📚 A-Level Maths: Horizontal Projectile Motion Problems | A-Level 数学：水平抛射运动问题
In A-Level Mathematics, projectile motion is a classic application of constant acceleration equations. A horizontal projectile is an object that is launched with an initial velocity that is entirely horizontal, then moves under the influence of gravity alone. Understanding how to split the motion into horizontal and vertical components is essential for solving many exam problems.
在 A-Level 数学中，抛射运动是匀加速运动方程的经典应用。水平抛射是指物体以完全水平方向的初速度被抛出，之后仅在重力作用下运动。理解如何将运动分解为水平与竖直两个方向的分量，是解决许多考试题目的关键。
1. Introduction to Horizontal Projectile Motion | 水平抛射运动简介
A horizontal projectile is launched with initial speed u horizontally from a point above the ground. The only force acting on the object after launch is gravity, so the vertical acceleration is constant at g ≈ 9.8 m/s² downward, while the horizontal acceleration is zero. This means the horizontal velocity remains constant throughout the motion.
水平抛射是指物体以初速率 u 从地面上方某点沿水平方向抛出。抛出后物体仅受重力作用，因此竖直方向加速度恒为向下的 g ≈ 9.8 m/s²，而水平方向加速度为零。这意味着水平速度在整个运动过程中保持不变。
Because the initial velocity has no vertical component, the problem is simpler than general projectile motion. We can analyse horizontal and vertical movements independently, then combine the results using time as the connecting variable.
由于初速度没有竖直分量，这类问题比一般的抛射运动更简单。我们可以将水平与竖直方向的运动分别分析，再以时间作为联系变量来综合结果。
2. Key Assumptions and Setting Up Axes | 关键假设与坐标系建立
To model horizontal projectile motion, we assume: air resistance is negligible; the object is a particle; gravity is constant; and the Earth’s surface is flat over the range considered. These assumptions allow us to use the standard constant acceleration equations.
为了建立水平抛射运动的模型，我们假设：空气阻力可忽略、物体视为质点、重力恒定、且在所考虑的水平距离内地面是平的。这些假设允许我们使用标准匀加速运动方程。
We choose a coordinate system with the origin at the point of launch. The positive x-axis points horizontally in the direction of motion, and the positive y-axis points vertically upward. The initial position is (0, 0), and the launch velocity is (u, 0).
我们选取坐标系，以抛出点为原点。x 轴正方向指向物体运动方向，y 轴正方向竖直向上。初始位置为 (0, 0)，初速度为 (u, 0)。
3. Equations of Motion in Horizontal and Vertical Directions | 水平与竖直方向的运动方程
For the horizontal direction, acceleration is zero. If the initial horizontal velocity is u, then at time t, the horizontal displacement x is given by:
水平方向加速度为零。若初速度的水平分量为 u，则 t 时刻水平位移 x 为：
x = ut
The horizontal velocity vₓ remains constant: vₓ = u.
水平速度 vₓ 保持不变：vₓ = u。
For the vertical direction, the initial velocity is 0, and the acceleration is -g (taking upward as positive). Thus at time t:
竖直方向初速度为 0，加速度为 -g（取向上为正）。因此 t 时刻有：
vᵧ = -gt
y = -½gt²
Here y is negative because the object moves downward. Many textbooks prefer to measure downward as positive, which avoids negative signs, but the choice must be consistent throughout.
此时 y 为负，因为物体向下运动。许多教材喜欢取向下为正，以避开负号，但全题必须自洽统一。
4. Finding Time of Flight | 求飞行时间
If the object is launched from a height H above the ground, then it hits the ground when y = -H. Using y = -½gt², we can solve for the time of flight T:
若物体从离地高度 H 处水平抛出，则落地时 y = -H。利用 y = -½gt²，可解得飞行时间 T：
-H = -½gT² → T = √(2H/g)
Notice that T depends only on the launch height and gravity, not on the initial horizontal speed. This is a key idea in horizontal projectile problems: the time of flight is independent of u.
注意 T 只取决于抛出高度和重力，与初速度的水平大小 u 无关。这是水平抛射问题的核心思想：飞行时间与 u 无关。
5. Finding Horizontal Range | 求水平射程
The horizontal range R is the horizontal distance travelled before landing. Since the horizontal velocity is constant, R is simply the product of the horizontal speed and the time of flight:
水平射程 R 是落地前水平方向走过的距离。由于水平速度恒定，R 就是水平速度与飞行时间的乘积：
R = uT = u√(2H/g)
This formula is useful when both H and u are given. For example, if u = 10 m/s and H = 19.6 m, then T = √(2×19.6/9.8) = 2 s, so R = 10×2 = 20 m.
