A-Level Mathematics: Solving Projectile Motion Using Vector Methods | A-Level 数学：向量方法求解抛体运动

📚 A-Level Mathematics: Solving Projectile Motion Using Vector Methods | A-Level 数学：向量方法求解抛体运动

Projectile motion is one of the most frequently tested topics in A-Level Mathematics. In this article, we explore how vector methods provide a clean, unified framework for solving projectile problems — from position and velocity to maximum height, time of flight, and range.

抛体运动是 A-Level 数学中最常考查的考点之一。本文将探讨如何利用向量方法为求解抛体问题提供一个简洁而统一的框架——从位置与速度，到最大高度、飞行时间和射程。

1. Why Use Vectors for Projectile Motion? | 为什么用向量解抛体运动？

Traditional scalar methods treat horizontal and vertical motion separately, requiring you to juggle two sets of equations. A vector approach combines both directions into a single equation of motion, reducing algebra errors and revealing the underlying structure of the motion.

传统标量方法将水平运动与垂直运动分开处理，需要同时处理两组方程。而向量方法将两个方向统一为一个运动方程，既能减少代数错误，又能揭示运动的内在结构。

In vector notation, the displacement of a projectile at time t is written as:

在向量记号中，抛体在时刻 t 的位移可写作：

r(t) = (u cosθ t) i + (u sinθ t − ½gt²) j

Here, i is the horizontal unit vector, j is the vertical unit vector, u is the initial speed, θ is the angle of projection above the horizontal, and g ≈ 9.8 m/s².

其中 i 为水平单位向量，j 为竖直单位向量，u 为初速度大小，θ 为相对于水平方向的抛射角，g ≈ 9.8 m/s²。

2. Setting Up the Vector Framework | 构建向量框架

We choose the origin O at the launch point, with the x-axis horizontal and the y-axis vertical. The initial velocity vector is:

我们以发射点 O 为原点，x 轴水平、y 轴竖直。初速度向量为：

u = u cosθ i + u sinθ j

The acceleration vector is constant throughout the flight, pointing vertically downward:

整个飞行过程中加速度向量恒定，方向竖直向下：

a = −g j

This single fact — no horizontal acceleration and constant vertical acceleration — makes the vector equation of motion particularly elegant:

正是这一事实——水平方向无加速度、竖直方向匀加速——使得向量运动方程格外优美：

r(t) = r₀ + ut + ½at²

For a launch from the origin, r₀ = 0, which simplifies matters considerably.

若从原点发射，则 r₀ = 0，问题会大大简化。

3. The Position Vector as a Function of Time | 位置向量关于时间的函数

Substituting u and a into the general equation gives:

将 u 和 a 代入一般方程得：

r(t) = (u cosθ t) i + (u sinθ t − ½gt²) j

The i-component x = u cosθ t describes horizontal displacement; it grows linearly with time because there is no horizontal acceleration. The j-component y = u sinθ t − ½gt² describes vertical displacement; it is quadratic in time due to gravity.

i 分量 x = u cosθ t 描述水平位移，由于不存在水平加速度，它随时间线性增长。j 分量 y = u sinθ t − ½gt² 描述竖直位移，由于重力作用，它随时间二次变化。

Always check the units: u cosθ t has units m/s × s = m, and each term in the j-component also has units of metres.

务必检查单位：u cosθ t 的单位为 m/s × s = m，j 分量中每一项的单位同样为米。

4. The Velocity Vector by Differentiation | 通过微分求速度向量

Differentiating the position vector with respect to time gives the velocity vector:

对位置向量关于时间求导，即得速度向量：

v(t) = dr/dt = u cosθ i + (u sinθ − gt) j

Notice that:

请注意：

The horizontal component vₓ = u cosθ is constant, since there is no force in the horizontal direction.
The vertical component vᵧ = u sinθ − gt decreases linearly with time, reflecting the constant downward pull of gravity.

水平分量 vₓ = u cosθ 恒定不变，因为水平方向不受力。
竖直分量 vᵧ = u sinθ − gt 随时间线性减小，体现重力的恒定向下作用。

At the highest point of the trajectory, vᵧ = 0, but vₓ remains nonzero — the projectile is still moving horizontally even at the instant it reaches its peak.

在轨迹最高点，vᵧ = 0，但 vₓ 仍不为零——即使到达最高点的瞬间，抛体仍在水平方向运动。

5. The Acceleration Vector | 加速度向量

Differentiating the velocity vector a second time yields the acceleration vector:

对速度向量再次求导，得到加速度向量：

a(t) = dv/dt = −g j

This confirms that the acceleration is constant, vertical, and directed downward. Many students lose marks by writing a = g as a scalar without a direction. In vector form, the minus sign encodes the downward direction automatically.

