Projectile Motion: Horizontal and Vertical Components | 抛体运动的水平与竖直分量

📚 Projectile Motion: Horizontal and Vertical Components | 抛体运动的水平与竖直分量

Projectile motion is one of the most elegant applications of Newtonian mechanics, and it all begins with a single, powerful idea: once an object is launched, its motion can be separated into two independent components — horizontal and vertical. This decomposition transforms a curved, two-dimensional path into two simple one-dimensional motion problems.

抛体运动是牛顿力学最优雅的应用之一，而这一切始于一个简单而有力的思想：物体一旦被抛出，其运动就可以分解为两个相互独立的分量——水平分量与竖直分量。这种分解将一条弯曲的二维轨迹转化为两个简单的一维运动问题。

1. The Principle of Independence | 运动的独立性原理

The horizontal and vertical motions of a projectile are independent of each other. This means that the force of gravity, which acts vertically downward, has no effect whatsoever on the horizontal velocity. Likewise, the horizontal motion does not influence the vertical acceleration.

抛体运动的水平与竖直运动是相互独立的。也就是说，作用于竖直向下的重力，对水平速度没有任何影响；同样地，水平运动也不会影响竖直加速度。

The only force we consider (ignoring air resistance) is gravity, which acts straight downward. Therefore:

在忽略空气阻力的情况下，我们所考虑的唯一天然力就是竖直向下的重力。因此：

Horizontal acceleration: aₓ = 0
Vertical acceleration: a_y = −g (taking upward as positive)

水平加速度：aₓ = 0
竖直加速度：a_y = −g （取向上为正方向）

aₓ = 0, a_y = −g

2. Setting Up the Equations | 建立基本方程

Suppose a projectile is launched with an initial speed u at an angle θ above the horizontal. The initial velocity components are:

设抛体以初速度 u、仰角 θ 射出。初速度分量为：

uₓ = u cos θ, u_y = u sin θ

Using the kinematic equations for constant acceleration, we can write the displacement components at time t:

利用匀变速直线运动方程，可写出 t 时刻的位移分量：

x = uₓ t = u cos θ · t

y = u_y t + ½ a_y t² = u sin θ · t − ½ g t²

These two equations tell us everything about the trajectory, time of flight, maximum height, and range.

这两个方程几乎告诉我们关于轨迹、飞行时间、最大高度和射程的全部信息。

3. The Trajectory Equation | 轨迹方程

By eliminating time t from the x and y equations, we obtain a relation between x and y that describes the parabolic path of the projectile.

从 x 和 y 方程中消去时间 t，即可得到描述抛体抛物线路径的 x 与 y 关系式。

From x = u cos θ · t, we have t = x / (u cos θ). Substituting into the y equation gives:

由 x = u cos θ · t，可得 t = x / (u cos θ)。代入 y 方程得：

y = x tan θ − (g x²) / (2 u² cos² θ)

This is a quadratic equation in x, confirming that the path is a parabola. The coefficient of x² is always negative, so the parabola opens downward.

这是关于 x 的二次方程，证实了抛体轨迹是一条抛物线。x² 的系数始终为负，因此抛物线开口向下。

4. Time of Flight | 飞行时间

The time of flight is the total time the projectile remains in the air. It is determined entirely by the vertical motion, since the projectile returns to its launch height when y = 0 (for level ground).

飞行时间是指抛体在空中停留的总时间。它完全由竖直运动决定，因为当 y = 0（水平地面）时，抛体回到抛出高度。

Setting y = 0 in the vertical equation:

令竖直方程中 y = 0：

0 = u sin θ · T − ½ g T²

This gives two solutions: T = 0 (the launch instant) and

这给出两个解：T = 0（抛出瞬间）以及

T = (2 u sin θ) / g

Notice that the time of flight depends only on the vertical component of the initial velocity and on g. A projectile launched straight up (θ = 90°) has no horizontal motion but has the longest time of flight for a given speed.

注意，飞行时间只取决于初速度的竖直分量和 g。竖直上抛（θ = 90°）没有水平运动，但在相同初速度下拥有最长的飞行时间。

5. Maximum Height | 最大高度

At the highest point of the trajectory, the vertical velocity becomes zero. The projectile momentarily stops rising and is about to fall. This instant occurs at half the time of flight (for level-ground launches), i.e., t = u sin θ / g.

在轨迹的最高点，竖直速度变为零。抛体瞬间停止上升，即将下落。这一时刻出现在飞行时间的一半处（对于水平地面发射），即 t = u sin θ / g。

Using the vertical displacement equation:

利用竖直位移方程：

H = (u² sin² θ) / (2g)

Key observations:

关键要点：

Maximum height depends only on the vertical component u_y = u sin θ.
For a fixed launch speed, the height is greatest when θ = 90°.
The horizontal component uₓ plays no role in determining H.

最大高度仅取决于竖直分量 u_y = u sin θ。
对于固定初速度，当 θ = 90° 时高度最大。
水平分量 uₓ 在决定 H 时不起任何作用。

6. Horizontal Range | 水平射程

The horizontal range R is the total horizontal distance travelled when the projectile returns to its launch height. Since the horizontal velocity is constant, R = uₓ · T.

水平射程 R 是抛体回到抛出高度时水平方向走过的总距离。由于水平速度恒定，R = uₓ · T。

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

This result is rich in implications:

这个结果富含深意：

For a given launch speed u, the range is maximised when sin 2θ = 1, i.e., 2θ = 90°, or θ = 45°.
Any two angles that sum to 90° (e.g., 30° and 60°) produce the same range, provided air resistance is negligible.
Range depends on both uₓ and u_y, unlike time of flight and maximum height.