当 H 与 u 都给定时，此公式很实用。例如若 u = 10 m/s，H = 19.6 m，则 T = √(2×19.6/9.8) = 2 s，所以 R = 10×2 = 20 m。
If instead you are asked for the speed needed to hit a target, rearrange the same equation to find u:
若反过来要求满足射程所需的初速度，则可重排上述方程求 u：
u = R / √(2H/g) = R√(g/(2H))
6. Finding Velocity and Speed at Any Time | 求任意时刻的速度与速率
At time t, the horizontal velocity component is vₓ = u, and the vertical component is vᵧ = -gt. The velocity vector is therefore (vₓ, vᵧ).
在 t 时刻，速度的水平分量为 vₓ = u，竖直分量为 vᵧ = -gt。因此速度矢量为 (vₓ, vᵧ)。
We can also write the speed v and the direction of motion relative to the horizontal. The speed is the magnitude of the velocity vector:
我们还可以写出速率 v 以及运动方向与水平方向的夹角。速率是速度矢量的大小：
v = √(vₓ² + vᵧ²) = √(u² + g²t²)
If θ is the angle below the horizontal at which the object is moving, then:
若 θ 表示物体运动方向与水平方向向下的夹角，则：
tan θ = |vᵧ| / vₓ = gt / u
Remember that as time increases, the vertical speed grows, so the angle θ increases from 0 toward 90°, but it only reaches 90° asymptotically because the horizontal speed never decreases.
注意随着时间增加，竖直速度不断增大，因此 θ 从 0° 逐渐增大并趋近 90°，但永远不会达到 90°，因为水平速度始终不变。
7. Finding Position and Trajectory Equation | 求位置与轨迹方程
The position of the projectile at time t is (x, y), where x = ut and y = -½gt². To find the trajectory equation, eliminate t from these two equations.
物体在 t 时刻的位置为 (x, y)，其中 x = ut，y = -½gt²。要得到轨迹方程，需要从这两个方程中消去 t。
From x = ut we have t = x/u. Substituting into y = -½gt² gives:
由 x = ut 得 t = x/u。将其代入 y = -½gt² 得：
y = -gx² / (2u²)
This is a quadratic equation in x, so the trajectory is a parabola, as expected for projectile motion. The equation can be written in the form y = kx², where k is negative, confirming a downward-opening parabola.
这是一个关于 x 的二次方程，因此轨迹为抛物线，这与抛射运动的已知规律一致。方程也可写成 y = kx² 的形式，其中 k 为负，说明抛物线开口向下。
When asked to find the position at a specific time, simply use x = ut and y = -½gt² directly. Always check that y is negative relative to the launch point when using the upward-positive convention.
当要求某时刻的位置时，直接使用 x = ut 和 y = -½gt² 即可。取向上为正时，务必检查 y 相对于抛出点为负值。
8. Worked Example 1: Cliff Problem | 例题1：悬崖问题
A ball is kicked horizontally from the top of a cliff with a speed of 15 m/s. The cliff is 45 m high. Find: (a) the time of flight; (b) the horizontal distance from the base of the cliff where the ball lands; (c) the speed of the ball just before it lands. Take g = 10 m/s².
一个球从 45 m 高的悬崖顶以 15 m/s 的速度水平踢出。求：(a) 飞行时间；(b) 球落地时距悬崖底部的水平距离；(c) 球刚落地前的速率。取 g = 10 m/s²。
Solution (a): Use the vertical motion. The initial vertical velocity is 0, displacement is -45 m, and acceleration is -10 m/s².
解 (a)：利用竖直方向运动。初速度竖直分量为 0，位移为 -45 m，加速度为 -10 m/s²。
-45 = -½ × 10 × T² → T² = 9 → T = 3 s
Solution (b): Horizontal motion is uniform, so R = uT = 15 × 3 = 45 m.
解 (b)：水平方向为匀速运动，故 R = uT = 15 × 3 = 45 m。
Solution (c): Just before landing, vₓ = 15 m/s and vᵧ = -gT = -30 m/s. Therefore the speed is:
解 (c)：落地前瞬间，vₓ = 15 m/s，vᵧ = -gT = -30 m/s。因此速率为：
v = √(15² + 30²) = √(225 + 900) = √1125 = 15√5 ≈ 33.5 m/s
Notice that we did not need the mass of the ball; projectile motion is independent of mass under these assumptions.
