Projection at Any Angle | 任意角度斜抛运动

📚 Projection at Any Angle | 任意角度斜抛运动

In Edexcel A Level Mechanics, projectile motion describes a particle moving freely under gravity after being projected. When the initial velocity makes any angle θ with the horizontal, the most important step is to split the motion into independent horizontal and vertical components.

在 Edexcel A Level 力学中，抛体运动描述物体被抛出后在重力作用下的自由运动。当初速度与水平方向成任意角度 θ 时，最关键的一步是把运动分解为互相独立的水平分量和竖直分量。

1. Setting up the Model | 建立模型

A projectile is modelled as a particle moving in a vertical plane. Air resistance is ignored, so the only force acting after projection is gravity, which gives a constant downward acceleration of g = 9.8 m s⁻².

抛体被建模为在竖直平面内运动的质点。忽略空气阻力，因此抛出后唯一作用的力是重力，它产生恒定向下的加速度 g = 9.8 m s⁻²。

Choose a coordinate system with the positive x-direction horizontal and the positive y-direction vertically upward. The launch point is usually taken as the origin, and the initial speed is denoted by u, with the launch angle θ measured above the horizontal.

选择坐标系，使 x 轴正方向水平，y 轴正方向竖直向上。通常把抛出点作为原点，初速度大小记为 u，发射角 θ 从水平方向向上测量。

aₕ = 0, aᵥ = −g

2. Resolving the Initial Velocity | 分解初速度

The initial velocity vector must be resolved into horizontal and vertical components. For a launch speed u at angle θ to the horizontal, these components are found using trigonometry.

初速度矢量必须分解为水平分量和竖直分量。对于以速度 u 与水平方向成 θ 角抛出，这些分量可以用三角函数求出。

uₕ = u cos θ, uᵥ = u sin θ

Here uₕ is the constant horizontal component of velocity, and uᵥ is the initial vertical component of velocity. The angle θ is usually given between 0° and 90°, but the same component method also works for projection below the horizontal if θ is treated as negative.

这里 uₕ 是速度的水平分量，uᵥ 是初速度的竖直分量。角度 θ 通常介于 0° 到 90° 之间，但如果物体向水平面下方抛出，可以把 θ 视为负角，仍使用相同的分解方法。

3. Horizontal Motion | 水平运动

Because air resistance is ignored, there is no horizontal acceleration. The horizontal velocity remains constant throughout the flight, so the horizontal displacement after time t is simply speed multiplied by time.

由于忽略空气阻力，水平方向没有加速度。整个飞行过程中水平速度保持不变，因此经过时间 t 后的水平位移就是速度乘以时间。

x = u cos θ × t

This equation is very useful for linking horizontal distance and time. It shows that the horizontal motion is uniform, no matter what happens in the vertical direction.

这个公式在联系水平距离和时间时非常有用。它说明无论竖直方向如何变化，水平运动都是匀速的。

4. Vertical Motion | 竖直运动

The vertical motion has constant acceleration −g, so the standard constant-acceleration equations can be applied with displacement y, initial velocity u sin θ, and acceleration −g.

竖直运动具有恒定的加速度 −g，因此可以应用标准的匀加速运动公式，其中位移为 y，初速度为 u sin θ，加速度为 −g。

vᵥ = u sin θ − gt
y = u sin θ × t − ½ gt²
vᵥ² = (u sin θ)² − 2gy

The sign of y depends on the chosen positive direction. If upward is positive, a projectile above the starting point has positive y, while one below the starting point has negative y.

y 的正负取决于所选的正方向。如果规定向上为正，则物体在出发点上方时 y 为正，在出发点下方时 y 为负。

5. Time of Flight on Level Ground | 水平地面上的飞行时间

When a projectile is launched from level ground and lands at the same vertical height, the total time of flight is found by setting the vertical displacement equal to zero.

