📚 Vertical Motion Under Gravity | 重力作用下的竖直运动
plication of the constant-acceleration (SUVAT) equations. When an object moves vertically near the Earth' 9.8 m s⁻² downwards. This allows us to model free fall, vertical projection, and motion returning to the ground using a single set of four equations.
在Edexcel A Level数学中，重力作用下的竖直运动是匀加速直线运动公式（SUVAT）的核心应用。当物体在地球表面附近竖直运动且忽略空气阻力时，其加速度恒定：g = 9.8 m s⁻²，方向向下。因此可以用四个运动学公式建立自由下落、竖直上抛和回到地面的模型。
1. The Basic Model and g | 基本模型与重力加速度 g
tude of acceleration due to gravity is g = 9.8 m s⁻² unless the question states otherwise.
在标准模型中，唯一作用的力是重力，因此加速度恒定。重力加速度的大小为 g = 9.8 m s⁻²，除非题目另有说明。
This value is always directed vertically downwards. Air resistance is ignored unless the question explicitly includes it.
该加速度始终竖直向下。除非题目明确说明，否则忽略空气阻力。
The object can be a ball, stone, particle or any projectile. Its mass does not affect the acceleration.
物体可以是球、石块、质点或任何抛射体。质量不会影响加速度。
2. Choosing a Sign Convention | 选择正方向
Before using SUVAT, choose a positive direction for displacement, velocity and acceleration. The most common choice is upwards positive.
在使用SUVAT公式前，必须先为位移、速度和加速度选定正方向。最常见的选择是向上为正。
If upwards is positive, then acceleration a = -9.8 m s⁻² because gravity acts downwards.
如果向上为正，则加速度 a = -9.8 m s⁻²，因为重力方向向下。
hen a = +9.8 m s⁻². Sticking to one convention throughout the whole question is essential.
如果向下为正，则 a = +9.8 m s⁻²。在整道题中必须始终坚持同一个正方向。
3. The Four SUVAT Equations | 四个运动学公式
For constant acceleration a, the four equations link s, u, v, a and t.
对于恒定加速度a，四个公式将位移s、初速度u、末速度v、加速度a和时间t联系起来。
v = u + at
s = ut + ½at²
v² = u² + 2as
s = ½(u + v)t
In vertical motion, replace a by -g or +g depending on your sign convention. Each equation uses displacement s, not total distance travelled.
在竖直运动中，根据正方向将 a 替换为 -g 或 +g。每个公式使用的是位移 s，而不是运动的总路程。
4. Free Fall from Rest | 从静止自由下落
ocity is u = 0. If downwards is positive and the object falls a distance s, then s = ½gt².
当物体从静止释放时，初速度 u = 0。若向下为正，物体下落距离 s，则 s = ½gt²。
fter falling for time t is v = gt. For example, after 2 seconds, v = 9.8 × 2 = 19.6 m s⁻¹.
下落 t 秒后的速度为 v = gt。例如，2秒后 v = 9.8 × 2 = 19.6 m s⁻¹。
The distance fallen in 1 s, 2 s, 3 s follows the pattern 4.9 m, 19.6 m, 44.1 m because s is proportional to t².
1秒、2秒、3秒内下落的距离分别为4.9 m、19.6 m、44.1 m，遵循比例规律，因为 s 与 t² 成正比。
5. Vertical Projection Upwards: Time to Maximum Height | 竖直上抛：到达最高点的时间
If a ball is thrown upwards with initial speed u, then at maximum height its velocity v is instantaneously zero.
如果球以初速度u向上抛出，在最高点时速度v瞬时为零。
= u + at with upwards positive gives 0 = u - gt, so the time to maximum height is t = u/g.
取向上为正，利用 v = u + at 得 0 = u - gt，因此到达最高点的时间为 t = u/g。
This time does not depend on the mass or the height of the point of release; it depends only on u and g.
该时间与质量或抛出点高度无关，只取决于u和g。
6. Maximum Height and Total Time of Flight | 最大高度与总飞行时间
m height above the point of release, use v² = u² + 2as with v = 0. This gives h = u² / 2g.
求最高点高于抛出点的最大高度时，令 v = 0，用 v² = u² + 2as，得 h = u² / 2g。
returns to the same level from which it was thrown, the total time of flight is T = 2u/g.
如果物体回到抛出时的同一水平面，总飞行时间为 T = 2u/g。
The time up equals the time down, because the motion is symmetric when air resistance is ignored.
上升时间等于下落时间，因为忽略空气阻力时运动具有对称性。
7. Projection Downwards and Returning to the Ground | 向下抛出与回到地面
If an object is thrown downwards with initial speed u, its initial velocity and gravity act in the same direction.
如果物体以初速度u向下抛出，初速度与重力方向相同。
th downwards positive, the final speed after falling distance s is given by v² = u² + 2gs.
