Vertical Motion Under Gravity | 重力作用下的竖直运动

📚 Vertical Motion Under Gravity | 重力作用下的竖直运动

In Edexcel A Level Mechanics, vertical motion under gravity is modelled as motion with constant acceleration due to gravity, provided air resistance is ignored. This topic tests your ability to apply the SUVAT equations, manage sign conventions, and interpret displacement, velocity, and time correctly.

在 Edexcel A Level 力学中，忽略空气阻力时，重力作用下的竖直运动可视为加速度恒定的运动。本主题考察学生运用 SUVAT 方程、处理符号约定以及正确理解位移、速度和时间的能力。

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1. Modelling assumptions and acceleration due to gravity | 模型假设与重力加速度

In vertical motion problems on Edexcel Mechanics papers, a particle or object is usually treated as a point mass. Air resistance is ignored unless the question explicitly states otherwise. The acceleration due to gravity, g, is therefore constant and directed vertically downwards.

在 Edexcel 力学试卷中，物体通常被看作质点。除非题目明确说明，否则忽略空气阻力。因此重力加速度 g 恒定，方向竖直向下。

The standard value used in Edexcel exams is g = 9.8 m s⁻², though some questions may instruct you to use g = 10 m s⁻² or 9.81 m s⁻². Always check the question.

Edexcel 考试中通常使用 g = 9.8 m s⁻²，但有些题目可能要求使用 g = 10 m s⁻² 或 9.81 m s⁻²。务必仔细审题。

Since acceleration is constant, the object’s mass does not affect its motion when air resistance is neglected. A heavy ball and a light ball released from the same height will reach the ground at the same time in this ideal model.

由于加速度恒定，忽略空气阻力时物体的质量不影响运动。同一高度释放的重球和轻球在理想模型中会同时落地。

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2. Sign conventions and displacement | 符号约定与位移

You must choose a positive direction before using SUVAT equations. Most students choose upwards as positive. With this convention, acceleration is a = −g because gravity acts downwards.

在使用 SUVAT 方程前必须选定正方向。大多数学生选择向上为正。按照这一约定，加速度 a = −g，因为重力方向向下。

Displacement s, initial velocity u, and final velocity v are then positive when they point upwards and negative when they point downwards. For example, a ball thrown upwards with initial speed 20 m s⁻¹ has u = +20 m s⁻¹ if upwards is positive.

此时，位移 s、初速度 u 和末速度 v 向上为正、向下为负。例如，若选向上为正，以 20 m s⁻¹ 初速度上抛的小球有 u = +20 m s⁻¹。

If you choose downwards as positive, then a = +g and all signs reverse. The key is consistency: once a sign convention is chosen, every vector quantity must follow it.

如果选择向下为正，则 a = +g，所有符号相反。关键是保持一致性：正方向一旦确定，所有矢量都必须遵守。

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3. The SUVAT equations for vertical motion | 竖直运动的 SUVAT 方程

For constant acceleration, the four SUVAT equations link displacement s, initial velocity u, final velocity v, acceleration a, and time t. In vertical motion, replace a with ±g according to your sign convention.

对于匀加速运动，四个 SUVAT 方程将位移 s、初速度 u、末速度 v、加速度 a 和时间 t 联系起来。在竖直运动中，根据符号约定用 ±g 代替 a。

Equation	Meaning
v = u + at	final velocity / 末速度
s = ut + ½at²	displacement with initial velocity / 有初速度的位移
v² = u² + 2as	avoids time / 不含时间
s = ½(u + v)t	average velocity form / 平均速度形式

In the table, s is displacement, not distance. If an object returns to its starting point, its displacement is zero even though it has travelled a positive distance.

表中 s 是位移而非路程。如果物体回到出发点，位移为零，但它实际走过的路程为正。

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4. Falling from rest | 从静止下落

When an object is released from rest at height h above the ground, its initial velocity is u = 0. Taking downwards as positive is often convenient, giving a = +g.

当物体从地面上方高度 h 处由静止释放时，初速度 u = 0。通常选择向下为正更方便，此时 a = +g。

The time to fall is found from s = ut + ½at². With u = 0 and s = h, the equation becomes h = ½gt², so t = √(2h/g).

下落时间可由 s = ut + ½at² 求得。代入 u = 0 和 s = h，得到 h = ½gt²，因此 t = √(2h/g)。

The final speed just before hitting the ground is v = gt or v² = 2gh, so v = √(2gh). This result is independent of mass.

落地前瞬间的速度为 v = gt 或由 v² = 2gh 得 v = √(2gh)。该结果与质量无关。

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5. Projection vertically upwards | 竖直上抛

If an object is projected vertically upwards with speed U, taking upwards as positive gives u = +U and a = −g. The velocity decreases uniformly as the object rises.