当抛体从水平地面发射并落回同一竖直高度时，飞行总时间可通过令竖直位移等于零来求得。

Set y = 0 in the vertical displacement equation: u sin θ × t − ½ gt² = 0. Factorising gives t(u sin θ − ½ gt) = 0, so the non-zero solution is the time of flight.

在竖直位移方程中令 y = 0：u sin θ × t − ½ gt² = 0。因式分解得到 t(u sin θ − ½ gt) = 0，因此非零解就是飞行时间。

T = 2u sin θ / g

This is the total time from launch to landing on level ground. It depends only on the initial vertical component of velocity.

这是从发射到落在水平地面上的总时间。它只取决于初速度的竖直分量。

6. Maximum Height | 最大高度

The projectile reaches its maximum height when the vertical component of velocity is zero. This is because the particle stops moving upward at the highest point before starting to fall.

当竖直速度分量为零时，抛体达到最大高度。因为在最高点物体停止向上运动，随后开始下落。

Using vᵥ = u sin θ − gt and setting vᵥ = 0 gives t = u sin θ / g. This is half of the total time of flight on level ground.

利用 vᵥ = u sin θ − gt 并令 vᵥ = 0，得到 t = u sin θ / g。这正好是水平地面上总飞行时间的一半。

Substituting this time into the vertical displacement equation gives the maximum height H.

将这个时间代入竖直位移方程，可得最大高度 H。

H = u² sin² θ / (2g)

The maximum height is proportional to the square of the initial speed and to sin² θ. A steeper launch angle, up to 90°, gives a larger maximum height.

最大高度与初速度的平方以及 sin² θ 成正比。发射角越接近 90°，最大高度越大。

7. Horizontal Range and Optimum Angle | 水平射程与最佳角度

The horizontal range R is the total horizontal distance travelled when the projectile returns to its original vertical level. It is found by multiplying the constant horizontal velocity by the time of flight.

水平射程 R 是抛体回到原来竖直高度时通过的总水平距离。它等于恒定的水平速度乘以飞行时间。

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

This shows that for a fixed initial speed u, the range is largest when sin 2θ = 1, which gives 2θ = 90°. Therefore the maximum range on level ground occurs at a launch angle of 45°.

这表明当初速度 u 一定时，射程在 sin 2θ = 1 时最大，即 2θ = 90°。因此水平地面上的最大射程出现在发射角为 45° 时。

The range formula also shows a symmetry: angles θ and 90° − θ give the same range for the same initial speed.

射程公式还显示出对称性：对于相同的初速度，角度 θ 与 90° − θ 给出相同的射程。

8. The Trajectory Equation | 轨迹方程

The path of a projectile can be described by eliminating time t between the horizontal and vertical displacement equations. This gives a quadratic relationship between y and x.

通过消去水平位移方程和竖直位移方程中的时间 t，可以描述抛体的运动路径。这样得到 y 与 x 之间的二次关系。

From x = u cos θ × t, we get t = x / (u cos θ). Substituting into the vertical displacement equation gives the trajectory equation.

由 x = u cos θ × t 得到 t = x / (u cos θ)。将其代入竖直位移方程，就得到轨迹方程。

y = x tan θ − g x² / (2u² cos² θ)

This equation is useful for finding whether a projectile clears a wall or reaches a particular point. It is a downward-opening parabola, reflecting the effect of gravity.

这个方程用于判断抛体是否能越过墙壁或到达某一点。它是一条开口向下的抛物线，反映了重力的影响。

9. Velocity at Any Time | 任意时刻的速度

The velocity of the projectile at any instant is the vector sum of its horizontal and vertical components. The horizontal component remains u cos θ, while the vertical component changes at rate −g.

抛体在任意时刻的速度是其水平分量和竖直分量的矢量合成。水平分量保持为 u cos θ，竖直分量则以速率 −g 变化。

vₕ = u cos θ, vᵥ = u sin θ − gt

The speed v at that time is the magnitude of the velocity vector, and the direction of motion is the angle the velocity makes with the horizontal.

