A-Level Mathematics: Analysis of Projectile Motion at Arbitrary Angles | A-Level 数学：任意角度抛射运动分析

📚 A-Level Mathematics: Analysis of Projectile Motion at Arbitrary Angles | A-Level 数学：任意角度抛射运动分析

Projectile motion is one of the most elegant applications of kinematics in A-Level Mathematics. When a particle is launched at an arbitrary angle to the horizontal, its motion can be fully described by combining uniform horizontal velocity with uniformly accelerated vertical motion under gravity. This article provides a complete and exam-focused analysis of projectile motion at any angle of projection.

抛射运动是 A-Level 数学中运动学最优雅的应用之一。当质点以任意角度向水平方向抛出时，其运动可通过水平方向的匀速运动与重力作用下竖直方向匀加速运动的叠加来完整描述。本文将对任意投射角下的抛射运动进行完整且紧扣考点的分析。

1. Fundamental Assumptions and Kinematic Equations | 基本假设与运动学方程

We model the projectile as a particle moving in a vertical plane. We assume that air resistance is negligible, the only force acting is gravity (downward), and the gravitational acceleration g is constant, typically taken as g = 9.81 m s⁻². Under these assumptions, the horizontal acceleration is zero while the vertical acceleration is −g (taking upward as positive).

我们将抛射体建模为在竖直平面内运动的质点。假设空气阻力可忽略，唯一作用的力为重力（向下），重力加速度 g 为常数，通常取 g = 9.81 m s⁻²。在这些假设下，水平加速度为零，竖直加速度为 −g（取向上为正）。

The three constant-acceleration kinematic equations that govern projectile motion are:

支配抛射运动的三个匀加速运动学方程为：

v = u + at

s = ut + ½at²

v² = u² + 2as

Here, u is the initial velocity component, v is the final velocity component, a is the acceleration component, s is the displacement component, and t is the time elapsed.

其中 u 为初速度分量，v 为末速度分量，a 为加速度分量，s 为位移分量，t 为经过的时间。

2. Resolving the Initial Velocity | 初始速度的分解

Suppose a particle is projected from the origin with initial speed u at an angle θ above the horizontal. To analyse the motion, we resolve the initial velocity into two perpendicular components:

设质点从原点以初速度 u、仰角 θ（高于水平方向）抛出。为分析其运动，我们将初速度分解为两个互相垂直的分量：

Horizontal component: uₓ = u cos θ

Vertical component: uᵧ = u sin θ

The horizontal component uₓ remains constant throughout the flight because there is no horizontal acceleration. The vertical component uᵧ changes at a constant rate due to gravity.

水平分量 uₓ 在整个飞行过程中保持不变，因为没有水平加速度。竖直分量 uᵧ 由于重力作用以恒定速率变化。

3. Parametric Equations of Motion | 运动的参数方程

Taking the launch point as the origin, the horizontal displacement x and vertical displacement y at time t are given by the parametric equations:

以发射点为原点，时刻 t 的水平位移 x 和竖直位移 y 由以下参数方程给出：

x = (u cos θ)t

y = (u sin θ)t − ½gt²

The first equation arises from uniform horizontal motion, while the second comes from applying s = ut + ½at² vertically with a = −g. These two equations are the foundation for deriving every other result in projectile motion.

第一个方程来自水平方向的匀速运动，第二个方程来自竖直方向应用 s = ut + ½at²，其中 a = −g。这两个方程是推导抛射运动所有其他结果的基础。

4. Time of Flight | 飞行时间

The time of flight is the total time the projectile remains in the air before returning to the same vertical level as the launch point (assuming level ground). At landing, the vertical displacement y = 0. Setting y = 0 in the vertical equation:

飞行时间是指抛射体在落回与发射点同一水平高度之前在空中停留的总时间（假设地面水平）。落地时，竖直位移 y = 0。在竖直方程中令 y = 0：

0 = (u sin θ)T − ½gT²

Factoring out T gives T[(u sin θ) − ½gT] = 0. The solution T = 0 corresponds to the launch instant, and the non-zero solution gives the time of flight:

提取公因式 T 得 T[(u sin θ) − ½gT] = 0。解 T = 0 对应发射瞬间，非零解即为飞行时间：

T = 2u sin θ / g

Note that the time of flight depends on the vertical component of the initial velocity only — a projectile launched at a steeper angle stays airborne longer, provided the launch speed is the same.

