📚 Vertical Motion Under Gravity | 重力作用下的竖直运动
Understanding the motion of objects moving vertically under the influence of gravity is a fundamental part of Edexcel A-Level Physics. Whether an object is dropped, thrown upwards, or projected downwards, its motion can be described using the SUVAT equations with a constant acceleration due to gravity, g. This article covers the essential concepts, equations, sign conventions, graphical analysis, and common exam traps to help you master vertical motion problems.
理解物体在重力作用下竖直运动的规律是 Edexcel A-Level 物理的重要基础。无论是自由落体、竖直上抛还是竖直下抛，其运动都可以通过 SUVAT 方程来描述，且加速度恒定为重力加速度 g。本文涵盖核心概念、公式、符号规则、图像分析以及常见考试陷阱，助你彻底攻克竖直运动问题。
1. Introduction to Vertical Motion Under Gravity | 重力作用下竖直运动简介
Vertical motion under gravity refers to the movement of objects along a straight vertical line where the only acceleration is caused by the Earth's gravitational pull. In the absence of air resistance, all objects fall with the same constant acceleration, independent of their mass. This idealised motion is a direct application of uniformly accelerated motion equations, often called the SUVAT equations, making it a cornerstone topic in mechanics.
重力作用下的竖直运动是指物体仅在地球引力产生的恒定加速度下沿竖直方向的运动。忽略空气阻力时，所有物体都以相同的加速度下落，与质量无关。这种理想化运动是匀加速运动方程（常称 SUVAT 方程）的直接应用，因此成为力学中的基石内容。
2. Acceleration Due to Gravity (g) | 重力加速度 (g)
Near the Earth' ² unless instructed otherwise. Remember that g is a vector: it always acts downwards towards the centre of the Earth. This downward direction will be crucial when assigning signs in your equations.
在地球表面附近，重力加速度用 g 表示，其大小约为 9.81 m s⁻²。在 Edexcel 考试中，除非题目另有说明，应使用 g = 9.81 m s⁻²。请记住 g 是矢量：它始终竖直向下指向地心。这一方向在你对方程赋予正负号时至关重要。
3. The SUVAT Equations Adapted for Vertical Motion | 适用于竖直运动的 SUVAT 方程
The five standard SUVAT equations for constant acceleration are equally valid for vertical motion. The key is to replace the generic acceleration a with the appropriate value according to your chosen sign convention. The equations are:
五个标准匀加速 SUVAT 方程在竖直运动中同样适用。关键在于根据你所选的正方向，将加速度 a 替换为相应的数值。方程组如下：
v = u + a t
s = u t + ½ a t²
s = v t − ½ a t²
v² = u² + 2 a s
s = ½ (u + v) t
where u is initial velocity, v is final velocity, s is displacement, t is time, and a is acceleration. For vertical motion, a will either be +g, −g, or a value including g depending on the positive direction you set.
其中 u 为初速度，v 为末速度，s 为位移，t 为时间，a 为加速度。对于竖直运动，a 将根据你设定的正方向取 +g、−g 或其他含 g 的数值。
4. Sign Conventions and Choosing the Positive Direction | 符号约定与正方向选择
One of the most common mistakes in vertical motion problems is inconsistent sign usage. You must decide at the start which direction you will call positive and stick to it throughout the calculation. The two most common choices are:
竖直运动中最常见的错误之一就是符号使用不一致。你必须在解题之初决定哪一个方向为正，并在整个计算中严格遵守。两种最常用的选择是：
Upwards as positive: Velocities directed upwards are positive, downwards are negative. Displacement measured from the starting point is positive if above the start, negative if below.
设向上为正: 则加速度 a = −g = −9.81 m s⁻²。向上的速度为正，向下的速度为负。从起点测量的位移若在起点上方为正，下方为负。
Downwards as positive: Velocities directed downwards are positive, and displacements below the start are positive. This choice often simplifies free-fall calculations.
设向下为正: 则加速度 a = +g = +9.81 m s⁻²。向下的速度为正，起点下方的位移为正。这一选择常使自由落体计算更为简便。
Always state your sign convention at the beginning of your solution to avoid confusion and to earn method marks.
务必在解答开头声明你的符号约定，以避免混淆并获得方法分。
5. Free Fall from Rest | 静止开始的自由落体
ime t is s = ½ g t², and the velocity after falling a distance s is v² = 2 g s. This square relationship means that doubling the time quadruples the fall distance, a key idea often tested in multiple-choice questions.
若物体从静止释放，初速度 u = 0。选取向下为正可使方程简化：a = +g。经时间 t 后的位移为 s = ½ g t²，下落距离 s 后的速度为 v² = 2 g s。这种平方关系意味着时间加倍会使下落距离增至四倍，这一关键概念常出现在选择题中。
e: s = ½ × 9.81 × 3.0² ≈ 44.1 m, and final velocity v = 9.81 × 3.0 = 29.4 m s⁻¹ downwards.