该时刻的速率 v 是速度矢量的大小，运动方向是速度与水平方向所成的角。

v = √(vₕ² + vᵥ²)
tan α = vᵥ / vₕ

At the highest point, the vertical component is zero, so the speed is simply u cos θ. This is the minimum speed during the flight for a projectile launched above the horizontal.

在最高点，竖直分量为零，所以速率就是 u cos θ。对于从水平面以上抛出的物体，这是飞行过程中的最小速率。

10. Projection from a Height | 从高处抛出

If a projectile is launched from a point above the landing point, such as from a cliff or a building, the vertical displacement is not zero at landing. The time of flight must be found by solving a quadratic equation.

如果抛体从高于落点的位置发射，例如从悬崖或建筑物上，落地时竖直位移并不为零。必须通过解二次方程来求飞行时间。

Suppose the landing point is h metres below the launch point. With upward positive, the vertical displacement is y = −h, so the equation becomes u sin θ × t − ½ gt² = −h.

假设落点比发射点低 h 米。以向上为正，竖直位移为 y = −h，因此方程变为 u sin θ × t − ½ gt² = −h。

This can be rearranged into standard quadratic form ½ gt² − u sin θ × t − h = 0. The positive root gives the time of flight, and the horizontal range is then x = u cos θ × t.

该式可整理为标准二次方程 ½ gt² − u sin θ × t − h = 0。取正的根即为飞行时间，然后水平射程为 x = u cos θ × t。

11. Worked Example | 例题解析

A ball is projected from level ground with speed 20 m s⁻¹ at an angle of 30° to the horizontal. Calculate the time of flight, the maximum height, and the horizontal range.

一个球以 20 m s⁻¹ 的初速度与水平方向成 30° 角从水平地面抛出。计算飞行时间、最大高度和水平射程。

First resolve the initial velocity: uₕ = 20 cos 30° ≈ 17.3 m s⁻¹, and uᵥ = 20 sin 30° = 10 m s⁻¹.

首先分解初速度：uₕ = 20 cos 30° ≈ 17.3 m s⁻¹，uᵥ = 20 sin 30° = 10 m s⁻¹。

The time of flight is T = 2uᵥ / g = 2 × 10 / 9.8 ≈ 2.04 s. The maximum height is H = uᵥ² / (2g) = 10² / (2 × 9.8) ≈ 5.10 m.

飞行时间为 T = 2uᵥ / g = 2 × 10 / 9.8 ≈ 2.04 s。最大高度为 H = uᵥ² / (2g) = 10² / (2 × 9.8) ≈ 5.10 m。

The horizontal range is R = uₕ × T = 17.3 × 2.04 ≈ 35.3 m. This agrees with the range formula R = u² sin 60° / g.

水平射程为 R = uₕ × T = 17.3 × 2.04 ≈ 35.3 m。这与射程公式 R = u² sin 60° / g 一致。

12. Exam Tips and Common Mistakes | 考试提示与常见错误

Always start by resolving the initial velocity into components and writing down the horizontal and vertical equations separately. This makes projectile problems much easier to organise.

解题时始终先把初速度分解为分量，并分别写出水平方向和竖直方向的方程。这样处理抛体问题会清晰很多。

Be careful with signs. If upward is positive, then vertical acceleration is −g, and any displacement below the starting point is negative. Confusing signs is one of the most common errors in Edexcel projectile questions.

注意正负号。如果规定向上为正，那么竖直加速度为 −g，出发点以下的位移为负。正负号混淆是 Edexcel 抛体问题中最常见的错误之一。

Use the horizontal motion to find time when vertical information is lacking, or use the vertical motion to find time when horizontal information is lacking. Time links the two independent components together.

当缺少竖直信息时，用水平运动求时间；当缺少水平信息时，用竖直运动求时间。时间把两个独立分量联系在一起。

Do not assume the speed at the highest point is zero. Only the vertical component is zero there; the horizontal component remains unchanged.

不要以为最高点的速度为零。最高点只有竖直分量为零，水平分量保持不变。

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Projection at Any Angle | 任意角度斜抛运动