the vertical component of velocity is momentarily zero, so vy tion is still −g at that instant: it is the velocity, not the acceleration, that vanishes.
在轨迹最高点，竖直分速度瞬时为零，即 vy = 0。关键点在于：那一刻加速度仍然是 −g，消失的是速度而不是加速度。
The vertical component of the initial velocity is u sin θ, and the vertical displacement y is measured from the point of projection, positive upwards if that is your convention. If the particle lands below the projection point, y is negative.
初速度的竖直分量是 u sin θ，竖直位移 y 从抛出点量起，若取向上为正则为正。如果物体落在抛出点下方，则 y 为负值。
4. Using SUVAT in Each Direction | 两个方向分别使用 SUVAT
ame set of equations twice: once horizontally with a = 0, and once vertically with a = ±g.
SUVAT 的五个量是 s、u、v、a、t。在抛体问题中，你要把同一组公式用两次：一次用于水平方向且 a = 0，一次用于竖直方向且 a = ±g。
| Quantity | Horizontal | Vertical |
| acceleration | 0 | −g (upward positive) |
| initial velocity | u cos θ | u sin θ |
| displacement | x | y |
| time | t | t (same value) |
The two directions share the same value of t, and this shared variable is the bridge between them. The standard strategy is to use the vertical direction to find t, then substitute that t into the horizontal equation to find the distance travelled.
两个方向共用同一个 t，这个共同变量正是两者之间的桥梁。标准策略是：先用竖直方向求出 t，再把 t 代入水平方程求水平距离。
ement from horizontal data with a = 0, even though they describe the same moving particle.
绝不要把两个方向的分量混用。竖直位移必须用竖直数据配合 a = −g 求得，水平位移必须用水平数据配合 a = 0 求得，尽管它们描述的是同一个运动物体。
5. Position Vector and the Trajectory Equation | 位置矢量与轨迹方程
Because position can be written with vectors, the position of the projectile at time t is often given in terms of the unit vectors i and j, where i is horizontal and j is vertical.
由于位置可以用向量表达，抛体在时刻 t 的位置常以单位向量 i 与 j 表示，其中 i 为水平方向，j 为竖直方向。
r = (u cos θ × t) i + (u sin θ × t − ½ g t²) j
The velocity vector at time t follows by differentiating each component, or by writing the two component formulas together in a single vector expression.
时刻 t 的速度向量可由对每个分量求导得到，也可直接把两个分量公式合并成一个向量表达式。
v = (u cos θ) i + (u sin θ − g t) j
Eliminating t between the horizontal and vertical displacement equations gives the Cartesian equation of the path, which is always a parabola when the ground is horizontal and flat.
从水平与竖直位移方程中消去 t，即得轨迹的直角坐标方程；在地面水平平坦的情况下，这条轨迹必定是抛物线。
y = x tan θ − g x² ÷ (2 u² cos² θ)
this equation and solving for the non-zero root gives the range on level ground directly.
由于 x² 的系数为负，这条抛物线开口向下。令 y = 0 并解出非零根，便可直接得到水平地面上的射程。
6. Time of Flight, Range and Greatest Height | 飞行时间、射程与最大高度
These three special results are worth knowing, but only for projection from and landing on the same horizontal level. In other situations you must return to the SUVAT equations.
这三个特殊结论值得掌握，但仅在抛出点与落点位于同一水平面时成立。其他情形必须回到 SUVAT 公式去解。
At the highest point vy to reach the top. The greatest height then follows from the vertical displacement formula.
在最高点 vy = 0，由此得到到达最高点所需时间；再把该时间代入竖直位移公式即得最大高度。
ttop = u sin θ ÷ g, H = u² sin² θ ÷ (2 g)
The total time of flight on level ground is twice the time to the top, by symmetry of the parabolic path, and the horizontal range is the constant horizontal velocity multiplied by that total time.
在水平地面上，由于抛物线轨迹具有对称性，总飞行时间是到达最高点时间的两倍，水平射程等于恒定水平速度乘以总飞行时间。
T = 2 u sin θ ÷ g, R = u² sin 2θ ÷ g
occurs at θ = 45°, and that complementary angles such as 30° and 60° give the same range.
