A-Level Physics: Projectile Motion in Two Dimensions | A-Level 物理：二维运动中的抛体问题

📚 A-Level Physics: Projectile Motion in Two Dimensions | A-Level 物理：二维运动中的抛体问题

Projectile motion is one of the most important applications of two-dimensional kinematics. In A-Level Physics, you are expected to treat the motion of an object launched into the air as the superposition of two independent motions: uniform horizontal motion and uniformly accelerated vertical motion under gravity.

抛体运动是二维运动学最重要的应用之一。在 A-Level 物理中，你需要将物体抛入空中后的运动视为两个独立运动的叠加：水平方向的匀速直线运动和竖直方向受重力作用的匀加速直线运动。

1. Fundamental Assumptions | 基本假设

To analyse projectile motion at A-Level standard, we make several simplifying assumptions. First, air resistance is neglected, meaning the only force acting on the projectile is its weight. Second, the acceleration due to gravity is constant, with a magnitude of approximately 9.81 m s⁻² and directed vertically downward. Third, the Earth’s curvature and rotation are ignored over the scale of the motion.

在 A-Level 标准下分析抛体运动时，我们作若干简化假设。首先，忽略空气阻力，即物体仅受重力作用。其次，重力加速度恒定，大小约为 9.81 m s⁻²，方向竖直向下。第三，在运动尺度内忽略地球曲率和自转的影响。

These assumptions allow us to write the horizontal and vertical components of motion independently. In the horizontal direction, there is no acceleration, so the horizontal velocity remains constant. In the vertical direction, the acceleration is constant and equal to g, which means the vertical velocity changes uniformly with time.

这些假设使我们能够将水平方向和竖直方向的运动独立写出。在水平方向，没有加速度，因此水平速度保持不变。在竖直方向，加速度恒定且等于 g，因此竖直速度随时间均匀变化。

2. Resolving Initial Velocity | 初速度的分解

Consider a projectile launched with initial speed u at an angle θ above the horizontal. The initial velocity can be resolved into two perpendicular components using trigonometry:

考虑一个以初速度 u、与水平方向成 θ 角抛出的物体。利用三角函数，可以将初速度分解为两个互相垂直的分量：

uₓ = u cos θ

uᵧ = u sin θ

Here, uₓ is the horizontal component of the initial velocity, and uᵧ is the vertical component. The horizontal component remains constant throughout the motion because no horizontal force acts on the projectile. The vertical component changes linearly with time due to the constant acceleration g.

其中，uₓ 为初速度的水平分量，uᵧ 为初速度的竖直分量。由于物体在水平方向不受力，水平分量在整个运动过程中保持不变；而竖直分量因恒定加速度 g 而随时间线性变化。

It is essential to choose a consistent sign convention. A common choice is to take upward as positive and downward as negative. Under this convention, the vertical acceleration is aᵧ = −g. Some textbooks choose downward as positive, in which case aᵧ = +g. Always state your convention clearly in exam answers.

选择一致的符号约定至关重要。通常取向上为正、向下为负，此时竖直加速度 aᵧ = −g。部分教材取向下为正，此时 aᵧ = +g。在考试作答中务必明确说明你的约定。

3. Equations of Motion for Each Component | 各分量的运动方程

For the horizontal direction, since acceleration is zero, the displacement after time t is:

对于水平方向，由于加速度为零，经过时间 t 后的位移为：

x = uₓ t = (u cos θ) t

For the vertical direction, using the SUVAT equations with acceleration aᵧ = −g:

对于竖直方向，利用加速度 aᵧ = −g 的 SUVAT 方程组：

vᵧ = u sin θ − g t

y = (u sin θ) t − ½ g t²

vᵧ² = (u sin θ)² − 2 g y

These three vertical equations are the standard SUVAT equations adapted to projectile motion. The third equation is especially useful when time is not given and you need to find the vertical velocity at a specific height y.

以上三个竖直方向的方程是 SUVAT 方程组在抛体运动中的标准形式。第三个方程在时间未知、而需要求特定高度 y 处的竖直速度时特别有用。

4. The Trajectory Equation | 轨迹方程

By eliminating time between the horizontal and vertical equations, we can obtain the equation of the trajectory, which describes the path of the projectile in the x–y plane. From x = uₓ t, we have t = x / (u cos θ). Substituting this into the vertical displacement equation gives:

通过消去水平方程和竖直方程中的时间，可以得到轨迹方程，该方程描述抛体在 x–y 平面内的路径。由 x = uₓ t，得 t = x / (u cos θ)。将其代入竖直位移方程，得到：

y = x tan θ − (g x²) / (2 u² cos² θ)

This equation is quadratic in x, confirming that the trajectory is a parabola. The first term x tan θ gives the straight-line projection along the initial direction, while the second term represents the downward deviation caused by gravity.

