📚 Projectile Motion | 抛体运动
Projectile motion is a form of motion experienced by an object that is launched into the air and moves under the influence of gravity alone. In IB Physics, we analyse this two-dimensional motion by resolving it into independent horizontal and vertical components, which dramatically simplifies calculations.
抛体运动是物体被抛入空中并仅在重力作用下运动的一种运动形式。在 IB 物理中，我们将这种二维运动分解为相互独立的水平分量和垂直分量，从而大大简化计算。
1. Introduction to Projectile Motion | 抛体运动导论
ce so that the only acceleration is the constant downward acceleration due to gravity, g = 9.81 m s⁻².
抛体是指一旦获得初速度后，仅在重力和自身惯性的作用下沿曲线路径运动的任何物体。经典例子包括以一定角度投出的球、从悬崖发射的炮弹或飞行中的跳远运动员。关键的一点是我们忽略空气阻力，因此唯一的加速度就是恒定向下的重力加速度，g = 9.81 m s⁻²。
The path of a projectile is a parabola if we ignore air resistance and assume the Earth's surface is flat over the scale of the motion. This topic is foundational because it combines kinematics in two perpendicular directions and reinforces vector decomposition.
在忽略空气阻力并假设运动范围内地球表面是平面的情况下，抛体的轨迹是一条抛物线。这个主题是基础性的，因为它结合了两个垂直方向上的运动学知识，并强化了矢量分解。
2. Assumptions and Idealizations | 假设与理想化
To model projectile motion effectively, we make several key assumptions. Firstly, the only force acting on the projectile after launch is gravity; this means air resistance is negligible. Secondly, the acceleration due to gravity is constant in magnitude (9.81 m s⁻²) and always directed vertically downward. Thirdly, the Earth's curvature is ignored, so the ground is treated as flat. Finally, we treat the projectile as a point mass, ignoring any rotation or complex aerodynamic effects.
为了有效建立抛体运动模型，我们作出几个关键假设。首先，抛体在发射后仅受重力作用；这意味着空气阻力可以忽略。其次，重力加速度大小恒定（9.81 m s⁻²），方向始终垂直向下。第三，忽略地球曲率，地面视为平面。最后，我们将抛体视为质点，忽略任何旋转或复杂的空气动力学效应。
These assumptions allow the motion to be separated into two independent one-dimensional motions: horizontal motion with constant velocity and vertical motion with constant acceleration. This independence is the cornerstone of solving projectile problems in the IB syllabus.
这些假设使得运动可以分解为两个独立的一维运动：水平方向的匀速运动和竖直方向的匀加速运动。这种独立性是 IB 大纲中解决抛体问题的基石。
3. Resolving Initial Velocity | 分解初速度
component is uᵧ = u sin θ. These components are treated separately throughout the motion.
对于以初速度 u、与水平面成角度 θ 发射的抛体，第一步总是将 u 分解为水平和垂直分量。水平分量为 uₓ = u cos θ，垂直分量为 uᵧ = u sin θ。在整个运动过程中，这两个分量将被分别处理。
a special case that still follows the same principles but with simpler initial conditions.
如果抛体从某一高度水平抛出，则 θ = 0°，因此 uₓ = u，uᵧ = 0。这是一种特殊情况，遵循相同的原理，但初始条件更简单。
4. Horizontal Motion | 水平运动
splacement x after time t is simply given by the equation for constant velocity: x = uₓ t.
在水平方向上，由于重力沿竖直方向作用，因此没有加速度。所以，速度的水平分量在整个飞行过程中保持不变：vₓ = uₓ = u cos θ。时间 t 后的水平位移 x 直接用匀速运动方程给出：x = uₓ t。
This constancy means that the time of flight directly determines the horizontal range. Because the horizontal motion is uniform, the projectile covers equal horizontal distances in equal time intervals.
这种不变性意味着飞行时间直接决定了水平射程。由于水平运动是匀速的，抛体在相等的时间间隔内通过的水平距离相等。
5. Vertical Motion | 垂直运动
governed by constant downward acceleration g. We use the standard SUVAT equations with a = – upward as positive). The vertical component of velocity at time t is given by vᵧ = u sin θ – g t. The vertical displacement y is given by y = (u sin θ) t – ½ g t².
抛体的垂直运动受恒定向下的加速度 g 支配。我们采用标准的 SUVAT 方程，并设 a = –g（若取向上为正）。时间 t 时的垂直速度分量为 vᵧ = u sin θ – g t。垂直位移 y 由公式 y = (u sin θ) t – ½ g t² 给出。
ht, and the maximum height. The vertical motion is symmetric about the highest point when launch and landing are at the same height.
这些方程可用于求解到达最高点（此时 vᵧ = 0）所需的时间、总飞行时间以及最大高度。当发射点与落点高度相同时，垂直运动关于最高点对称。
6. Equations of Motion for Projectile | 抛体运动方程
Combining the horizontal and vertical descriptions, the full set of kinematic equations for a projectile launched from the origin (0,0) with speed u at angle θ is:
结合水平和垂直描述，从原点 (0,0) 以速度 u、角度 θ 发射的抛体运动全套运动学方程为：
x = (u cos θ) t
y = (u sin θ) t – ½ g t²
vₓ = u cos θ
vᵧ = u sin θ – g t
These four equations are sufficient to solve any standard projectile problem. Note that the acceleration is exclusively in the y-direction, and its magnitude is g. The time parameter t links the two independent motions.
