📚 Vertical Motion Under Gravity | 重力作用下的竖直运动
When a particle moves vertically under the influence of gravity alone, it experiences a constant acceleration directed towards the centre of the Earth. This is one of the most fundamental applications of the constant-acceleration (SUVAT) equations in A-Level Mathematics.
当质点仅在重力作用下做竖直运动时，它受到一个指向地心的恒定加速度。这是 A-Level 数学中恒定加速度（SUVAT）方程最基础的应用之一。
1. The SUVAT Equations | 基本运动学方程
For vertical motion under gravity, the acceleration is constant and equal to g (approximately 9.8 m/s² near the Earth's surface). The five SUVAT equations relate displacement s, initial velocity u, final velocity v, acceleration a, and time t.
对于重力作用下的竖直运动，加速度恒定且等于 g（在地球表面附近约为 9.8 m/s²）。五个 SUVAT 方程描述了位移 s、初速度 u、末速度 v、加速度 a 和时间 t 之间的关系。
v = u + at
s = ut + ½at²
s = ½(u + v)t
v² = u² + 2as
s = vt − ½at²
When applying these to vertical motion, we replace a with ±g depending on our chosen sign convention.
将这些方程应用于竖直运动时，我们根据所选的正方向约定，将 a 替换为 ±g。
2. Sign Conventions | 正方向约定
Choosing a consistent sign convention is the single most important step in solving vertical motion problems. The standard convention is to take upwards as positive, which makes a = −g (since gravity acts downwards).
选择一致的正方向约定是解决竖直运动问题最重要的一步。标准约定是取向上为正方向，这样 a = −g（因为重力向下作用）。
| Quantity | Sign when moving up | Sign when moving down |
| Displacement s (above start) | s > 0 | s < 0 (below start) |
| Velocity v | v > 0 | v < 0 |
| Initial velocity u | u > 0 (thrown upward) | u < 0 (thrown downward) |
| Acceleration a | a = −9.8 m/s² (always downward) | |
Alternatively, if you choose downwards as positive, then a = +g. Either convention works, provided you stay consistent throughout the problem.
或者，如果你选择向下为正方向，则 a = +g。两种约定都可行，只要在整个问题中保持一致即可。
3. Maximum Height and Time to Reach It | 最大高度与到达时间
At maximum height, the velocity of the particle is instantaneously zero. Using v = u + at with v = 0, we find the time to reach maximum height:
在最大高度处，质点的速度瞬时为零。利用 v = u + at，令 v = 0，我们可求出到达最大高度的时间：
tmax = u/g
Substituting this into s = ut + ½at² with a = −g gives the maximum height:
将其代入 s = ut + ½at²，其中 a = −g，得到最大高度：
H = u²/(2g)
For example, if a ball is thrown upwards with an initial speed of 20 m/s, the maximum height is H = 20²/(2 × 9.8) = 400/19.6 ≈ 20.4 m, and it takes t = 20/9.8 ≈ 2.04 seconds to get there.
例如，如果一个小球以 20 m/s 的初速度向上抛出，最大高度为 H = 20²/(2 × 9.8) = 400/19.6 ≈ 20.4 m，到达该高度所需时间为 t = 20/9.8 ≈ 2.04 秒。
4. Time of Flight | 飞行时间
The total time of flight for a particle projected upwards from its start point and returning to the same level is twice the time to reach maximum height:
从出发点竖直上抛并返回同一水平面的质点的总飞行时间是到达最大高度时间的两倍：
T = 2u/g
This symmetry arises because the motion is perfectly symmetric: the upward journey and downward journey take equal times and cover equal distances, with the velocities at corresponding points being equal in magnitude but opposite in direction.
这种对称性源于运动完全对称：上升阶段和下降阶段所需时间相等、经过的距离相等，且在对应点处的速度大小相等、方向相反。
If the particle lands below (or above) the point of projection, the time of flight must be found by solving a quadratic equation using s = ut + ½at².
如果质点落在抛出点下方（或上方），则必须通过 s = ut + ½at² 解二次方程来求飞行时间。
5. Velocity on Return | 返回时的速度
When a particle returns to its starting level, its speed is the same as its initial speed, but the velocity is in the opposite direction. This can be verified using v² = u² + 2as with s = 0:
当质点返回起点水平时，其速率与初速度大小相同，但方向相反。令 s = 0，利用 v² = u² + 2as 可验证：
v² = u² + 2g × 0 = u² → v = ±u
The negative root v = −u corresponds to the particle moving downward on its return.
负根 v = −u 对应质点返回时向下运动的情况。
6. Displacement-Time Graphs | 位移-时间图像
The displacement-time graph for a particle thrown vertically upward is a parabola. With upwards as positive, the graph rises to a maximum at t = u/g, then falls back to s = 0 at t = 2u/g, and continues to negative displacements if the particle falls below the start point.
竖直上抛质点的位移-时间图像是一条抛物线。取向上为正方向时，图像在 t = u/g 处升至最大值，然后在 t = 2u/g 处回落至 s = 0；如果质点下落到起点以下，图像继续延伸至负位移区域。
Key features of the displacement-time graph:
位移-时间图像的关键特征：
- The gradient at any point gives the velocity. The gradient is zero at maximum height, confirming v = 0 there.
- 曲线上任意一点的斜率表示该时刻的速度。在最大高度处斜率为零，证实该处 v = 0。
- The curve is concave downward since the acceleration is negative (a = −g).
- 曲线向下凹，因为加速度为负（a = −g）。
- The graph crosses the time axis (s = 0) at the time of return, t = 2u/g.
- 图像与时间轴相交（s = 0）的时刻即为返回时间 t = 2u/g。
7. Velocity-Time Graphs | 速度-时间图像
The velocity-time graph for vertical motion under gravity is a straight line with gradient −g (if upwards is positive). It starts at v = u, crosses v = 0 at t = u/g (maximum height), and becomes increasingly negative.
重力作用下竖直运动的速度-时间图像是一条斜率为 −g 的直线（若取向上为正方向）。直线从 v = u 开始，在 t = u/g 处穿过 v = 0（即最大高度），随后变为负值并持续减小。
For the velocity-time graph:
关于速度-时间图像：
- The gradient is constant and equals a = −9.8 m/s².
- 斜率恒定，等于 a = −9.8 m/s²。
- The area under the graph between t = 0 and t = T equals the displacement.
- 图像在 t = 0 至 t = T 之间与时间轴围成的面积等于位移。
- The area above the axis equals the maximum height; the area below the axis equals the distance fallen below the start.
- 时间轴上方的面积等于最大高度；时间轴下方的面积等于起点下方下落的距离。
8. Worked Example 1: Thrown Vertically Upward | 例题 1：竖直上抛
A ball is thrown vertically upward with a speed of 25 m/s from the ground. Take g = 9.8 m/s² and upwards as positive. Find (a) the maximum height reached, (b) the total time of flight, and (c) the velocity after 4 seconds.
一个小球以 25 m/s 的初速度从地面竖直上抛。取 g = 9.8 m/s²，向上为正方向。求 (a) 达到的最大高度，(b) 总飞行时间，(c) 4 秒后的速度。
(a) Maximum height:
(a) 最大高度：
H = u²/(2g) = 25²/(2 × 9.8) = 625/19.6 ≈ 31.9 m
(b) Total time of flight:
(b) 总飞行时间：
Using s = ut + ½at², with s = 0 (returns to ground):
利用 s = ut + ½at²，令 s = 0（回到地面）：
0 = 25t + ½(−
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