这证实了加速度是恒定的、竖直向下的。许多学生因将 a = g 写成不含方向的标量而失分。在向量形式中，负号自动体现了向下方向。

If additional forces such as air resistance were present, the acceleration vector would no longer be constant, and the equations would become significantly more complex. For A-Level purposes, we assume a vacuum.

如果存在空气阻力等额外力，加速度向量将不再恒定，方程会复杂得多。在 A-Level 考试中，我们默认在真空中运动。

6. Maximum Height and Time to Peak | 最大高度与到达最高点的时间

At the maximum height, the vertical component of velocity is zero. Setting vᵧ = 0:

在最大高度处，速度的竖直分量为零。令 vᵧ = 0：

u sinθ − gt = 0 ⇒ t_peak = u sinθ ⁄ g

Substitute this time into the vertical component of r(t):

将此时间代入 r(t) 的竖直分量：

H = u sinθ (u sinθ ⁄ g) − ½g(u sinθ ⁄ g)² = u² sin²θ ⁄ (2g)

Thus the maximum height depends only on the initial speed and the angle of projection, not on the mass of the projectile. This is a classic result worth memorising:

因此，最大高度仅取决于初速度和抛射角，而与抛体质量无关。这是一个值得记忆的经典结论：

H = u² sin²θ ⁄ (2g)

7. Time of Flight and Range | 飞行时间与射程

The projectile returns to the ground when the j-component of r(t) is zero. Since t = 0 corresponds to the launch instant, the nonzero root gives the time of flight:

当 r(t) 的 j 分量为零时，抛体回到地面。由于 t = 0 对应发射瞬间，非零根即给出飞行时间：

u sinθ t − ½gt² = 0 ⇒ t(u sinθ − ½gt) = 0 ⇒ T = 2u sinθ ⁄ g

Substitute T into the i-component x = u cosθ t to obtain the range:

将 T 代入 i 分量 x = u cosθ t，即得射程：

R = u cosθ × (2u sinθ ⁄ g) = u² × 2sinθ cosθ ⁄ g = u² sin(2θ) ⁄ g

The range is maximised when sin(2θ) = 1, i.e., when θ = 45°. For any other range value below the maximum, there are two complementary projection angles that give the same range, since sin(2θ) = sin(180° − 2θ).

当 sin(2θ) = 1 时射程最大，即 θ = 45°。对于小于最大值的任何射程，存在两个互补的抛射角给出相同射程，因为 sin(2θ) = sin(180° − 2θ)。

8. The Cartesian Equation of the Trajectory | 轨迹的笛卡尔方程

To eliminate time t, use the horizontal component x = u cosθ t, so that t = x ⁄ (u cosθ). Substituting into y:

为消去时间 t，利用水平分量 x = u cosθ t，得 t = x ⁄ (u cosθ)。代入 y：

y = u sinθ (x ⁄ (u cosθ)) − ½g (x ⁄ (u cosθ))² = x tanθ − gx² ⁄ (2u² cos²θ)

This is a quadratic in x, confirming that the trajectory is a parabola. Exam questions sometimes ask you to derive this equation or to use it to check whether a projectile passes through a given point.

这是关于 x 的二次方程，证实轨迹为抛物线。考试有时要求推导此方程，或用它判断抛体是否经过给定点。

If the projectile starts from a height h above the ground, the equation becomes y = h + x tanθ − gx² ⁄ (2u² cos²θ). Be alert to such variations in exam questions.

若抛体从离地高度 h 处发射，方程为 y = h + x tanθ − gx² ⁄ (2u² cos²θ)。考试中要警惕此类变化。

9. Worked Example 1: Full Vector Analysis | 实例1：完整的向量分析

A particle is projected from the origin with initial speed 25 m/s at an angle of 40° above the horizontal. Taking g = 9.8 m/s², find:

一质点从原点以 25 m/s 的初速度、仰角 40° 被抛出。取 g = 9.8 m/s²，求：

(a) the position vector after 2 seconds;

(a) 2 秒后的位置向量；

r(2) = (25 cos40° × 2) i + (25 sin40° × 2 − ½ × 9.8 × 2²) j

= (50 × 0.7660) i + (50 × 0.6428 − 19.6) j = 38.3 i + 12.5 j m

(b) the velocity vector after 2 seconds;

(b) 2 秒后的速度向量；

v(2) = 25 cos40° i + (25 sin40° − 9.8 × 2) j

= 19.2 i + (16.1 − 19.6) j = 19.2 i − 3.5 j m/s

The negative j-component shows the particle is already descending after 2 seconds.

j 分量为负，说明 2 秒后质点已开始下降。

H = u² sin²θ ⁄ (2g) = 625 × sin²40° ⁄ (2 × 9.8) = 625 × 0.4132 ⁄ 19.6 ≈ 13.2 m

(d) the time of flight;

(d) 飞行时间；

T = 2u sinθ ⁄ g = 2 × 25 × 0.6428 ⁄ 9.8 ≈ 3.28 s

(e) the range.

