📚 Solving Projectile Motion Problems with Vector Methods | 用向量方法解抛体运动问题
Projectile motion is a classic application of vectors in mechanics. Instead of treating horizontal and vertical motion as two separate problems, we can write displacement, velocity, and acceleration as single vector quantities. This vector method is concise, powerful, and directly reflects the two-dimensional structure of the motion.
抛体运动是力学中向量应用的经典例子。与其把水平方向和竖直方向上的运动当成两个独立的问题，我们可以把位移、速度和加速度写成单一的向量量。这种向量方法简洁有力，并直接反映二维运动的数学结构。
1. Why Use Vectors? | 为什么要用向量？
When a projectile moves under gravity alone, its acceleration is constant in both magnitude and direction. We may write this acceleration as a = -g j, where j is the upward unit vector. Since acceleration is never horizontal during projectile motion, a vector formulation automatically prevents us from accidentally trying to use a horizontal acceleration.
当抛体只在重力作用下运动时，其加速度的大小和方向都恒定。我们可以把这个加速度写成 a = -g j，其中 j 是竖直向上的单位向量。因为在抛体运动过程中加速度永远不会水平，所以使用向量形式可以自动避免误用水平加速度。
The vector approach also makes it easier to combine the independent x and y equations. We use one equation for the whole position vector and one equation for the whole velocity vector.
向量方法还使得把相互独立的 x 方程和 y 方程合并起来变得更容易。我们只需要一个位置向量方程和一个速度向量方程。
2. The Vector Equation of Motion | 运动向量方程
For a particle with initial position vector r₀ and initial velocity vector v₀, moving with constant acceleration a, the position at time t is given by the vector equation
对于初始位置向量为 r₀、初速度向量为 v₀、并以恒定加速度 a 运动的质点，其在时间 t 的位置由以下向量方程给出
r(t) = r₀ + v₀t + ½a t²
Here r(t) is the displacement vector measured from the origin, v₀t is the displacement caused by the initial velocity, and ½a t² is the displacement caused by the constant acceleration.
这里 r(t) 是从原点测得的位移向量，v₀t 是由初速度产生的位移，½a t² 是由恒定加速度产生的位移。
Similarly, the velocity vector at time t is
类似地，时间 t 的速度向量为
v(t) = v₀ + a t
This vector equation contains both the horizontal and vertical velocity components in one expression.
这个向量方程把水平方向和竖直方向的速度分量同时包含在一个表达式中。
3. Resolving the Initial Velocity | 分解初速度
Suppose a projectile is launched from the origin with speed v₀ at an angle θ above the horizontal. The initial velocity vector can be resolved into components:
假设抛体从原点以速率 v₀、与水平方向夹角 θ 向上发射。初速度向量可以分解为如下分量：
v₀ = (v₀cosθ)i + (v₀sinθ)j
The horizontal component v₀cosθ is constant throughout the motion because there is no horizontal acceleration. The vertical component v₀sinθ changes with time because of gravity.
水平分量 v₀cosθ 在整个运动过程中保持不变，因为没有水平加速度。竖直分量 v₀sinθ 会因重力作用而随时间变化。
We can therefore separate the vector equation into two scalar equations:
因此我们可以把向量方程拆成两个标量方程：
x = v₀cosθ t
y = v₀sinθ t – ½g t²
These are not independent in the vector method; they are simply the i and j components of the same vector equation.
在向量方法中，这两个方程并不是互不相干的；它们只是同一个向量方程的 i 分量和 j 分量。
4. Position and Velocity Vectors of a Projectile | 抛体的位移向量与速度向量
For a projectile launched from the origin, the position vector at time t is
对于从原点发射的抛体，其在时间 t 的位移向量为
r(t) = (v₀cosθ)t i + (v₀sinθ t – ½g t²) j
The acceleration vector is always a = -g j, so the velocity vector at time t is
加速度向量始终为 a = -g j，所以时间 t 的速度向量为
v(t) = v₀cosθ i + (v₀sinθ – g t) j
Notice that the i-component of velocity never changes, because the unit vector i is horizontal. The j-component is a linear function of time.
注意速度的 i 分量永远不会改变，因为单位向量 i 是水平的。j 分量则是时间的一次函数。
5. Time of Flight and Range | 飞行时间与射程
When the projectile returns to the same height from which it was launched, we set y = 0. Using the vertical component of the position vector,
当抛体回到与发射高度相同的水平面时，我们令 y = 0。利用位移向量的竖直分量：
v₀sinθ t – ½g t² = 0
Factorising gives
因式分解得
t (v₀sinθ – ½g t) = 0
The first solution t = 0 is the launch instant. The non-zero solution is the time of flight:
第一个解 t = 0 是发射时刻。非零解就是飞行时间：
T = 2v₀sinθ / g
The horizontal range follows by substituting T into x = v₀cosθ t:
水平射程可通过把 T 代入 x = v₀cosθ t 得到：
R = v₀cosθ × (2v₀sinθ / g) = v₀² sin2θ / g
This is why the range is maximised when sin2θ = 1, that is when θ = 45°.
