📚 Vertical Motion under Gravity | 重力作用下的竖直运动
Vertical motion under gravity is a core topic in Edexcel A-Level Mechanics. It deals with objects moving solely under the influence of gravity, neglecting air resistance. The acceleration is constant and equal to the gravitational acceleration g. Using the SUVAT equations, we can predict heights, times, and velocities for particles projected upwards, dropped from rest, or thrown downwards from a height. This article revisits the key concepts, sign conventions, and common problem types you will encounter in the exam.
重力作用下的竖直运动是Edexcel A-Level力学中的一个核心主题。它处理仅在重力影响下运动的物体，忽略空气阻力。加速度恒定且等于重力加速度 g。借助五个匀加速直线运动方程（SUVAT方程），我们可以预测向上抛射、从静止下落或从一定高度向下抛出的物体的高度、时间和速度。本文回顾关键概念、符号约定以及考试中常见的典型问题类型。
1. Introduction to Vertical Motion under Gravity | 重力下的竖直运动简介
When a particle moves vertically near the Earth's surface and air resistance is negligible, the only force acting is its weight. This produces a constant downward acceleration of magnitude g, usually taken as 9.8 m s−2 unless specified otherwise. The motion is therefore an example of uniformly accelerated linear motion, making the standard kinematic equations applicable.
当一个质点在地球表面附近竖直运动且空气阻力可忽略时，唯一作用的力就是它的重量。这产生一个向下的恒定加速度，大小通常取 g = 9.8 m s−2，除非题目另有说明。因此，这种运动是匀加速直线运动的一个实例，标准的运动学方程完全适用。
In exam questions, objects are often modelled as particles, meaning we ignore their size and rotational effects. Gravity is treated as uniform, so g is constant in both magnitude and direction. Horizontal motion does not affect the vertical analysis, which keeps the topic nicely focused.
在考题中，物体通常被模型化为质点，这意味着我们忽略其大小和转动影响。重力被视为均匀的，因此 g 的大小和方向处处恒定。水平运动不影响竖直方向的分析，这让这个主题重点明确。
2. SUVAT Equations and Their Application | SUVAT方程及其应用
The five fundamental SUVAT equations for constant acceleration are the backbone of vertical motion problems. They link initial velocity u, final velocity v, acceleration a, displacement s, and time t. You must be completely fluent with these formulas and be able to select the appropriate one depending on the given information.
五个基本的匀加速直线运动方程是解决竖直运动问题的核心工具。它们联系了初速度 u、末速度 v、加速度 a、位移 s 和时间 t。你必须非常熟练地掌握这些公式，并能根据已知信息选择合适的方程。
v = u + a t
速度与时间关系 | relation between velocity and time
s = u t + ½ a t2
位移与时间关系 | displacement-time relation
v2 = u2 + 2 a s
速度与位移关系 | velocity-displacement relation
s = ½(u + v) t
平均速度求位移 | average velocity formula
In vertical motion problems, the acceleration a is replaced by g or −g depending on the chosen sign convention. It is good practice to write down the known quantities first, noting u, v, a, s, and t, then choose the equation that does not contain the unknown you are trying to find. Most mistakes come from mixing up signs rather than choosing the wrong equation.
在竖直运动问题中，加速度 a 会根据所选的符号约定替换为 g 或 −g。一种好的习惯是先列出已知量 u、v、a、s 和 t，记下它们，然后选出不含所求未知量的方程。大多数错误源于符号混用，而不是选错公式。
3. Sign Convention for Gravitational Acceleration | 重力加速度的符号约定
rd direction as positive. Then the acceleration due to gravity is always a = −g = −9.8 m s−2. Quantities like initial velocity, final velocity, and displacement then carry signs according to their direction relative to the chosen positive sense.
