📚 Projection at any angle | 任意角度斜抛运动
In the study of mechanics, the motion of a particle projected at an angle to the horizontal is a classic application of kinematics. This topic, often referred to as projectile motion, assumes that the only force acting on the particle after projection is gravity (ignoring air resistance). By treating the horizontal and vertical components of motion independently, students can derive key quantities such as time of flight, maximum height, and horizontal range. Mastery of projection at any angle forms the basis for solving more complex problems in A-Level Mathematics and Physics.
在力学学习中，质点以与水平方向成一定角度抛出的运动是运动学的经典应用。这一内容通常被称为抛体运动，它假设质点在抛出后仅在重力作用下运动（忽略空气阻力）。通过独立处理运动的水平分量和竖直分量，学生可以推导出飞行时间、最大高度和水平射程等关键量。掌握任意角度斜抛运动是解决A-Level数学和物理中更复杂问题的基础。
1. Resolving the Initial Velocity | 分解初速度
When an object is projected with initial speed u at an angle θ to the horizontal, the first step is to resolve the velocity vector into horizontal and vertical components. The horizontal component remains constant as there is no acceleration in that direction, while the vertical component is subject to gravitational acceleration g acting downwards. The resolution yields ux = u cos θ and uy = u sin θ, taking upward as the positive vertical direction.
ux = u cos θ uy = u sin θ
当物体以初速率 u 与水平面成 θ 角抛出时，第一步是将速度矢量分解为水平和竖直分量。水平分量因该方向无加速度而保持不变，竖直分量则受到向下的重力加速度 g 作用。若取竖直向上为正方向，分解可得 ux = u cos θ 和 uy = u sin θ。
2. Horizontal and Vertical Motion Equations | 水平和竖直运动方程
Since air resistance is neglected, the horizontal motion is uniform with a constant velocity u cos θ. The vertical motion is governed by constant acceleration g downwards. Using the SUVAT equations, the displacement and velocity components at time t are given by:
x = (u cos θ) t vx = u cos θ
y = (u sin θ) t − ½ g t² vy = u sin θ − g t
在忽略空气阻力的情况下，水平方向为匀速运动，速度恒为 u cos θ。竖直方向则受到向下的恒定加速度 g 作用。利用匀加速运动公式，在时刻 t 的位移和速度分量如下：
水平位移与速度不变，竖直位移受重力影响呈二次函数关系，竖直速度随时间线性减小。这些方程是推导所有斜抛运动特征量的基础。
3. Time of Flight | 飞行时间
The time of flight is the total time the projectile remains in the air, from launch until it returns to the original vertical level (usually the ground). We set the vertical displacement y = 0 and solve for t (excluding the trivial solution t = 0). This gives:
T = (2 u sin θ) / g
飞行时间是指抛体从抛出到落回原竖直高度（通常为地面）的总时间。令竖直位移 y = 0 并求解 t（舍去平凡解 t = 0），得到飞行时间 T。注意飞行时间完全由初速度的竖直分量决定，水平分量不影响在空中停留的时长。
4. Maximum Height | 最大高度
At the peak of its trajectory, the vertical component of velocity becomes zero. Using the SUVAT equation vy² = uy² − 2 g H with final vertical velocity 0, the maximum height H above the launch level is:
H = (u² sin² θ) / (2g)
在轨迹最高点，竖直速度分量为零。利用方程 vy² = uy² − 2 g H，代入末速度零，可解得距抛出水平面的最大高度 H。最大高度与初速度竖直分量的平方成正比，角度越接近竖直，高度越大。
5. Horizontal Range | 水平射程
The horizontal range R is the total horizontal distance travelled during the flight. Since horizontal velocity is constant, R = u cos θ × T. Substituting the expression for time of flight yields:
R = (u² sin 2θ) / g
水平射程 R 指飞行期间经过的水平总距离。由于水平速度恒定，有 R = u cos θ × T。代入飞行时间表达式即得上述结果。射程同时依赖于初速率和两倍角的正弦值，是斜抛运动重要的优化量。
6. Equation of the Trajectory | 轨迹方程
To describe the shape of the path, we eliminate time t from the parametric equations. Solving x = u cos θ t for t and substituting into the y equation gives the parabolic trajectory:
y = x tan θ − (g x²) / (2 u² cos² θ)
为描述路径的形状，从参数方程中消去时间 t。将 t = x / (u cos θ) 代入竖直位移方程，即得到抛物线形式的轨迹方程。该式表明抛体路径总是一条开口向下的抛物线，二次项系数与初速度和角度有关。
7. Effect of Projection Angle on Range | 投射角对射程的影响
The range formula R = (u² sin 2θ)/g reveals that for a fixed initial speed, the range depends on sin 2θ. Maximum range occurs when sin 2θ = 1, i.e. 2θ = 90° so θ = 45°. Moreover, angles that are complementary (e.g. 30° and 60°) produce the same range because sin(2×30°) = sin(2×60°). This symmetry is useful in problems comparing different launch conditions.
射程公式 R = (u² sin 2θ)/g 表明当初速率固定时，射程取决于 sin 2θ。最大射程在 sin 2θ = 1 时取得，即 2θ = 90°，因此最佳投射角为 θ = 45°。此外，互余角（如30°与60°）会给出相同的射程，因为 sin(2×30°) = sin(2×60°)。这一对称性在比较不同抛出条件的问题中非常实用。
8. Symmetry of the Parabolic Path | 抛物线路径的对称性
Projectile motion is symmetric about the vertical line through the highest point. The time taken to reach the maximum height equals half the total time of flight, and the time to descend back to the launch level is identical. The speed at any height on the way up is equal to the speed at the same height on the way down, although the vertical velocity components have opposite signs. The horizontal component remains unchanged throughout.
抛体运动关于经过最高点的竖直线对称。到达最大高度所需的时间等于总飞行时间的一半，落回原高度的下降时间与上升时间相等。在同一位势高度，上升时的速率与下降时的速率大小相等，但竖直速度分量符号相反。水平速度分量在整个运动过程中保持不变。理解这一对称性有助于快速解决许多比较性问题。
9. Worked Example | 例题解析
A ball is projected with an initial speed of 20 m s⁻¹ at an angle of 30° above the horizontal. Taking g = 9.8 m s⁻², find the time of flight, the maximum height, and the horizontal range.
一球以初速率20 m s⁻¹、与水平面成30°角向上抛出。取 g = 9.8 m s⁻²，求飞行时间、最大高度和水平射程。
First, resolve the initial velocity: ux = 20 cos 30° ≈ 17.32 m s⁻¹, uy = 20 sin 30° = 10 m s⁻¹. Time of flight T = (2 × 20 sin 30°) / 9.8 = 40 × 0.5 / 9.8 ≈ 2.04 s. Maximum height H = (20² sin² 30°) / (2 × 9.8) = (400 × 0.25) / 19.6 ≈ 5.10 m. Horizontal range R = (20² sin 60°) / 9.8 = (400 × 0.8660) / 9.8 ≈ 35.3 m. The results are consistent with the derived formulas.
首先分解初速度：ux = 20 cos 30° ≈ 17.32 m s⁻¹，uy = 20 sin 30° = 10 m s⁻¹。飞行时间 T =
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