例如，一球从悬崖静止下落，经 3.0 s 着地。取向下为正，g = 9.81 m s⁻²：s = ½ × 9.81 × 3.0² ≈ 44.1 m，末速度 v = 9.81 × 3.0 = 29.4 m s⁻¹ 向下。
6. Vertical Projection Upwards | 竖直上抛运动
l slow down, reach a maximum height where its velocity becomes zero, and then fall back down. The time to reach the maximum height is given by t = u / g, and the maximum height above the launch point is s = u² / (2g). Because the motion is symmetrical (ignoring air resistance), the total time of flight is 2u/g, and the speed upon returning to the launch height equals the initial speed.
当物体竖直向上抛出时，其初始运动与重力方向相反。取向上为正，则 a = −g = −9.81 m s⁻²。物体将减速上升，到达最高点时速度为零，随后落回初始高度。到达最高点的时间为 t = u / g，最高点离抛出点的高度为 s = u² / (2g)。由于运动具有对称性（忽略空气阻力），全程飞行时间为 2u/g，落回原高度时的速度大小等于初速度大小。
s = (20)²/(2×9.81) ≈ 20.4 m; time to max height t = 20/9.81 ≈ 2.04 s. Total time = 4.08 s.
一物体以 20 m s⁻¹ 竖直上抛。设向上为正：最大高度 s = (20)²/(2×9.81) ≈ 20.4 m；升至最高点时间 t = 20/9.81 ≈ 2.04 s。总时间约 4.08 s。
7. Vertical Projection Downwards | 竖直下抛运动
a = +g. The equations then become v = u + g t and s = u t + ½ g t². The motion is straightforward but you must carefully distinguish projection downwards from free fall; in projection downwards the object already has an initial velocity, so its displacement and final velocity are greater for the same time interval compared to a dropped object.
若物体以初速度 u 向下抛出，重力加速度将使其速度持续增大。选取向下为正，则 a = +g。方程变为 v = u + g t 及 s = u t + ½ g t²。该运动很直接，但必须仔细区分下抛与自由落体：下抛已有初速度，因此在相同时间内其位移和末速度均大于静止释放的物体。
t + ½×9.81× t², giving t ≈ 2.0 s. The impact velocity is v = 5.0 + 9.81×2.0 ≈ 24.6 m s⁻¹.
一块石头以 5.0 m s⁻¹ 的速度向下抛出，初始高度 30 m。取向下为正，解方程 30 = 5.0 t + ½×9.81× t² 得 t ≈ 2.0 s，撞击速度 v ≈ 24.6 m s⁻¹。
8. Motion Graphs for Vertical Motion | 竖直运动的运动学图像
Sketching and interpreting displacement-time (s-t), velocity-time (v-t), and acceleration-time (a-t) graphs is an essential skill. For an object projected upwards with up as positive:
绘制并解读位移-时间 (s-t)、速度-时间 (v-t) 和加速度-时间 (a-t) 图像是必备技能。对于向上为正的竖直上抛：
v-t graph: A straight line with negative gradient −g, starting at +u, crossing the time axis at the instant of maximum height, and ending at −u when it returns to the launch point. The area under the graph gives displacement.
v-t 图像: 一条斜率为负的直线（梯度 = −g），起点为 +u，在最高点时刻穿过时间轴，落回起抛点时速度为 −u。图像下方面积表示位移。
s-t graph: A parabola opening downwards, with a maximum at the apex. The gradient at any point gives instantaneous velocity.
s-t 图像: 开口向下的抛物线，在最高点处取极大值。曲线上任意点的斜率代表瞬时速度。
a-t graph: nstant acceleration throughout the motion, even at the highest point where velocity is zero.
a-t 图像: 一条水平线 a = −g，表明整个运动过程中加速度恒定，即使在速度为零的最高点也是如此。
Being able to link these graphs to the physical stages of flight will help you answer graphical analysis questions confidently.
能够将这些图像与飞行的物理阶段对应起来，将帮助你自信地回答图像分析类考题。
9. Solving Problems with Displacement and Direction | 含位移与方向的解题
A common exam challenge is treating displacement rather than distance. Since displacement is a vector, when an object goes up and then back down past its starting point, the net displacement can be negative (if up is positive) even though the distance travelled is large. For instance, if a ball is thrown up from a cliff edge and eventually lands in the sea below, the final displacement is negative (below the starting point). Always insert the correct sign for s when using SUVAT equations.
