Projectile Motion Essentials | 抛体运动要点解析

📚 Projectile Motion Essentials | 抛体运动要点解析

Projectile motion is the two-dimensional motion of an object launched into the air and continuing to move under the influence of gravity alone. In CIE A-Level Physics, this topic tests your ability to resolve vectors, apply the constant-acceleration equations independently to horizontal and vertical motion, and interpret the resulting parabolic trajectory.

抛体运动是物体被抛出后，仅在重力作用下进行的二维运动。在CIE A-Level物理中，该主题考查你分解矢量、将匀变速运动学公式分别应用于水平与垂直方向运动，以及理解所得抛物线轨迹的能力。

1. What Is Projectile Motion? | 什么是抛体运动

A projectile is any object given an initial velocity and then allowed to move freely under gravity. After the launch force is removed, the only force considered is the weight of the object, which produces a constant downward acceleration g = 9.81 m s⁻². If air resistance is ignored, the path followed by a projectile is a parabola.

抛体是指被赋予初速度后，在重力作用下自由运动的任何物体。一旦发射力消失，唯一考虑的力就是物体自身的重力，它产生恒定的向下加速度 g = 9.81 m s⁻²。若忽略空气阻力，抛体所经过的路径是一条抛物线。

This is why projectile motion is not a new set of equations but a natural application of the uniform acceleration equations, also called SUVAT equations, to two perpendicular directions.

因此抛体运动并不是一套全新的公式，而是将匀变速运动学方程（也称为SUVAT方程）应用于两个相互垂直方向的结果。

2. Assumptions and Idealisation | 假设与理想化

In A-Level projectile problems, several simplifying assumptions are made unless the question states otherwise:

在A-Level抛体问题中，除非题目特别说明，通常需要作以下简化假设：

Air resistance is negligible, so the horizontal acceleration is zero and the vertical acceleration is constant.

空气阻力可以忽略，因此水平方向加速度为零，垂直方向加速度恒定。
Gravity is uniform over the flight path, so g = 9.81 m s⁻² everywhere during the motion.

重力在运动全程处处均匀，因此运动过程中 g = 9.81 m s⁻² 保持恒定。
The Earth is treated as flat over the range of the motion, so the horizontal ground is a straight line.

在运动范围内地球可视为平面，因此水平地面是一条直线。
Rotation and curvature of the Earth are not considered.

不考虑地球的自转与曲率。

These assumptions allow us to use the kinematic equations safely. In a real situation with air resistance, the path would be non-parabolic and the range would be reduced.

这些假设使我们可以安全地使用运动学公式。在真实有空气阻力的情况下，轨迹不再是抛物线，射程也会减小。

3. Independence of Horizontal and Vertical Motion | 水平与垂直运动的独立性

The most important idea in projectile motion is that horizontal and vertical motion are independent. Newton’s second law tells us that forces in perpendicular directions do not affect each other. Gravity acts only vertically, so the horizontal velocity remains unchanged throughout the flight, while the vertical velocity changes uniformly with acceleration g.

抛体运动最重要的思想是：水平运动与垂直运动相互独立。牛顿第二定律告诉我们，垂直方向的力不会影响水平方向的运动。重力只沿垂直方向作用，因此水平速度在整个飞行过程中保持不变，而垂直速度则以加速度 g 均匀变化。

However, the two directions are linked by a common quantity: time. The same time interval t applies to both horizontal and vertical motion. This shared time is the key to solving most projectile problems.

但这两个方向由一个共同的物理量联系：时间。同一时间 t 同时适用于水平运动和垂直运动。这个共用的时间就是解决大多数抛体问题的关键。

4. Resolving the Initial Velocity | 分解初速度

Consider a projectile launched from ground level with initial speed v₀ at an angle θ above the horizontal. The initial velocity must be resolved into components:

考虑一个从地面以初速度 v₀、与水平方向夹角 θ 向上抛出的物体。初速度需分解为两个分量：

v₀ₓ = v₀ cos θ

v₀ᵧ = v₀ sin θ

Here v₀ₓ is the horizontal component and v₀ᵧ is the initial vertical component. Always check which angle is given: if the angle is to the vertical, you must swap the sine and cosine functions.