该方程关于 x 是二次的，证实轨迹为抛物线。第一项 x tan θ 表示沿初速度方向的直线投影，第二项表示由重力引起的向下偏离。

In exam problems, the trajectory equation is useful when you are given a point on the path (x, y) and asked to verify whether the projectile passes through that point, or to find the initial speed u required to reach a specific target.

在考试题目中，当给出轨迹上某一点 (x, y)、要求判断物体是否经过该点，或求到达某一目标所需的初速度 u 时，轨迹方程非常有用。

5. Time of Flight | 飞行时间

The time of flight is the total time the projectile remains in the air. If the projectile lands at the same vertical level from which it was launched, we can set y = 0 in the vertical displacement equation:

飞行时间是指抛体在空中停留的总时间。如果抛体落回与出发点相同的高度，令竖直位移方程中的 y = 0：

0 = (u sin θ) T − ½ g T²

Factoring out T gives two solutions: T = 0 (the launch instant) and the non-zero solution:

提取 T 后得到两个解：T = 0（发射时刻）以及非零解：

T = (2 u sin θ) / g

This formula shows that the time of flight depends on the vertical component of the initial velocity and the gravitational acceleration. It does not depend on the horizontal component. If the projectile lands at a different height, you must solve the full quadratic equation for t.

该公式表明，飞行时间取决于初速度的竖直分量和重力加速度，而与水平分量无关。如果抛体落点高度不同，则必须求解完整的二次方程来得到 t。

6. Maximum Height | 最大高度

The maximum height is reached when the vertical velocity becomes zero, i.e., vᵧ = 0. Using the equation vᵧ² = (u sin θ)² − 2 g h, we set vᵧ = 0 and solve for h:

当竖直速度为零时，抛体达到最大高度，即 vᵧ = 0。利用方程 vᵧ² = (u sin θ)² − 2 g h，令 vᵧ = 0 并求解 h：

h_max = (u² sin² θ) / (2 g)

Alternatively, the time to reach maximum height is t = (u sin θ)/g, which is exactly half of the total time of flight for a projectile returning to the same height. Substituting this time into y = (u sin θ)t − ½ g t² gives the same result.

另一种方式：到达最大高度的时间为 t = (u sin θ)/g，这恰好是返回同一高度时总飞行时间的一半。将该时间代入 y = (u sin θ)t − ½ g t² 可得到相同结果。

The maximum height increases with the square of the initial speed and with the square of the sine of the launch angle. For a fixed initial speed, the maximum height is greatest when θ = 90°, i.e., vertical launch.

最大高度随初速度的平方和发射角正弦值的平方增大。对于固定的初速度，当 θ = 90° 即竖直上抛时，最大高度最大。

7. Horizontal Range | 水平射程

The horizontal range R is the horizontal distance travelled by the projectile before returning to its original launch height. Using x = uₓ T and substituting the time of flight T = (2 u sin θ)/g:

水平射程 R 是指抛体回到原发射高度前所经过的水平距离。利用 x = uₓ T，并代入飞行时间 T = (2 u sin θ)/g：

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

This compact result is extremely useful. It shows that the range depends on the product of the horizontal and vertical components of velocity, which is proportional to sin 2θ.

这个简洁结果非常实用。它表明射程取决于速度水平分量与竖直分量的乘积，该乘积正比于 sin 2θ。

For a fixed initial speed u, the range is maximum when sin 2θ = 1, which gives 2θ = 90°, or θ = 45°. This is a classic result: the maximum horizontal range is achieved at a launch angle of 45°.

对于固定的初速度 u，当 sin 2θ = 1 时射程最大，即 2θ = 90°，也就是 θ = 45°。这是经典结论：水平射程最大的发射角为 45°。

Furthermore, because sin 2θ = sin(180° − 2θ), two different launch angles θ and (90° − θ) produce the same range for the same initial speed. For example, angles of 30° and 60° give identical ranges, although their flight times and maximum heights differ.

此外，由于 sin 2θ = sin(180° − 2θ)，对于相同的初速度，两个不同的发射角 θ 和 (90° − θ) 会产生相同的射程。例如，30° 和 60° 的射程相同，但飞行时间和最大高度不同。

8. Projectile Launched from a Height | 从高处抛出的抛体

Many exam questions involve a projectile launched horizontally or at an angle from a cliff, building, or other elevated position. In such cases, the launch height h is not zero, and the final vertical displacement is negative relative to the launch point.

