Horizontal Projection | 水平抛体运动

📚 Horizontal Projection | 水平抛体运动

In A-Level Mechanics, horizontal projection is a core projectile motion model. An object is given an initial velocity in the horizontal direction only, with no initial vertical velocity, and then moves freely under gravity. This model appears frequently in Edexcel Mechanics papers, usually involving a particle launched from a height or from the edge of a cliff.

在 A-Level 力学中，水平抛体运动是抛体运动的核心模型。物体只获得水平方向的初速度，竖直方向初速度为零，然后在重力作用下自由运动。该模型在 Edexcel 力学考试中经常出现，通常涉及从高处或悬崖边缘抛出的质点。

1. Definition of Horizontal Projection | 水平抛体运动的定义

Horizontal projection means an object is launched with an initial velocity that is entirely horizontal. The vertical component of the initial velocity is zero, so the object begins to fall immediately under gravity while continuing to move forward at a constant horizontal speed.

水平抛体运动是指物体以完全水平的初速度抛出。初速度的竖直分量为零，因此物体在重力作用下立即开始下落，同时以恒定的水平速度继续向前运动。

Typical examples include a ball rolling off a table, a stone thrown horizontally from a cliff, or a dart fired horizontally from a blowpipe. In all these cases, the path is a half-parabola from the launch point to the ground.

典型例子包括从桌面滚下的小球、从悬崖水平抛出的石块，或从吹管水平射出的飞镖。在所有这些情况中，轨迹是从抛出点到地面的半抛物线。

2. Modelling Assumptions | 建模假设

We model the object as a particle, so its size and rotation are ignored. Air resistance is usually neglected unless the question states otherwise. The only acceleration acting on the object is the acceleration due to gravity, g = 9.8 m s⁻², directed vertically downwards.

我们将物体建模为质点，因此其大小和旋转忽略不计。除非题目另有说明，通常忽略空气阻力。作用在物体上的唯一加速度是重力加速度 g = 9.8 m s⁻²，方向竖直向下。

For horizontal projection questions, it is convenient to choose the positive y-direction as downwards. This makes the vertical acceleration positive, so the vertical displacement formula becomes y = ½ g t² without negative signs. If you choose upwards as positive, you must include a negative sign in front of g.

对于水平抛体问题，通常选择向下为正方向。这样竖直加速度为正，竖直位移公式变为 y = ½ g t²，不会出现负号。如果选择向上为正方向，则必须在 g 前加上负号。

3. Horizontal Motion: Constant Velocity | 水平运动：匀速直线运动

Since there is no horizontal force, the horizontal acceleration is zero. Therefore the horizontal velocity remains constant throughout the motion. The horizontal displacement after time t is given by:

由于没有水平力，水平加速度为零。因此水平速度在整个运动过程中保持不变。经过时间 t 后的水平位移由下式给出：

x = u t

Here u is the initial horizontal speed and x is the horizontal distance travelled. This equation is independent of the vertical motion, so you can use it directly once the time of flight is known.

其中 u 是初始水平速度，x 是水平方向经过的距离。该方程与竖直运动无关，因此只要知道飞行时间，就可以直接使用它。

4. Vertical Motion: Constant Acceleration | 竖直运动：匀加速运动

The vertical motion starts from rest and accelerates downwards at g. The vertical velocity at time t is:

竖直运动从静止开始，以加速度 g 向下加速。时间 t 时的竖直速度为：

v = g t

The vertical displacement after time t is:

时间 t 后的竖直位移为：

y = ½ g t²

These two equations are exactly the SUVAT equations for an object released from rest under gravity, with initial vertical velocity 0 and acceleration g downwards.

这两个方程正是物体在重力作用下从静止释放的 SUVAT 方程，其中竖直初速度为 0，向下加速度为 g。

5. Displacement at Time t | 任意时刻 t 的位移

At any time t, the object has two displacement components. The horizontal displacement is x = u t, and the vertical displacement is y = ½ g t². The position vector from the launch point is therefore:

在任意时刻 t，物体有两个位移分量。水平位移为 x = u t，竖直位移为 y = ½ g t²。因此从抛出点出发的位置矢量为：

Position = (u t, ½ g t²)

This pair of parametric equations describes a parabola. Eliminating t between x = u t and y = ½ g t² gives the trajectory equation:

这对参数方程描述了一条抛物线。消去 x = u t 和 y = ½ g t² 中的 t，可得轨迹方程：

y = (g / 2u²) x²

This is useful for showing that the path is parabolic and for solving problems where a projectile must clear an obstacle of a given height at a given horizontal distance.

该方程可用于证明轨迹是抛物线，也可用于解决抛体必须越过给定水平距离处给定高度障碍物的问题。

6. Velocity at Time t | 任意时刻 t 的速度

The horizontal component of velocity remains u, and the vertical component is v = g t. The resultant speed is the magnitude of the velocity vector:

速度的水平分量保持为 u，竖直分量为 v = g t。合速度是速度矢量的大小：

Speed = √(u² + v²) = √(u² + g² t²)

The direction of the velocity is measured from the horizontal. The angle θ below the horizontal satisfies:

速度方向以水平方向为基准测量。与水平方向向下形成的角度 θ 满足：

tan θ = v / u = g t / u

In Edexcel questions, you may be asked for the speed and angle just before hitting the ground, so you must combine the horizontal and vertical components using Pythagoras and trigonometry.