注意飞行时间仅取决于初速度的竖直分量——在发射速度相同的情况下，投射角越陡，抛射体在空中停留的时间越长。

5. Maximum Height | 最大高度

At the highest point of its trajectory, the vertical velocity of the projectile becomes zero. Using v² = u² + 2as with v = 0, u = u sin θ, a = −g, and s = H (the maximum height):

在轨迹最高点，抛射体的竖直速度变为零。利用 v² = u² + 2as，其中 v = 0，u = u sin θ，a = −g，s = H（最大高度）：

0 = (u sin θ)² − 2gH

Rearranging gives the maximum height:

整理得最大高度：

H = u² sin² θ / (2g)

Alternatively, the time taken to reach the peak is T/2 = u sin θ / g, and substituting this into the vertical displacement equation yields the same result. The maximum height increases with sin² θ, so a vertical launch (θ = 90°) achieves the greatest height for a given speed.

另一种方法：到达最高点所需时间为 T/2 = u sin θ / g，将其代入竖直位移方程同样得到该结果。最大高度随 sin² θ 增大而增大，因此在给定速度下，垂直发射（θ = 90°）能达到最高高度。

6. Horizontal Range | 水平射程

The horizontal range R is the horizontal distance travelled by the projectile when it returns to the launch level. Since horizontal motion is uniform, R = (u cos θ) × T. Substituting the time of flight:

水平射程 R 是抛射体回到发射高度时在水平方向行进的距离。由于水平运动为匀速运动，R = (u cos θ) × T。代入飞行时间：

R = (u cos θ) × (2u sin θ / g) = 2u² sin θ cos θ / g

Using the double-angle identity sin 2θ = 2 sin θ cos θ, the range simplifies to:

利用二倍角公式 sin 2θ = 2 sin θ cos θ，射程化简为：

R = u² sin 2θ / g

This compact expression shows that the range depends on the launch speed and the angle through the term sin 2θ.

这一简洁表达式表明射程通过 sin 2θ 项依赖于发射速度和发射角。

7. Equation of Trajectory | 轨迹方程

By eliminating the parameter t from the parametric equations, we obtain the Cartesian equation of the trajectory. From x = (u cos θ)t, we have t = x / (u cos θ). Substituting into y = (u sin θ)t − ½gt²:

从参数方程中消去参数 t，得到轨迹的笛卡尔方程。由 x = (u cos θ)t，得 t = x / (u cos θ)。代入 y = (u sin θ)t − ½gt²：

y = x tan θ − gx² / (2u² cos² θ)

This is the equation of a downward-opening parabola. It allows us to determine the height y at any horizontal position x without knowing the time. This form is particularly useful when solving problems that specify a point the projectile must pass through.

这是开口向下的抛物线方程。它使我们无需知道时间就能确定任意水平位置 x 处的高度 y。当题目指定抛射体必须经过某一点时，这种形式尤为有用。

8. Maximum Range and Complementary Angles | 最大射程与互补角

For a fixed launch speed u, the range R = u² sin 2θ / g is maximised when sin 2θ = 1, i.e. when 2θ = 90°, giving θ = 45°. Thus, the maximum possible range on level ground is:

对于固定发射速度 u，当 sin 2θ = 1 时，即 2θ = 90°，θ = 45° 时，射程 R = u² sin 2θ / g 取得最大值。因此，水平地面上的最大射程为：

Rₘₐₓ = u² / g

Since sin 2θ = sin(180° − 2θ) = sin 2(90° − θ), two complementary angles θ and 90° − θ produce the same range. For example, launch angles of 30° and 60° give identical ranges, although the 60° launch has a longer time of flight and greater maximum height.

由于 sin 2θ = sin(180° − 2θ) = sin 2(90° − θ)，两个互补角 θ 和 90° − θ 产生相同的射程。例如，30° 和 60° 的发射角产生相同的射程，但 60° 发射的飞行时间更长、最大高度更高。

9. Worked Example | 典型例题

Problem: A particle is projected from ground level with a speed of 20 m s⁻¹ at an angle of 35° above the horizontal. Find: (a) the time of flight, (b) the maximum height, (c) the horizontal range, and (d) the velocity of the particle after 1.5 seconds.

例题：一质点以 20 m s⁻¹ 的速率、与水平方向成 35° 的仰角从地面发射。求：(a) 飞行时间；(b) 最大高度；(c) 水平射程；(d) 1.5 秒后质点的速度。

Solution (a): Using T = 2u sin θ / g with g = 9.81 m s⁻²:

解 (a)：利用 T = 2u sin θ / g，取 g = 9.81 m s⁻²：

T = 2 × 20 × sin 35° / 9.81 = 40 × 0.5736 / 9.81 ≈ 2.34 s

Solution (b): Using H = u² sin² θ / (2g):

解 (b)：利用 H = u² sin² θ / (2g)：

H = 400 × (0.5736)² / (2 × 9.81) = 400 × 0.3290 / 19.62 ≈ 6.71 m

Solution (c): Using R = u² sin 2θ / g:

解 (c)：利用 R = u² sin 2θ / g：

R = 400 × sin 70° / 9.81 = 400 × 0.9397 / 9.81 ≈ 38.3 m

Solution (d): The horizontal velocity component is constant: vₓ = 20 cos 35° = 16.38 m s⁻¹. The vertical component after t = 1.5 s is vᵧ = 20 sin 35° − 9.81 × 1.5 = 11.47 − 14.72 = −3.25 m s⁻¹. The magnitude is √(16.38² + 3.25²) ≈ 16.7 m s⁻¹, and the direction below the horizontal is tan⁻¹(3.25 / 16.38) ≈ 11.2°.