许多考试题目涉及从悬崖、建筑物或其他高处水平或倾斜抛出的物体。在这种情形下，发射高度 h 不为零，最终竖直位移相对于发射点为负值。

For example, a ball is kicked horizontally from a cliff of height H with initial speed u. The horizontal motion gives x = u t. The vertical motion starts with uᵧ = 0, so y = −½ g t². To find the time to reach the ground, set y = −H:

例如，一个球以初速度 u 从高度为 H 的悬崖水平踢出。水平运动给出 x = u t；竖直运动初速度 uᵧ = 0，所以 y = −½ g t²。求落地时间时令 y = −H：

−H = −½ g t² → t = √(2H/g)

Notice that this time is independent of the horizontal speed. A faster ball travels farther horizontally but takes the same time to fall. This is a direct consequence of the independence of horizontal and vertical motions.

注意，该时间与水平速度无关。球速越快，水平飞得越远，但下落所需时间相同。这是水平与竖直运动独立性的直接结果。

Quantity 物理量	Formula 公式	Key condition 关键条件
Horizontal displacement 水平位移	x = u cos θ · t	aₓ = 0
Vertical displacement 竖直位移	y = u sin θ · t − ½ g t²	aᵧ = −g
Time of flight 飞行时间	T = 2u sin θ / g	Returns to same height 回到同一高度
Maximum height 最大高度	h = u² sin² θ / (2g)	vᵧ = 0
Horizontal range 水平射程	R = u² sin 2θ / g	Lands at launch height 落回发射高度

9. Worked Example | 例题精讲

A projectile is launched from ground level with an initial speed of 20 m s⁻¹ at an angle of 35° above the horizontal. Calculate (a) the time of flight, (b) the maximum height, (c) the horizontal range. Take g = 9.81 m s⁻².

一个抛体从地面以 20 m s⁻¹ 的初速度、与水平方向成 35° 的仰角射出。计算 (a) 飞行时间，(b) 最大高度，(c) 水平射程。取 g = 9.81 m s⁻²。

First resolve the initial velocity:

首先分解初速度：

uₓ = 20 cos 35° = 16.38 m s⁻¹

uᵧ = 20 sin 35° = 11.47 m s⁻¹

(a) Using T = 2uᵧ / g:

(a) 利用 T = 2uᵧ / g：

T = (2 × 11.47) / 9.81 = 2.34 s

(b) Using h = uᵧ² / (2g):

(b) 利用 h = uᵧ² / (2g)：

h = 11.47² / (2 × 9.81) = 6.71 m

R = 16.38 × 2.34 = 38.3 m

This worked example illustrates the standard step-by-step method: resolve, then apply the appropriate equation for each quantity.

此例题展示了标准的分步解法：先分解速度，然后对每个待求量应用相应公式。

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

One frequent mistake is using the total speed instead of the vertical component in vertical motion equations. For example, in the equation vᵧ² = uᵧ² − 2g y, you must use uᵧ = u sin θ, not u. Another common error is forgetting that at the highest point, the vertical velocity is zero but the horizontal velocity is still u cos θ.

一个常见错误是在竖直运动方程中使用总速度而非竖直分量。例如，在方程 vᵧ² = uᵧ² − 2g y 中，必须使用 uᵧ = u sin θ，而不是 u。另一个常见错误是忘记在最高点竖直速度为零，但水平速度仍为 u cos θ。

Always draw a clear diagram showing the launch point, the trajectory, and the landing point. Label all known quantities and choose a coordinate system. Write down the sign convention explicitly. If air resistance is not mentioned, assume it is negligible.

务必画出清晰的示意图，标明发射点、轨迹和落点。标注所有已知量并选择坐标系。明确写出符号约定。若题目未提及空气阻力，则默认忽略。

When solving for time using the quadratic formula, there may be two positive solutions. Choose the physically meaningful root based on the context. For instance, if a ball is thrown upward from a cliff, the time to reach ground level is the larger positive root.

使用二次公式求时间时，可能有两个正解。应根据具体情况选择有物理意义的根。例如，从悬崖向上抛球，到达地面所需的时间应取较大的正根。

Finally, always check the units of your final answer. In the CIE exam, marks are often awarded for correct units as well as correct numerical values.

最后，务必检查最终答案的单位。在 CIE 考试中，正确的单位和正确的数值同样会被给分。

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A-Level Physics: Projectile Motion in Two Dimensions | A-Level 物理：二维运动中的抛体问题