注意计算结果与球的质量无关；在这些假设下，抛射运动与质量无关。
9. Worked Example 2: Combination with Motion on Ground | 例题2：结合地面运动的题目
An arrow is fired horizontally from a platform at height 20 m with speed 40 m/s. At the same moment a car on the ground starts from rest at a point directly below the platform and accelerates uniformly to the right at 5 m/s². Does the arrow land in the car? If not, how far horizontally is the arrow from the car when the arrow lands?
一支箭从离地 20 m 的平台以 40 m/s 的速度水平射出。同一时刻，一辆停在该平台正下方地面上的汽车从静止开始以 5 m/s² 的加速度向右做匀加速运动。箭能否落进汽车？若不能，箭落地时与汽车的水平距离是多少？
First find the time of flight of the arrow:
先求箭的飞行时间：
20 = ½ × 10 × T² → T = 2 s
Horizontal displacement of the arrow: x_arrow = 40 × 2 = 80 m.
箭的水平位移：x_arrow = 40 × 2 = 80 m。
Horizontal displacement of the car from rest with a = 5 m/s²:
汽车从静止做 a = 5 m/s² 的匀加速运动，其水平位移为：
x_car = ½ × 5 × 2² = 10 m
The car has only travelled 10 m, while the arrow travelled 80 m. The arrow lands 70 m to the right of the car. Since the arrow falls much faster than the car accelerates, it misses.
汽车仅行驶了 10 m，而箭飞行了 80 m。箭落在汽车右侧 70 m 处。因为箭落地所需时间很短，而汽车加速度不够大，所以箭没有落进汽车。
This example shows how to connect projectile motion with other kinematics using the same time variable. Always calculate the time from vertical motion first, then apply it to horizontal motion.
这个例子说明了如何用同一个时间变量将抛射运动与其他运动学问题联系起来。务必先用竖直运动求出时间，再将其应用到水平运动中。
10. Common Mistakes and How to Avoid Them | 常见错误与避免方法
Many students make predictable errors in horizontal projectile problems. Here are the most common ones and how to avoid them.
学生在水平抛射问题中常犯一些可预测的错误。以下是最常见的错误及避免方法。
| Mistake / 错误 | Correction / 改正 |
| Using the initial horizontal speed u in vertical equations. | Vertical motion has zero initial velocity; use the vertical component only. |
| Forgetting that the time of flight depends only on height and g. | Always calculate T from y = -½gt², never from horizontal motion. |
| Using the same acceleration for both directions. | Horizontal acceleration is 0; vertical acceleration is -g. |
| Confusing speed and velocity. | Speed is the magnitude; velocity includes direction components. |
| Forgetting that y is negative when upward is positive. | Use displacement sign conventions consistently. |
It is also helpful to draw a diagram of the situation. Show the launch point, the ground, the axes, and label all known values. A clear diagram reduces sign errors and helps you plan the solution.
画示意图也很有帮助。画出抛出点、地面、坐标轴，并标出所有已知量。清晰的示意图能减少符号错误，并帮助你规划解题步骤。
11. Exam Tips and Summary | 考试提示与总结
In the exam, always start by resolving the motion into horizontal and vertical components. Write down what you know: u, H, g, and the target quantity. Then decide which equation to use. The time of flight is the bridge between the two directions.
考试中，首先将运动分解为水平与竖直两个方向。写下已知量：u、H、g 以及要求的目标量。然后决定使用哪个方程。飞行时间是联系两个方向的桥梁。
For horizontal projection, the following three equations are the core toolkit:
对于水平抛射，以下三个方程是核心工具：
-
Horizontal displacement: x = ut
水平位移：x = ut
-
Vertical displacement: y = -½gt²
竖直位移：y = -½gt²
-
Vertical velocity: vᵧ = -gt
竖直速度：vᵧ = -gt
Remember to use g = 9.8 m/s² unless the question tells you to take g = 10. Always include units in your final answer. Check whether the answer is sensible: a very long range with a small height might indicate a mis-sign in the vertical equation.
除非题目要求取 g = 10，否则均使用 g = 9.8 m/s²。最终答案要带单位。检查结果是否合理：如果高度很小而射程非常大，可能说明竖直方程中出现了符号错误。
By mastering horizontal projectile motion, you also build a strong foundation for angled projectiles. The same separation technique appears throughout mechanics and is frequently tested in A-Level papers.
掌握水平抛射运动，也为学习斜抛运动打下坚实基础。这种分解技巧贯穿整个力学章节，并且在 A-Level 考试中经常出现。
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