在 Edexcel 考试题中，可能会要求计算落地前瞬间的速度大小和方向，因此必须使用勾股定理和三角函数合成水平与竖直分量。

7. Time of Flight | 飞行时间

The time of flight is found from the vertical motion. If the object falls through a vertical height h, then:

飞行时间由竖直运动求出。如果物体下落了竖直高度 h，则：

h = ½ g t²

Solving for t gives:

解出 t 得：

t = √(2h / g)

Notice that the time of flight depends only on the vertical height and the acceleration due to gravity. It does not depend on the horizontal speed u. This is a key idea often tested in multiple-choice and structured questions.

注意飞行时间只取决于竖直高度和重力加速度，与水平速度 u 无关。这是选择题和结构化题目中经常考查的关键概念。

8. Horizontal Range | 水平射程

The range is the horizontal distance travelled from launch to landing. Since horizontal velocity is constant, the range R is:

射程是从抛出点到落地点的水平距离。由于水平速度恒定，射程 R 为：

R = u t

Substituting the time of flight t = √(2h / g) gives:

代入飞行时间 t = √(2h / g)，得：

R = u √(2h / g)

This shows that doubling the horizontal speed doubles the range, while doubling the height increases the range by a factor of √2, not 2, because time of flight depends on the square root of height.

这表明水平速度加倍会使射程加倍，而高度加倍只会使射程变为原来的 √2 倍，而不是 2 倍，因为飞行时间与高度的平方根有关。

9. Worked Example | 例题解析

A ball is projected horizontally at 6 m s⁻¹ from a cliff 20 m high. Take g = 9.8 m s⁻². Find the time of flight, the horizontal range, and the speed just before the ball hits the ground.

一小球以 6 m s⁻¹ 的水平速度从 20 m 高的悬崖上抛出。取 g = 9.8 m s⁻²。求飞行时间、水平射程以及小球落地前瞬间的速度大小。

Step 1: Use vertical motion to find time.

步骤 1：利用竖直运动求时间。

20 = ½ × 9.8 × t²

t² = 20 / 4.9 ≈ 4.0816

t ≈ 2.02 s

Step 2: Find the range using x = u t.

步骤 2：利用 x = u t 求射程。

x = 6 × 2.02 ≈ 12.1 m

Step 3: Find the vertical velocity at impact.

步骤 3：求落地瞬间的竖直速度。

v = g t = 9.8 × 2.02 ≈ 19.8 m s⁻¹

Step 4: Combine components for resultant speed.

步骤 4：合成两个分量求合速度。

Speed = √(6² + 19.8²) = √(36 + 392.04) = √428.04 ≈ 20.7 m s⁻¹

The direction is about 73.1° below the horizontal, found from tan θ = 19.8 / 6.

速度方向约为水平方向向下 73.1°，由 tan θ = 19.8 / 6 求得。

10. Common Misconceptions | 常见误区

One common mistake is to think that the horizontal velocity increases because the object speeds up overall. In fact, the horizontal velocity remains constant, and only the vertical velocity increases. The overall speed increases because of the growing vertical component.

一个常见误区是认为水平速度会增大，因为物体总体上在加速。实际上，水平速度保持不变，只有竖直速度在增大。合速度之所以增大，是因为竖直分量不断增加。

Another mistake is using the resultant speed as the horizontal velocity in x = u t. The horizontal distance equation must use the horizontal component only. Similarly, do not mix horizontal and vertical quantities in the same SUVAT equation.

另一个误区是在 x = u t 中把合速度当作水平速度使用。水平距离方程必须只使用水平分量。同样，不要在同一个 SUVAT 方程中混用水平量和竖直量。

Finally, many students forget to state the direction of the final velocity when a question asks for velocity rather than speed. Velocity is a vector, so you must give both magnitude and direction.

最后，很多学生在题目要求速度而不是速率时忘记说明末速度的方向。速度是矢量，因此必须同时给出大小和方向。

11. Summary Table | 公式总结表

Quantity 物理量	Formula 公式
Horizontal displacement 水平位移	x = u t
Vertical displacement 竖直位移	y = ½ g t²
Vertical velocity 竖直速度	v = g t
Time of flight 飞行时间	t = √(2h / g)
Range 射程	R = u √(2h / g)
Resultant speed 合速度大小	√(u² + g² t²)
Direction 方向	tan θ = g t / u

This table assumes downward is positive and air resistance is negligible. Always check the sign convention required by the problem before applying these formulas.

该表假设向下为正方向且空气阻力可忽略。在应用这些公式前，请务必检查题目所要求的正方向规定。

12. Exam Tips and Final Check | 考试技巧与最后检查

In Edexcel Mechanics questions, start by drawing a diagram and clearly marking the initial velocity, the height, and the positive x and y directions. Then separate the motion into horizontal and vertical components before using SUVAT equations.

在 Edexcel 力学题中，首先画出示意图，清楚地标出初速度、高度以及 x 和 y 正方向。然后将运动分解为水平和竖直两个分量，再使用 SUVAT 方程。

Check that your final answer has the correct units: time in seconds, distances in metres, speeds in m s⁻¹, and angles in degrees unless radians are specified by the question. Always state the direction of any vector quantity.

检查最终答案的单位是否正确：时间用秒，距离用米，速度用 m s⁻¹，角度用度，除非题目指定弧度。对于任何矢量量，始终说明其方向。

If a question gives the horizontal distance and asks for the height, use x = u t to find t first, then substitute into y = ½ g t². If it gives the height and asks for the horizontal distance, find t from the vertical motion first, then use x = u t.

如果题目给出水平距离并要求高度，先用 x = u t 求出 t，再代入 y = ½ g t²。如果给出高度并要求水平距离，则先从竖直运动求出 t，再代入 x = u t。

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Horizontal Projection | 水平抛体运动