这就是为什么当 sin2θ = 1，即 θ = 45° 时射程最大。
6. Maximum Height | 最大高度
At the highest point of a projectile path, the vertical component of velocity is zero. Using the j-component of the velocity vector,
在抛体轨迹的最高点，速度的竖直分量为零。利用速度向量的 j 分量：
v₀sinθ – g t = 0
Therefore the time to reach maximum height is
因此到达最大高度所需的时间为
t = v₀sinθ / g
Substituting this into y = v₀sinθ t – ½g t² gives the maximum height:
将其代入 y = v₀sinθ t – ½g t²，得到最大高度：
H = v₀² sin²θ / (2g)
At this instant the velocity vector is still horizontal and has magnitude v₀cosθ; the speed is not zero.
在这一瞬间速度向量仍然是水平的，其大小为 v₀cosθ；速度并不是零。
7. Worked Example 1: Using Component Form | 例题 1：利用分量形式求解
A particle is projected from the origin with initial velocity (12i + 5j) m/s. Taking g = 10 m/s², find the time of flight, the horizontal range, and the maximum height.
一个质点从原点以初速度 (12i + 5j) m/s 发射。取 g = 10 m/s²，求飞行时间、水平射程和最大高度。
Using the vector equation of motion, the position vector is
利用运动向量方程，位移向量为
r(t) = 12t i + (5t – 5t²) j
When the projectile lands, y = 0, so 5t – 5t² = 0. Thus t = 1 s. The horizontal range is x = 12 × 1 = 12 m.
当抛体落地时 y = 0，所以 5t – 5t² = 0，因此 t = 1 s。水平射程为 x = 12 × 1 = 12 m。
For maximum height, the vertical component of velocity is 5 – 10t = 0, so t = 0.5 s. Substituting gives y = 5(0.5) – 5(0.25) = 1.25 m.
对于最大高度，速度的竖直分量为 5 – 10t = 0，所以 t = 0.5 s。代入得 y = 5(0.5) – 5(0.25) = 1.25 m。
8. Worked Example 2: Velocity Vector at a Given Time | 例题 2：求指定时刻的速度向量
Using the same projectile as Example 1, find the velocity vector and the speed when t = 2 s.
沿用例题 1 中的抛体，求 t = 2 s 时的速度向量和速率。
The velocity vector is
速度向量为
v(t) = 12i + (5 – 10t)j
At t = 2 s,
当 t = 2 s 时，
v(2) = 12i + (5 – 20)j = 12i – 15j m/s
The speed is the magnitude of the velocity vector:
速率就是速度向量的大小：
|v| = √(12² + 15²) = √369 ≈ 19.2 m/s
Because the j-component is negative, the projectile is moving downwards. The angle below the horizontal is arctan(15/12) ≈ 51.3°.
因为 j 分量为负，所以抛体正在向下运动。其方向与水平线向下的夹角为 arctan(15/12) ≈ 51.3°。
9. The Vector Equation of the Trajectory | 轨迹的向量方程
We can eliminate t between the horizontal and vertical components to find the equation of the path. From x = v₀cosθ t,
我们可以消去水平分量和竖直分量之间的 t，以求出轨迹方程。由 x = v₀cosθ t 得
t = x / (v₀cosθ)
Substituting into y = v₀sinθ t – ½g t² gives
代入 y = v₀sinθ t – ½g t² 得
y = x tanθ – (g x²) / (2v₀² cos²θ)
Since 1/cos²θ = 1 + tan²θ, this can also be written as
因为 1/cos²θ = 1 + tan²θ，所以也可以写成
y = x tanθ – (g x² / 2v₀²)(1 + tan²θ)
This form is useful when a target point (x, y) is given and we need to find the required launch angle θ.
当给定目标点 (x, y) 而需要求所需发射角 θ 时，这一形式非常有用。
10. Common Pitfalls and Exam Tips | 常见错误与应试技巧
Always set the sign of g consistently. If upward is positive j, then the acceleration is -g j. Using a positive g in the vertical displacement equation will produce incorrect heights and times of flight.
始终统一设定 g 的正负号。如果向上为正 j 方向，那么加速度就是 -g j。如果在竖直位移方程中把 g 当成正值，就会得到错误的高度和飞行时间。
Do not mix horizontal and vertical components. The horizontal part has constant velocity; the vertical part has constant acceleration. The vector equation keeps these components independent but neatly contained in one expression.
不要把水平分量和竖直分量混在一起。水平方向是匀速运动，竖直方向是匀加速运动。向量方程让这些分量相互独立，同时又整齐地包含在一个表达式中。
At maximum height, the vertical velocity is zero, but the total velocity is not zero. The projectile still has a horizontal velocity v₀cosθ.
在最高点，竖直速度为 0，但合速度不是 0。抛体仍然具有水平速度 v₀cosθ。
Remember that time of flight is a positive scalar. When solving t(v₀sinθ – ½g t) = 0, discard the t = 0 solution unless it is the launch moment.
记住飞行时间是一个正的标量。在解 t(v₀sinθ – ½g t) = 0 时，要舍去 t = 0 的解，除非讨论的是发射时刻。
Check units carefully. Standard projectile motion questions use metres, seconds, and metres per second squared. Always state final answers with the correct units.
仔细检查单位。标准的抛体运动问题使用米、秒和米每二次方秒。最终答案必须写出正确的单位。
By understanding projectile motion as one vector equation, you gain a coherent method that works for all problems involving constant acceleration. Practise writing r(t), v(t), and a(t) as vectors first, and the scalar equations will follow naturally.
通过把抛体运动理解为一个向量方程，你可以获得一种统一的方法，它适用于所有涉及恒定加速度的问题。练习先写出 r(t)、v(t) 和 a(t)，标量方程自然会随之得出。
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