选择一个一致的符号约定是解决竖直运动问题最关键的一步。力学中的标准方法是取向上为正方向。此时重力加速度始终为 a = −g = −9.8 m s−2。初速度、末速度和位移等物理量都依据它们相对于所选正方向来取正负号。
for an upward throw, and displacements downwards are positive. Both methods work, but the 'up as positive' convention aligns naturally with the graph shapes you will sketch.
如果物体向上抛出，其初速度 u 为正。到达最高点时速度 v = 0，但加速度仍为 −g，这就是物体减速的原因。下落时速度变为负值。如果你取向下为正，则 a = +g，这时向上抛射的初速度为负，向下的位移为正。两种方法都可行，但“向上为正”的约定会使你画出的速度–时间图形状更为直观。
Always clearly state your sign convention at the start of a solution. This single habit will prevent a huge number of errors and show the examiner that you understand the vector nature of the quantities involved.
务必在解题开始时清楚写明你的符号约定。这个小小的习惯可以避免大量错误，并向考官展示你理解了相关量的矢量性质。
4. Projected Vertically Upwards from the Ground | 从地面竖直上抛
The particle reaches its maximum height when v = 0, giving the time to the highest point tmax = u / g.
考虑一个质点从地面以初速率 u 竖直向上抛出。取向上为正，则在抛出点 s = 0，u > 0，a = −g。任意时刻 t 的速度为 v = u − g t。当 v = 0 时质点到达最高点，由此得到到达最高点的时间 tmax = u / g。
The maximum height H can be found using v2 = u2 + 2 a s. Substituting v = 0, a = −g gives 0 = u2 − 2 g H, so H = u2 the particle returns to the ground is obtained by setting s = 0 in s = u t − ½ g t2 2 u / g (landing). This shows the total time is exactly double the time to reach the peak.
最大高度 H 可用 v2 = u2 + 2 a s 求得。代入 v = 0、a = −g，得 0 = u2 − 2 g H，所以 H = u2 / (2 g)。质点落回地面的总飞行时间可通过令 s = 0 代入 s = u t − ½ g t2 求得。解 t (u − ½ g t) = 0，得 t = 0（抛出时刻）和 t = 2 u / g（落地时刻）。这表明总飞行时间恰好是到达最高点时间的两倍。
e displacement is zero, but the distance travelled is 2 H. Distinguishing between displacement and distance is critical in suvat applications.
落回地面时的速度大小与初速率相等，但方向相反：v = u − g (2 u / g) = −u。注意此时位移为零，但路程为 2 H。在匀变速公式的应用中，区分位移和路程至关重要。
5. Object Dropped or Projected Vertically Downwards | 自由落体与竖直下抛
al velocity u = 0. Taking downward as positive can be convenient here, but if you stick to 'up as positive' acement while falling will be negative. The speed acquired after falling a vertical height |s| is found from v2 e s = −h if the object drops by a distance h. The magnitude of velocity is √(2 g h).
如果物体从某一高度仅由静止释放，其初速度 u = 0。此时取向下为正可能较方便，但若坚持“向上为正”，加速度仍为 a = −g，下落过程中的位移为负值。下落竖直高度 |s| 后获得的速度可由 v2 = 0 + 2 (−g) (−h) = 2 g h 求出，其中 s = −h，因为物体下降距离 h。速度大小为 √(2 g h)。
. Using v = u − g t with u negative produces a more negative v, meaning faster downward motion.
当物体以初速率 u 竖直向下抛出时（向下），u 的正负号应与你选定的正方向一致。如果你保持向上为正，则 u 取负值，加速度仍为 −g。由于速度与加速度同号，速率增加。使用 v = u − g t，代入负的 u 会得到更负的 v，表示向下的速度更快。
Many students prefer to switch to 'down as positive' for dropped objects to avoid negative signs. While acceptable, you must be explicit about your choice. The exam will accept any consistent sign convention, but mixing conventions within a single solution will lead to incorrect results.