(e) 射程。

R = u² sin(2θ) ⁄ g = 625 × sin80° ⁄ 9.8 = 625 × 0.9848 ⁄ 9.8 ≈ 62.8 m

10. Worked Example 2: Two Possible Angles | 实例2：两个可能的抛射角

A projectile is launched from ground level with speed 20 m/s and lands 30 m away on level ground. Find the two possible angles of projection.

一抛体从地面以 20 m/s 的初速度发射，落在水平地面上 30 m 处。求两个可能的抛射角。

Using the range formula:

利用射程公式：

30 = 20² × sin(2θ) ⁄ 9.8
sin(2θ) = 30 × 9.8 ⁄ 400 = 0.735

2θ = arcsin(0.735) = 47.3° or 132.7°
θ = 23.7° or 66.4°

Both angles are valid: the low, flat trajectory (23.7°) and the high, arched trajectory (66.4°) cover the same horizontal distance. This result reinforces the complementary-angle property of projectile motion.

两个角度均有效：低平轨迹（23.7°）和高抛弧线轨迹（66.4°）覆盖相同的水平距离。这一结果印证了抛体运动的互补角性质。

11. Common Exam Pitfalls and Tips | 常见考试陷阱与提示

Students frequently make the following mistakes. Avoid them to secure full marks.

以下是学生常犯的错误，避开它们方能拿到满分。

Pitfall \| 陷阱	How to Avoid It \| 如何避免
Setting v = 0 at maximum height — forgetting the horizontal component \| 在最高点令 v = 0 — 忘记水平分量	Only the j-component of velocity is zero; vₓ = u cosθ remains. \| 只有速度的 j 分量为零；vₓ = u cosθ 仍然存在。
Sign errors: writing +gt instead of −gt \| 符号错误：写成 +gt 而非 −gt	Consistently take upward as positive; gravity gives −gj. \| 始终取向上为正；重力提供 −gj。
Omitting the factor ½ in ½gt² \| 遗漏 ½gt² 中的因子 ½	Memorise the vector equation r = ut + ½at²; test with t = 1 s. \| 牢记向量方程 r = ut + ½at²；用 t = 1 s 检验。
Using the wrong trigonometric ratio \| 用错三角函数	Resolve along axes: horizontal uses cosθ, vertical uses sinθ. \| 沿轴分解：水平用 cosθ，竖直用 sinθ。
Ignoring nonzero initial position \| 忽略非零初始位置	Use r(t) = r₀ + ut + ½at² when launching from height. \| 从高处发射时使用 r(t) = r₀ + ut + ½at²。

Always draw a clear diagram, label the origin, axes, initial velocity vector, and the target point before beginning any calculation.

开始任何计算前，一定要画出清晰的示意图，标出原点、坐标轴、初速度向量和目标点。

12. Summary | 总结

Vector methods give a compact, powerful approach to projectile motion. The three key vectors are:

向量方法为抛体运动提供了简洁而有力的处理途径。三个关键向量为：

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Pitfall \| 陷阱	How to Avoid It \| 如何避免
Setting v = 0 at maximum height — forgetting the horizontal component \| 在最高点令 v = 0 — 忘记水平分量	Only the j-component of velocity is zero; vₓ = u cosθ remains. \| 只有速度的 j 分量为零；vₓ = u cosθ 仍然存在。
Sign errors: writing +gt instead of −gt \| 符号错误：写成 +gt 而非 −gt	Consistently take upward as positive; gravity gives −gj. \| 始终取向上为正；重力提供 −gj。
Omitting the factor ½ in ½gt² \| 遗漏 ½gt² 中的因子 ½	Memorise the vector equation r = ut + ½at²; test with t = 1 s. \| 牢记向量方程 r = ut + ½at²；用 t = 1 s 检验。
Using the wrong trigonometric ratio \| 用错三角函数	Resolve along axes: horizontal uses cosθ, vertical uses sinθ. \| 沿轴分解：水平用 cosθ，竖直用 sinθ。
Ignoring nonzero initial position \| 忽略非零初始位置	Use r(t) = r₀ + ut + ½at² when launching from height. \| 从高处发射时使用 r(t) = r₀ + ut + ½at²。

A-Level Mathematics: Solving Projectile Motion Using Vector Methods | A-Level 数学：向量方法求解抛体运动