解 (d)：水平速度分量恒定：vₓ = 20 cos 35° = 16.38 m s⁻¹。t = 1.5 s 后竖直分量为 vᵧ = 20 sin 35° − 9.81 × 1.5 = 11.47 − 14.72 = −3.25 m s⁻¹。速度大小为 √(16.38² + 3.25²) ≈ 16.7 m s⁻¹，方向在水平线下方，与水平方向夹角为 tan⁻¹(3.25 / 16.38) ≈ 11.2°。

10. Projection from a Height | 从高处发射的抛射运动

When a projectile is launched from a point above ground level, such as the top of a cliff or building, the vertical displacement at landing is not zero but a known negative value. If the launch point is h metres above the landing level, we set y = −h in the vertical displacement equation:

当抛射体从高于地面的点发射时，例如悬崖或建筑物顶部，落地时的竖直位移并非零，而是一个已知的负值。若发射点比落地点高 h 米，则在竖直位移方程中令 y = −h：

−h = (u sin θ)t − ½gt²

This quadratic in t must be solved using the quadratic formula. Only the positive root is physically meaningful. Once the time of flight is found, the horizontal range is simply R = (u cos θ)t. Notice that in this situation, the angle for maximum range is no longer 45° — it shifts to a smaller angle because a flatter launch gives the projectile more time to travel horizontally before hitting the lower ground.

这是关于 t 的二次方程，必须用求根公式求解，只有正根具有物理意义。求得飞行时间后，水平射程即为 R = (u cos θ)t。注意在此情形下，最大射程对应的角度不再是 45°——它会向更小的角度偏移，因为更平坦的发射使抛射体在撞击较低地面之前有更多时间水平行进。

11. Velocity and Direction at Any Instant | 任意时刻的速度与方向

At any time t, the velocity components of the projectile are:

在任意时刻 t，抛射体的速度分量为：

vₓ = u cos θ

vᵧ = u sin θ − gt

The magnitude of the velocity is |v| = √(vₓ² + vᵧ²), and its direction relative to the horizontal is given by tan α = vᵧ / vₓ, where α is positive above the horizontal and negative below it. At the highest point, vᵧ = 0 and the velocity is purely horizontal.

速度大小为 |v| = √(vₓ² + vᵧ²)，其相对于水平方向的角度由 tan α = vᵧ / vₓ 给出，其中 α 在水平线以上为正，以下为负。在最高点，vᵧ = 0，速度纯为水平方向。

12. Common Exam Pitfalls | 考试常见易错点

Forgetting to resolve velocity: Always split the initial velocity into u cos θ and u sin θ before applying kinematic equations.

忘记分解速度：在应用运动学方程之前，务必先将初速度分解为 u cos θ 和 u sin θ。
Incorrect sign convention: Be consistent with the positive direction. If upward is positive, then g must appear as −9.81 in vertical equations.

正方向约定错误：保持正方向的一致性。若取向上为正，则竖直方程中 g 必须是 −9.81。
Confusing range and displacement: The range formula R = u² sin 2θ / g applies only when the projectile lands at the same level as it was launched.

混淆射程与位移：射程公式 R = u² sin 2θ / g 仅在抛射体落回与发射点同一水平高度时适用。
Using degrees instead of radians (or vice versa): Always check the angle mode on your calculator. A-Level papers typically specify the required mode.

角度制与弧度制混淆：始终检查计算器的角度模式。A-Level 试卷通常会指明所需的模式。
Neglecting the negative root: When solving quadratic equations for time, discard negative roots as they have no physical meaning.

忽略负根：在解关于时间的一元二次方程时，应舍弃负根，因为负根没有物理意义。

Mastering projectile motion at arbitrary angles requires a clear understanding of vector resolution, the independence of horizontal and vertical motion, and the ability to apply kinematic equations correctly. With consistent practice, these skills become second nature, enabling you to tackle both standard questions and extension problems with confidence.

掌握任意角度抛射运动需要清晰理解矢量分解、水平运动与竖直运动的独立性，以及正确应用运动学方程的能力。通过持续练习，这些技能将变得得心应手，使你能够自信应对标准题型和拓展问题。

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A-Level Mathematics: Analysis of Projectile Motion at Arbitrary Angles | A-Level 数学：任意角度抛射运动分析