射程公式表明：在水平地面上初速度固定时，θ = 45° 时射程最大；而 30° 与 60° 这类互余角给出相同的射程。
7. Projection from a Height and Disguised Projectiles | 从高处抛出与伪装抛体
Many exam questions launch the particle from a point above the ground, such as a cliff, a platform or the top of a building. Then the vertical displacement from projection to landing is negative if upward is taken as positive.
许多考题把物体从地面之上的某点抛出，例如悬崖、平台或楼顶。此时若取向上为正，从抛出点到落地的竖直位移为负值。
equation produces a quadratic in t, which is solved by factorising or by the quadratic formula. The negative root is discarded because time cannot be negative.
把 s = −h 代入竖直方程会得到一个关于 t 的二次方程，可用因式分解或求根公式求解。负根要舍去，因为时间不可能为负。
e vertical initial velocity is zero, so the quadratic simplifies to h = ½ g t², giving t = √2 h ÷ g. This single line often unlocks the whole question.
当物体从高处水平抛出时，竖直初速度为零，二次方程简化为 h = ½ g t²，于是 t = √2 h ÷ g。这一行常常就是整道题的突破口。
A disguised projectile is simply a projectile in a different story: a ball rolled off a desk, a package dropped from a drone, or a stone thrown from a cliff top. Identify the initial speed, the angle of projection and the vertical displacement, and the physics is unchanged.
所谓伪装抛体只是换了情境的抛体：从桌面滚下的小球、从无人机上落下的包裹、从崖顶抛出的石子。只要识别出初速度、抛射角和竖直位移，物理过程完全一样。
8. Velocity, Speed and Direction at Any Instant | 任意时刻的速度、速率与方向
The velocity of a projectile is a vector with two components, so its speed is found by combining those components using Pythagoras, not by adding them.
抛体的速度是含两个分量的向量，因此速率要用勾股定理合成两个分量求得，而不是简单相加。
speed = √vₓ² + vy²
The direction of motion at any instant is the angle the velocity vector makes with the horizontal, found from the tangent of that angle.
任意时刻的运动方向是该速度向量与水平方向的夹角，可由该角的正切求出。
tan φ = vy ÷ vₓ
If the particle is moving downwards, vy is negative under the upward-positive convention, and you should describe the direction as an angle below the horizontal to make the answer unambiguous.
若物体正在向下运动，在取向上为正的约定下 vy 为负值，此时应把方向描述为水平向下的夹角，使答案明确无歧义。
On the way up, the speed decreases; at the highest point the speed equals u cos θ, its minimum value; on the way down the speed increases again and, on level ground, the landing speed equals the launch speed.
上升过程中速率减小；在最高点速率等于 u cos θ，为最小值；下降过程中速率再次增大，在水平地面上落地速率等于抛出速率。
9. Speed of Impact without Finding the Time | 不需求时间求落地速率
Sometimes a question asks only for the speed on landing, with no interest in the time of flight or the horizontal distance. In that case a shortcut based on energy, or on combining the component equations, saves considerable work.
有时题目只要求落地速率，并不关心飞行时间或水平距离。此时用基于能量、或由分量方程合并得到的捷径，可以省去大量计算。
The horizontal component is unchanged, and the vertical component satisfies vy h point. Adding the squared components and using sin² θ + cos² θ = 1 gives a clean result.
水平分量不变；当物体落点比抛出点低 h 时，竖直分量满足 vy² = (u sin θ)² + 2 g h。把两个分量的平方相加，并利用 sin² θ + cos² θ = 1，可得到简洁结果。
v² = u² + 2 g h, v = √u² + 2 g h
Notice that the launch angle has disappeared from this formula. Any two projectiles thrown from the same point with the same speed, whatever their angles, land at the same speed if they fall through the same vertical height.
注意该公式中抛射角已经消失。从同一点以相同速率抛出的任何两个抛体，无论角度如何，只要下落的竖直高度相同，落地速率就相同。
Remember that this shortcut applies to speed only, not to direction, and it assumes air resistance is still ignored. If the question also wants the angle of impact, you must calculate vy and vₓ separately.