许多学生更喜欢在处理下落物体时切换到“向下为正”以避免负号。这虽然可行，但你必须明确说明自己的选择。考官接受任何一致的符号约定，但若在同一解答中混合使用不同的约定，将导致错误的结果。
6. Projection from a Height Above Ground | 从一定高度抛射
e if thrown upward and negative if thrown downward. The acceleration is a = −g throughout.
考试中常见的问题是，质点从地面以上某一高度 H 的平台上、悬崖上或窗口处向上或向下抛射。取向上为正，将原点设在抛出点，则地面的位置为 s = −H。向上抛出时初速度 u 为正，向下抛出时 u 为负。整个过程中加速度始终为 a = −g。
of flight until the object hits the ground, substitute s = −H into s = u t − ½ g t2. This yields a quadratic equation in t: ½ g t2 ent form if u is negative. Solve using the quadratic formula, and select the positive root for time. The negative root represents a time before the launch and is typically disregarded.
要求出物体从抛出到落地的时间，将 s = −H 代入 s = u t − ½ g t2。这会得到一个关于 t 的二次方程：½ g t2 − u t − H = 0（若向上抛出），或者当 u 为负时形式略有不同。用求根公式求解，并选择时间的正根。负根代表抛出之前的时刻，通常舍去。
The impact velocity can be found directly using v2 = u2 ith time altogether, which is useful when time is not required. Always check that the sign of the final velocity matches the expected direction of impact (downwards, hence negative when upward is positive).
撞击地面的速度可直接使用 v2 = u2 + 2 a s，并代入 s = −H 来求得。这完全避开了时间的计算，当题目不要求时间时非常有用。务必检查末速度的符号是否与预期的撞击方向一致（向下，因此在向上为正时取负值）。
7. Velocity–Time Graphs for Vertical Motion | 竖直运动的速度–时间图
a particle projected vertically upwards and returning to the starting point, the velocity– , and continues to −u at t = 2u/g. The gradient is constant and equal to the acceleration.
对于竖直上抛并落回起点的质点，其速度–时间图是一条斜率为 −g 的直线。它从 +u 开始，在 t = u/g 时穿过时间轴（最高点时刻），并在 t = 2u/g 时达到 −u。斜率恒定并等于加速度。
, the net area is zero, consistent with the total displacement being zero. The total distance travelled is given by the sum of the absolute areas.
直线与时间轴围成的面积代表位移。轴上方的面积（正）相应于向上位移，轴下方的面积（负）相应于向下位移。从 t = 0 到 t = 2u/g 的整个飞行过程中，净面积为零，这与总位移为零是一致的。所经过的总路程等于各部分面积的绝对值之和。
Sketching a quick v–t graph can be a powerful check on your algebraic solution. It shows at a glance the symmetry of the motion and helps identify the time when the particle is at a particular height on the way up and again on the way down. For objects projected from a height, the graph will be cut off when the line reaches the impact velocity.
快速画出 v–t 图是检验代数解答的一个有力工具。它能一目了然地展示运动的对称性，并有助于识别物体在上升和下落途中经过某一高度的两个时刻。对于从一定高度抛出的物体，图形会在直线达到撞击速度时被截断。
8. Connected Particles in Vertical Motion | 竖直运动中的相连质点
Vertical motion problems often feature two particles connected by a light, inextensible string passing over a smooth pulley. When the system is released from rest, the heavier mass moves downwards and the lighter mass moves upwards. Both particles experience the same magnitude of acceleration a (since the string is inextensible) and the same tension T, but their directions of motion differ.
竖直运动问题中常出现两个由轻质不可伸长的绳子跨过光滑滑轮相连的质点。当系统从静止释放时，较重的质量向下运动，较轻的质量向上运动。两质点具有相同大小的加速度 a（因为绳子不可伸长）和相同的张力 T，但它们的运动方向不同。
Applying Newton's second law to each mass, with upward usually taken as positive, leads to simultaneous equations. For a mass m1 moving up, T − m1g = m1a, and for a mass m2 moving down (M > m), m2g − T = m2a. Solving gives the acceleration a = (
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