对于给定初速度 u，当 sin 2θ = 1 时射程最大，即 2θ = 90°，也就是 θ = 45°。
任意两个和为 90° 的角度（如 30° 和 60°）产生相同射程，前提是空气阻力可忽略。
射程同时依赖 uₓ 和 u_y，这与飞行时间和最大高度不同。

7. Velocity at Any Instant | 任意时刻的速度

The velocity of a projectile at any time t has both horizontal and vertical components. The horizontal component remains constant throughout the motion, while the vertical component changes linearly with time.

抛体在任意时刻 t 的速度同时具有水平分量和竖直分量。水平分量在整个运动过程中保持不变，而竖直分量随时间线性变化。

vₓ = u cos θ

v_y = u sin θ − g t

The magnitude and direction of the velocity are given by:

速度的大小和方向由下式给出：

v = √(vₓ² + v_y²), tan φ = v_y / vₓ

where φ is the angle of the velocity vector below or above the horizontal. At the maximum height, v_y = 0, so the velocity is purely horizontal and its magnitude equals u cos θ.

其中 φ 是速度矢量与水平方向的夹角。在最大高度处，v_y = 0，因此速度是纯水平的，其大小等于 u cos θ。

8. Worked Example | 例题精讲

Problem: A ball is kicked from ground level with an initial speed of 20 m/s at an angle of 30° above the horizontal. Taking g = 10 m/s², find (a) the time of flight, (b) the maximum height, and (c) the horizontal range.

题目：一球从地面以 20 m/s 的初速度、30° 的仰角踢出。取 g = 10 m/s²，求 (a) 飞行时间、(b) 最大高度、(c) 水平射程。

Solution: First find the components of the initial velocity:

解：先求出初速度分量：

uₓ = 20 cos 30° = 20 × 0.866 = 17.32 m/s

u_y = 20 sin 30° = 20 × 0.5 = 10 m/s

(a) Time of flight:

(a) 飞行时间：

T = (2 u_y) / g = (2 × 10) / 10 = 2 s

(b) Maximum height:

(b) 最大高度：

H = u_y² / (2g) = 10² / (2 × 10) = 100 / 20 = 5 m

R = uₓ × T = 17.32 × 2 = 34.64 m

Alternatively, using the range formula: R = (u² sin 2θ)/g = (400 × sin 60°)/10 = (400 × 0.866)/10 = 34.64 m. Both methods agree.

或者使用射程公式：R = (u² sin 2θ)/g = (400 × sin 60°)/10 = (400 × 0.866)/10 = 34.64 m。两种方法结果一致。

9. Common Mistakes and Exam Tips | 常见错误与考试提示

Students frequently lose marks on projectile motion questions because of a few recurring errors. Knowing these pitfalls in advance is half the battle.

学生在抛体运动题目中经常因一些反复出现的错误而失分。提前了解这些陷阱，就赢了一半。

Forgetting that v_y = 0 at maximum height: This is the most useful piece of information for finding H, yet many students try to solve for H without using it.
Using uₓ in the vertical equation: Never mix components. The vertical motion uses u_y and g, not uₓ.
Sign conventions: Be consistent. If upward is positive, then g = −9.8 m/s², and v_y decreases with time.
Assuming symmetric flight: Time up equals time down only when the projectile lands at the same height it was launched from.
Neglecting the angle of the final velocity: A projectile lands with the same speed it was launched with (for level ground), but its direction is the mirror image below the horizontal.

忘记最高点 v_y = 0：这是求 H 最有利的信息，但许多学生没有利用它去解 H。
在竖直方程中误用 uₓ：切勿混合分量。竖直运动只使用 u_y 和 g，与 uₓ 无关。
符号约定：务必一致。若向上为正，则 g = −9.8 m/s²，v_y 随时间减小。
假设对称飞行：上升时间等于下落时间，仅在抛体落回与抛出高度相同的水平面时才成立。
忽略末速度的方向：对于水平地面，抛体落地速率与初速率相同，但方向是水平线以下的镜像。

10. Summary Table | 公式速查表

The table below consolidates all the essential equations for projectile motion on level ground.

下表汇总了水平地面上抛体运动的所有关键公式。

Quantity (物理量)	Formula (公式)
Horizontal displacement (水平位移)	x = u cos θ · t
Vertical displacement (竖直位移)	y = u sin θ · t − ½ g t²
Time of flight (飞行时间)	T = (2 u sin θ) / g
Maximum height (最大高度)	H = (u² sin² θ) / (2g)
Horizontal range (水平射程)	R = (u² sin 2θ) / g
Trajectory (轨迹方程)	y = x tan θ − (g x²) / (2 u² cos² θ)

Mastering the horizontal and vertical decomposition of projectile motion is not just about memorising formulas — it is about understanding that the two directions are governed by independent dynamics, connected only through time. This insight will carry you through every projectile problem, from the simplest kick to a challenging examination question.

掌握抛体运动的水平与竖直分解，不仅仅是为了记住公式——更关键的是理解两个方向由独立的动力学支配，仅通过时间相连。这一洞察将带你攻克每一道抛体题目，从最简单的踢球到颇具挑战的考试难题。

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Projectile Motion: Horizontal and Vertical Components | 抛体运动的水平与竖直分量