其中 v₀ₓ 是水平分量，v₀ᵧ 是垂直初速度分量。请务必注意题目给出的是与水平方向的夹角，还是与竖直方向的夹角；如果是与竖直方向的夹角，需要交换正弦与余弦。

5. Kinematic Equations for Each Component | 两个方向的运动学公式

Using the sign convention that upward is positive, the vertical acceleration is a = -g. The motion in each direction is then:

取向上为正方向，垂直方向的加速度为 a = -g。则两个方向的运动方程为：

Horizontal: x = v₀ cos θ × t

Vertical: vᵧ = v₀ sin θ – g t

Vertical: y = v₀ sin θ × t – ½ g t²

The horizontal equation is simple because there is no horizontal acceleration. The vertical equation is exactly the SUVAT equation s = ut + ½at² with u = v₀ sin θ and a = -g.

水平方向方程非常简单，因为没有水平加速度。垂直方向方程实际上就是 s = ut + ½at²，其中 u = v₀ sin θ，a = -g。

6. Trajectory of a Projectile | 抛体的轨迹方程

To find the equation of the path, eliminate time from the horizontal and vertical equations. From x = v₀ cos θ × t, we get t = x / (v₀ cos θ). Substituting this into the vertical displacement equation gives:

为求出轨迹方程，需要从水平和垂直方程中消去时间 t。由 x = v₀ cos θ × t 得 t = x / (v₀ cos θ)，将其代入垂直位移方程，得到：

y = x tan θ – (g x²) / (2 v₀² cos² θ)

This is a quadratic expression in x, so the trajectory is parabolic. The first term x tan θ describes the straight line the projectile would follow if gravity were absent; the second term bends the path downward due to gravity.

这是关于 x 的二次表达式，因此轨迹是抛物线。第一项 x tan θ 表示没有重力时物体将沿直线运动；第二项则因重力使路径向下弯曲。

7. Time of Flight | 飞行时间

For a projectile launched from and returning to the same height, the total flight time T is found by setting y = 0 in the vertical displacement equation:

对于从同一高度抛出并落回同一高度的抛体，令垂直位移方程中 y = 0，即可求出总飞行时间 T：

0 = v₀ sin θ × T – ½ g T²

Factorising gives T = 0 or T = (2 v₀ sin θ) / g. The T = 0 solution corresponds to the launch instant, so the flight time is:

因式分解得 T = 0 或 T = (2 v₀ sin θ) / g。其中 T = 0 对应发射瞬间，所以飞行时间为：

T = (2 v₀ sin θ) / g

This result shows that flight time depends only on the vertical component of the initial velocity, not on the horizontal component.

可见飞行时间只取决于初速度的垂直分量，与水平分量无关。

8. Maximum Height | 最大高度

At the highest point of the trajectory, the vertical velocity becomes zero. Using vᵧ² = (v₀ sin θ)² – 2g y and setting vᵧ = 0 gives:

在轨迹最高点，垂直速度变为零。利用 vᵧ² = (v₀ sin θ)² – 2g y，并令 vᵧ = 0，可得：

H = (v₀ sin θ)² / (2g)

Alternatively, the time to reach maximum height is (v₀ sin θ)/g. Because vertical motion is symmetric, the projectile takes exactly half of the flight time to go up and half to come down.

也可以先求到达最高点所需的时间为 (v₀ sin θ)/g。由于垂直运动具有对称性，抛体上升所需时间正好是飞行时间的一半，下降所需时间也是另一半。

9. Horizontal Range | 水平射程

For a projectile returning to its launch height, the horizontal range R is the horizontal displacement at the total flight time T:

对于落回发射高度的抛体，水平射程 R 就是总飞行时间 T 内的水平位移：

R = (v₀ cos θ) × T = (v₀ cos θ) × (2 v₀ sin θ) / g

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

利用二倍角公式 2 sin θ cos θ = sin 2θ，可化简为：

R = (v₀² sin 2θ) / g

This compact formula is very useful in multiple-choice questions, but for full working you should still be prepared to derive it from the component equations.

这个简洁公式在选择题中非常有用，但如需写出完整过程，你仍应会从分量方程推导该式。

10. Angle for Maximum Range | 最大水平射程对应的角度

For a fixed initial speed, the range formula R = (v₀² sin 2θ)/g is maximised when sin 2θ is maximised. Since the maximum value of sin 2θ is 1, we need 2θ = 90°, so:

当初速度大小固定时，要使射程公式 R = (v₀² sin 2θ)/g 最大，需要 sin 2θ 取最大值。因为 sin 2θ 的最大值为 1，所以令 2θ = 90°，即：

θ = 45°

Another important point is symmetry: angles θ and 90° – θ give the same range. For example, 30° and 60° give equal ranges for the same initial speed, although the flight times and maximum heights are different.