常见的考试难点在于位移与路程的区别。位移是矢量，当物体上升再落回起点下方时，虽运动的路程很大，但净位移可为负（若上述方向为正）。例如，球从悬崖边缘上抛，最终落入下方海中，则末位移为负（在起点下方）。使用 SUVAT 方程时，一定要注意代入 s 的正确符号。
Using s = u t + ½ a t² gives −40 = 15 t − 4.905 t², leading to a quadratic equation for t.
例题：一球从高于海面 40 m 的悬崖上以 15 m s⁻¹ 上抛。取向上为正，a = −9.81 m s⁻²，入水时 s = −40 m （起点下方）。代入 s = u t + ½ a t² 得出 −40 = 15 t − 4.905 t²，解二次方程可求 t。
10. Two-Object Vertical Motion Scenarios | 双物体竖直运动场景
Problems involving two objects moving under gravity require careful time management. You may encounter a ball dropped from a height while another is projected upwards from the ground. To find when and where they meet, set up a common coordinate system and write displacement equations for each object. Equate their positions to find the meeting time, then substitute back to find the height. Remember that both experience the same gravitational acceleration.
涉及两个物体分别在重力作用下运动的问题需要仔细处理时间关系。你可能遇到一个球从高处释放，同时另一个球从地面向上抛出。为求它们何时何地相遇，需建立统一的坐标系，并分别写出两者的位移方程。令位置相等可求得相遇时间，回代可求高度。记住两者的重力加速度相同。
Example: Stone A is dropped from rest at height H. At the same instant, stone B is projected upwards from the ground with speed u. If up is positive for both: sA = H − ½ g t², sB = u t − ½ g t². Setting sA = sB yields H = u t, giving meeting time t = H/u.
例：石头 A 从高度 H 处静止释放，同一瞬间石头 B 以速度 u 从地面竖直上抛。若两者均取向上为正：sA = H − ½ g t²，sB = u t − ½ g t²。令 sA = sB 得 H = u t，相遇时间 t = H/u。
11. Common Misconceptions and Pitfalls | 常见误解与易错点
Avoid these typical errors:
请避免以下典型错误：
"Velocity is zero at the top, so acceleration is also zero." This is false; acceleration remains g throughout the motion, even at the instant of turnaround where velocity is zero.
"最高点速度为零，所以加速度也为零。" 这是错误的；整个运动过程中加速度恒为 g，即便在速度为零的转向瞬间也是如此。
"Heavier objects fall faster." In the absence of air resistance, all objects fall with the same acceleration. Do not let mass enter your SUVAT equations.
"较重的物体下落更快。" 在没有空气阻力时，所有物体下落加速度相同。不要让质量出现在 SUVAT 方程中。
The equation gives the magnitude of velocity, but direction must be inferred from context.
混淆速率与速度，尤其在应用 v² = u² + 2as 时容易出错。该方程给出速度的大小，方向需根据题意判断。
Using distance instead of displacement in SUVAT. If the object returns to its starting point, total displacement is zero, not the path length.
在 SUVAT 方程中用路程代替位移。若物体回到起点，总位移为零，而不是经过的路径长度。
12. Exam Tips and Summary | 考试技巧与总结
To excel in Edexcel A-Level Physics vertical motion questions:
要在 Edexcel A-Level 物理竖直运动题目中取得高分：
Always define your positive direction and label it clearly. Write "taking upwards as positive" at the start.
始终明确定义正方向，并在解答开头写清“设向上为正”。
List the known quantities u, v, s, a, t before selecting an equation. Not all will be needed, but identifying them prevents rash substitutions.
在选择方程前列出已知量 u, v, s, a, t。非全部都会用到，但罗列清楚可避免盲目代入。
Check for hidden information: 'dropped' means u = 0; 'reaches maximum height' means v = 0; 'returns to the same level' often implies displacement s = 0.
留意隐含信息：“静止释放” 意味着 u = 0；“到达最高点” 意味着 v = 0；“落回原高度” 常意味着位移 s = 0。
d otherwise, and be consistent with significant figures (usually 3 s.f. in final answers).
除非另有说明，使用 g = 9.81 m s⁻²，并保持有效数字一致（最终答案通常保留三位有效数字）。
When a problem involves heights both above and below the start, a clear sign convention will prevent sign errors in displacement.
当题目涉及起点上方和下方的高度时，清晰的符号约定可以避免位移的正负号错误。
Mastering vertical motion under gravity is about systematic application of the SUVAT equations with disciplined sign selection. With practice, you can handle any free-fall or projection problem that appears on your exam.
掌握重力作用下的竖直运动，关键在于系统地应用 SUVAT 方程并严格保持符号一致。通过练习，你将能够应对考试中出现的任何自由落体或抛体题目。
Published by TutorHao | Physics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图