取向下为正，下落距离s后的末速度满足 v² = u² + 2gs。
ve the ground and we need the time to hit the ground, use s = H, u = 0 and solve H = ½gt².
如果物体从离地高度H处落下，求落地时间时，令 s = H、u = 0，解 H = ½gt²。
Always check whether the displacement s is measured from the starting point or from the ground, because the SUVAT equations use displacement, not distance.
务必检查位移s是从起点还是从地面测量，因为SUVAT公式使用位移而非路程。
8. Velocity-Time Graphs for Vertical Motion | 竖直运动的速度-时间图
For an object thrown vertically upwards and caught at the same height, the velocity-time graph is a straight line with negative gradient -g.
对于竖直上抛后落回同一高度的物体，速度-时间图是一条斜率为 -g 的直线。
arts at +u, crosses the time axis at t = u/g (maximum height), and ends at -u at t = 2u/g.
直线从 +u 开始，在 t = u/g 处穿过时间轴（最高点），在 t = 2u/g 处到达 -u。
The area between the line and the time axis gives displacement. Positive area is upward displacement, and negative area is downward displacement.
速度图线与时间轴之间的面积表示位移。正面积为向上位移，负面积为向下位移。
9. Common Mistakes and Exam Tips | 常见错误与应试技巧
A common mistake is mixing sign conventions: using u positive but taking g positive when upwards is positive. This reverses the direction of acceleration.
常见错误是混淆正方向：当向上为正时，u取正，但g也取正，导致加速度方向颠倒。
Another mistake is treating the displacement from the ground as the same as the distance travelled. If a ball is thrown up from a height and then falls below the starting point, displacement and distance differ.
另一个错误是把离地位移与运动路程混为一谈。如果球从某一高度上抛后落到出发点以下，位移和路程不同。
Write down u, v, a, s, t first, assign signs, then choose the SUVAT equation that does not include the unknown variable you are not asked for.
先列出u、v、a、s、t，标好正负号，再选择不含题目未要求的未知量的SUVAT公式。
In Edexcel exams, show your substitution clearly. A neat diagram with an arrow showing positive direction can prevent sign errors.
在Edexcel考试中，要清晰展示代入过程。画一个带正方向箭头的简图可以避免符号错误。
10. Worked Example | 例题解析
Example: A ball is thrown vertically upwards with speed 19.6 m s⁻¹ from a point 1.5 m above the ground. Find the maximum height above the ground and the time until it hits the ground.
例题：一球以19.6 m s⁻¹的初速度从离地1.5 m处竖直上抛。求球离地面的最大高度以及从抛出到落地的时间。
Step 1: Take upwards as positive. Then u = +19.6, a = -9.8, and at maximum height v = 0.
步骤1：取向上为正。则 u = +19.6，a = -9.8，在最高点 v = 0。
.6 = 19.6 m above the point of release. Maximum height above ground = 19.6 + 1.5 = 21.1 m.
用 v² = u² + 2as：0 = 19.6² + 2(-9.8)s，因此 s = 19.6² / 19.6 = 19.6 m（高于抛出点）。离地最大高度 = 19.6 + 1.5 = 21.1 m。
displacement from the starting point is s = -1.5 m, because the ground is below the start.
求落地时间时，从起点算起的位移为 s = -1.5 m，因为地面在起点下方。
Use s = ut + ½at²: -1.5 = 19.6t - 4.9t². Rearranging gives 4.9t² - 19.6t - 1.5 = 0.
用 s = ut + ½at²：-1.5 = 19.6t - 4.9t²。整理得 4.9t² - 19.6t - 1.5 = 0。
, taking the positive root. The negative root is rejected because time cannot be negative.
解一元二次方程得 t ≈ 4.08 s，取正根。舍去负根，因为时间不能为负。
A common alternative is to split the motion into upward and downward stages, but the single displacement equation is quicker and less error-prone.
常见的替代方法是把运动分成上升和下降两段，但直接用一个位移方程更快且更不易出错。
11. Summary Checklist | 总结清单
Before attempting any vertical motion question, check the following: positive direction chosen, acceleration sign correct, initial velocity and displacement signs consistent, and the correct SUVAT equation selected.
在解答任何竖直运动问题前，请检查：已选择正方向、加速度符号正确、初速度和位移符号一致、选择了正确的SUVAT公式。
height t = u/g, maximum height h = u²/2g, total time of flight at the same level T = 2u/g.
关键结论：到达最高点时间 t = u/g，最大高度 h = u²/2g，同一水平面总飞行时间 T = 2u/g。
Practise past Edexcel mechanics questions, always showing units and rounding only at the final step.
练习Edexcel力学真题，始终标明单位，并且只在最后一步进行四舍五入。
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