如果物体以速率 U 竖直上抛，取向上为正，则 u = +U，a = −g。物体上升时速度均匀减小。

The velocity at any time t is v = U − gt. At the highest point, the instantaneous velocity is zero.

任意时刻的速度为 v = U − gt。在最高点，瞬时速度为零。

The displacement after time t is s = Ut − ½gt². This gives the height above the starting point, not necessarily the total distance travelled if the object has started falling again.

时间 t 后的位移为 s = Ut − ½gt²。这是相对于抛出点的高度；如果物体已经开始下落，它并不等于总路程。

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6. Maximum height and time to highest point | 最大高度与到达最高点时间

At maximum height, v = 0. Using v = u + at gives 0 = U − gt, so the time to reach maximum height is t = U/g.

在最大高度处，v = 0。由 v = u + at 得 0 = U − gt，所以到达最大高度所需时间 t = U/g。

The maximum height H is found from v² = u² + 2as: 0 = U² − 2gH, hence H = U²/(2g).

最大高度 H 可由 v² = u² + 2as 求得：0 = U² − 2gH，因此 H = U²/(2g)。

You can also find H by substituting t = U/g into s = Ut − ½gt². Both methods are common in exam solutions.

也可将 t = U/g 代入 s = Ut − ½gt² 求 H。两种方法在考试解答中都很常见。

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7. Total time of flight and landing speed | 总飞行时间与落地速度

For a particle projected vertically upwards from ground level and returning to the same level, the total displacement is zero. Using s = Ut − ½gt² and setting s = 0 gives t(U − ½gt) = 0.

对于从地面竖直上抛后又落回同一高度的物体，总位移为零。令 s = Ut − ½gt² 中 s = 0，得 t(U − ½gt) = 0。

The non-zero solution is t = 2U/g. This is twice the time to reach maximum height, showing the upward and downward journeys take equal times.

非零解为 t = 2U/g。这是到达最大高度时间的两倍，表明上升和下落所需时间相等。

The landing speed is found from v = U − gt with t = 2U/g: v = U − 2U = −U. The negative sign means the velocity is downwards, so the speed is equal to the initial speed U.

落地速度由 v = U − gt 代入 t = 2U/g 得 v = U − 2U = −U。负号表示速度方向向下，因此落地速率等于初速率 U。

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8. Speed-time graphs for vertical motion | 竖直运动的速度-时间图像

A speed-time graph is a useful way to visualise vertical motion. Speed is the magnitude of velocity, so it is always non-negative. For an object thrown upwards, the speed decreases linearly to zero at maximum height, then increases linearly as it falls.

速度-时间图像（速率为纵轴）是可视化竖直运动的有用工具。速率是速度的大小，始终为非负值。对于上抛物体，速率线性减小到最高点为零，然后下落时线性增大。

If you plot velocity against time instead, using upwards as positive, the graph is a straight line of gradient −g, crossing the time axis at t = U/g. The area between the line and the time axis has a signed meaning: area above gives upward displacement, area below gives downward displacement.

如果以速度（而非速率）对时间作图，取向上为正，则图像是一条斜率为 −g 的直线，在 t = U/g 处穿过时间轴。直线与时间轴之间的面积有正负含义：上方面积表示向上位移，下方面积表示向下位移。

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9. Displacement-time graphs and symmetry | 位移-时间图像与对称性

The displacement-time graph for vertical projection is a parabola. Taking upwards as positive and s = Ut − ½gt², the graph has a maximum at t = U/g and returns to s = 0 at t = 2U/g.

竖直上抛的位移-时间图像是抛物线。取向上为正，s = Ut − ½gt² 的图像在 t = U/g 处达到最大值，并在 t = 2U/g 处回到 s = 0。

The symmetry of the parabola reflects an important physical fact: at equal heights on the way up and down, the speed is the same, but the velocity direction is opposite.

抛物线的对称性反映了一个重要物理事实：在上升和下落经过同一高度时，速率相同，但速度方向相反。

For example, a ball thrown upwards passes a height h twice: once going up with velocity +√(U² − 2gh) and once coming down with velocity −√(U² − 2gh), if upwards is positive.

例如，上抛的小球两次经过高度 h：一次向上，速度为 +√(U² − 2gh)；一次向下，速度为 −√(U² − 2gh)（取向上为正）。

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10. Two-body vertical motion problems | 双物体竖直运动问题

Edexcel exam questions often involve two particles moving vertically, such as one dropped and one projected upwards from different heights. You should write separate SUVAT equations for each particle and link them through time, height, or meeting conditions.

Edexcel 考试题常涉及两个竖直运动的质点，例如一个从某高度下落，另一个从另一高度上抛。应分别为每个质点写 SUVAT 方程，并通过时间、高度或相遇条件将它们联系起来。

A common strategy is to define displacement from a fixed reference height, such as the ground or the lower particle’s starting point. Then set the two displacements equal when the

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