另一个重要现象是角度对称性：θ 与 90° – θ 的角度会得到相同的射程。例如在相同初速度下，30° 和 60° 的射程相等，但飞行时间和最大高度不同。

Launch Angle θ	sin 2θ	Relative Range (v₀²/g = 1)
15°	sin 30° = 0.500	0.500
30°	sin 60° = 0.866	0.866
45°	sin 90° = 1.000	1.000
60°	sin 120° = 0.866	0.866
75°	sin 150° = 0.500	0.500

11. Worked Example | 示例分析

Question: A ball is launched from ground level with a speed of 20 m s⁻¹ at an angle of 30° to the horizontal. Taking g = 9.81 m s⁻², find the time of flight, the maximum height reached, and the horizontal range.

例题：一个小球从地面以 20 m s⁻¹ 的速度、与水平方向成 30° 的角度抛出。取 g = 9.81 m s⁻²，求飞行时间、最大高度和水平射程。

Resolve the initial velocity. The vertical component is v₀ sin 30° = 20 × 0.5 = 10 m s⁻¹, and the horizontal component is v₀ cos 30° = 20 × 0.866 = 17.32 m s⁻¹.

先分解初速度。垂直分量为 v₀ sin 30° = 20 × 0.5 = 10 m s⁻¹，水平分量为 v₀ cos 30° = 20 × 0.866 = 17.32 m s⁻¹。

For the time of flight, use T = (2 v₀ sin θ)/g:

求飞行时间，使用 T = (2 v₀ sin θ)/g：

T = (2 × 10) / 9.81 = 2.04 s

For the maximum height, use H = (v₀ sin θ)²/(2g):

求最大高度，使用 H = (v₀ sin θ)²/(2g)：

H = 10² / (2 × 9.81) = 5.10 m

For the range, use R = (v₀² sin 2θ)/g:

求水平射程，使用 R = (v₀² sin 2θ)/g：

R = (20² × sin 60°) / 9.81 = (400 × 0.866) / 9.81 = 35.3 m

You could also find the range by multiplying the horizontal component by the flight time: 17.32 × 2.04 = 35.3 m, confirming the result.

也可以将水平分量乘以飞行时间：17.32 × 2.04 = 35.3 m，结果一致。

12. Common Mistakes and Exam Tips | 常见错误与考试技巧

The following points are frequently tested by CIE examiners and are common sources of lost marks:

以下是CIE考官经常考查、也是学生容易丢分的地方：

Forgetting to decompose the initial velocity: always split v₀ into v₀ cos θ and v₀ sin θ before applying the equations.

忘记分解初速度：应用公式前，务必把 v₀ 分解为 v₀ cos θ 和 v₀ sin θ。
Using the full initial speed in the vertical equation: the vertical equation requires only the vertical component v₀ sin θ, not v₀ itself.

在垂直方向方程中直接使用初速度大小：垂直方程应使用垂直分量 v₀ sin θ，而不是 v₀ 本身。
Ignoring the t = 0 solution when solving for the time of flight: remember that t = 0 is the launch moment, so choose the non-zero root.

求解飞行时间时漏掉 t = 0 这个解：t = 0 对应发射瞬间，应选择不为零的根。
Inconsistent sign convention: if upward is positive, then acceleration is -9.81 m s⁻². Do not mix upward velocities with a positive g in the same equation.

正方向选择不一致：若向上为正，则加速度应为 -9.81 m s⁻²。不要在同一个方程中把向上速度与正的 g 混用。
Assuming a projectile lands at the same height: if it is launched from a cliff or lands on higher ground, the simplified range formula cannot be used directly. Instead, write the vertical displacement equation and solve for time first.

默认抛体落回同一高度：如果从悬崖边抛出或落在更高处，简化射程公式不能直接使用。此时应先写出垂直位移方程，解出时间。
Remembering the default assumption: unless air resistance is explicitly mentioned, CIE questions expect you to ignore it and treat the path as parabolic.

记住默认假设：除非题目明确提到空气阻力，CIE题目默认忽略阻力，并将轨迹视为抛物线。

In exam answers, show your resolved components clearly, state the sign convention, and write down the equation before substituting numbers. This structure helps you earn method marks even if a final arithmetic error is made.

考试作答时应清楚写出分解后的分量、说明正方向，并在代入数值前先写出方程。这样的步骤化书写即使最后计算有误，也能获得方法分。

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Projectile Motion Essentials | 抛体运动要点解析