请记住这个捷径只适用于速率，不适用于方向，并且仍然假设忽略空气阻力。如果题目还要落地时的角度，就必须分别算出 vy 与 vₓ。
10. Exam Technique and Common Errors | 考试技巧与常见错误
Most marks in projectile questions are for method rather than for the final number, so write down the resolved components and the equation you are using before substituting values.
抛体题的分数大多给方法而不是最终数值，因此先写出分解后的分量和你所使用的方程，再代入数值。
Typical errors include: failing to state a sign convention; using g as positive while also treating upward displacement as positive; assuming the acceleration is zero at the highest point; forgetting that the particle is still moving horizontally at the top; and mixing components between the two directions.
典型错误包括：没有写出正方向约定；把 g 取为正的同时又把向上位移当正值；误以为最高点加速度为零；忘记物体在最高点仍有水平运动；以及把两个方向的分量混用。
Other frequent slips are using the full speed u instead of u cos θ or u sin θ, rounding too early in a multi-step calculation, and giving a negative time as an answer instead of rejecting that root.
其他常见失误有：用完整的初速度 u 代替 u cos θ 或 u sin θ；在多步计算中过早取近似值；以及把负的时间当作答案，而不是舍去该根。
Keep intermediate values to at least four significant figures and round only the final answer, usually to three significant figures. Always check that your answer is physically sensible: a projectile thrown at 20 m s⁻¹ cannot stay in the air for a minute.
中间值至少保留四位有效数字，只在最终答案处取近似，通常保留三位有效数字。始终检查答案在物理上是否合理：以 20 m s⁻¹ 抛出的物体不可能在空中停留一分钟。
11. Worked Examples | 典型例题
Example 1. izontal. Taking g = 9.8 m s⁻², find the time of flight, the greatest height and the range.
例 1. 小球从水平地面以 20 m s⁻¹ 的速率、与水平方向成 30° 抛出。取 g = 9.8 m s⁻²，求飞行时间、最大高度与射程。
Resolving: u sin 30° = 10 m s⁻¹ and u cos 30° = 17.3 m s⁻¹. Vertically, vy = 0 at the top, so ttop = 10 ÷ 9.8 = 1.02 s, giving T = 2.04 s and H = 10² ÷ (2 × 9.8) = 5.10 m.
分解速度：u sin 30° = 10 m s⁻¹，u cos 30° = 17.3 m s⁻¹。竖直方向上最高点 vy = 0，故 ttop = 10 ÷ 9.8 = 1.02 s，于是 T = 2.04 s，H = 10² ÷ (2 × 9.8) = 5.10 m。
check, the range formula gives u² sin 60° ÷ g = 400 × 0.866 ÷ 9.8 = 35.3 m, which agrees.
水平方向：x = u cos 30° × T = 17.3 × 2.04 = 35.3 m。用射程公式检验：u² sin 60° ÷ g = 400 × 0.866 ÷ 9.8 = 35.3 m，两者一致。
Example 2. A stone is thrown horizontally at 15 m s⁻¹ from the top of a cliff 25 m high. Find the time to reach the sea, the horizontal distance travelled and the speed of impact.
例 2. 石子以 15 m s⁻¹ 从 25 m 高的崖顶水平抛出。求落到海面所需时间、水平距离与落地速率。
downwards, so 25 = ½ × 9.8 × t², giving t = 2.26 s. Horizontally, x = 15 × 2.26 = 33.9 m.
竖直方向：u sin θ = 0，向下位移 y = 25 m，故 25 = ½ × 9.8 × t²，得 t = 2.26 s。水平方向：x = 15 × 2.26 = 33.9 m。
For the impact speed, vy = 9.8 × 2.26 = 22.1 m s⁻¹, so v = √15² + 22.1² = 26.7 m s⁻¹. The shortcut v = √15² + 2 × 9.8 × 25 = √715 = 26.7 m s⁻¹ confirms the answer.
求落地速率：vy = 9.8 × 2.26 = 22.1 m s⁻¹，故 v = √15² + 22.1² = 26.7 m s⁻¹。用捷径 v = √15² + 2 × 9.8 × 25 = √715 = 26.7 m s⁻¹ 验证，结果吻合。
12. Quick Reference Summary | 速查总结
The whole topic reduces to two independent one-dimensional problems joined by a single shared time. Fix a positive direction
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