这四个方程足以解决任何标准的抛体运动问题。注意，加速度仅存在于 y 方向，其大小为 g。时间参数 t 将两个独立的运动联系起来。
7. Time of Flight | 飞行时间
ertical displacement y = 0 (excluding the trivial t = 0 solution). Solving 0 = (u sin θ) t – ½ g t² gives t = (2 u sin θ) / g. This is the time from launch to landing.
对于落点与发射点在同一水平面的抛体，总飞行时间通过令垂直位移 y = 0（排除 t = 0 的平凡解）求得。解方程 0 = (u sin θ) t – ½ g t² 得到 t = (2 u sin θ) / g。这是从发射到落地的时间。
rent height, the time of flight is found by solving the quadratic equation y = (u sin θ) t – ½ g t² for t, taking the positive root that corresponds to the landing event.
如果抛体从高处发射并落在不同高度，飞行时间需通过解二次方程 y = (u sin θ) t – ½ g t² 求得 t，取与落地事件对应的正根。
An important consequence is that the time of flight depends only on the vertical component of the initial velocity and g; the horizontal component is irrelevant to how long the projectile stays in the air.
一个重要的推论是，飞行时间只取决于初速度的垂直分量和 g；水平分量与抛体在空中停留的时间无关。
8. Maximum Height | 最大高度
ts maximum height when the vertical component of velocity becomes zero. Using vᵧ = u sin θ – Substituting this time into the vertical displacement equation gives the maximum height H:
当速度的垂直分量为零时，抛体达到最大高度。利用 vᵧ = u sin θ – g t，令 vᵧ = 0 得到 tₘₐₓ = (u sin θ) / g。将此时间代入垂直位移方程，可得最大高度 H：
H = (u² sin² θ) / (2 g)
u, the maximum possible height is achieved when θ = 90° (straight upward), giving u²/(2g).
该表达式表明，最大高度与初速度垂直分量的平方成正比，与 g 成反比。对于给定的 u，当 θ = 90°（垂直向上）时，达到最大可能的高度，即 u²/(2g)。
9. Range of a Projectile | 射程
g). Using the trigonometric identity 2 sin θ cos θ = sin 2θ, we obtain the range equation:
假设发射点与落点高度相同，抛体的水平射程 R 是水平速度与总飞行时间的乘积：R = (u cos θ) × (2 u sin θ / g)。利用三角恒等式 2 sin θ cos θ = sin 2θ，我们得到射程方程：
R = (u² sin 2θ) / g
se ideal conditions. For any range less than the maximum, there are two complementary launch angles (θ and 90° – θ) that give the same range.
这一结果表明，当 sin 2θ = 1，即 2θ = 90° 时，射程最大，故在理想条件下，θ = 45° 时射程最远。对于任何小于最大射程的值，存在两个互补的发射角（θ 和 90° – θ），能产生相同的射程。
10. Trajectory Equation | 轨迹方程
nating the time variable t from the parametric equations x = (u cos θ)t and y = (u sin θ)t – ation of the parabolic path. From x, we have t = x / (u cos θ). Substituting into y gives:
通过从参数方程 x = (u cos θ)t 和 y = (u sin θ)t – ½ g t² 中消去时间变量 t，我们得到抛物线路径方程。由 x 得 t = x / (u cos θ)，代入 y 得到：
y = (tan θ) x – (g / (2 u² cos² θ)) x²
tic function, confirming the parabolic shape. The first term represents the straight line the projectile would follow in the absence of gravity, and the second term (negative) is the correction due to gravitational pull.
这是一个形如 y = ax + bx² 的二次函数，证实了抛物形状。第一项表示没有重力时抛体将沿直线运动的部分，第二项（负值）是重力作用引起的修正。
The trajectory equation is especially useful when solving problems that involve hitting a target at a specific coordinate or determining the condition for clearing an obstacle.
当需要解决涉及击中特定坐标目标或越过障碍物的条件时，轨迹方程尤其实用。
11. Symmetry and Special Cases | 对称性与特殊情况
When a projectile lands at the same height from which it was launched, the motion is perfectly symmetric about the vertical line through the highest point. The time taken to rise to the peak equals the time taken to fall back to the original level. The speed at any height during ascent equals the speed at the same height during descent, but the vertical velocity component reverses sign.
当抛体落回与发射点相同的高度时，运动关于通过最高点的垂直线完全对称。上升到最高点所用的时间等于落回原高度所用的时间。上升过程中任一高度的速率等于下降过程中相同高度的速率，但垂直速度分量符号相反。
itial vertical velocity is zero, and the time of flight is found from the height using t = √2h/g rtical projection (θ = 90°), which reduces to simple one-dimensional motion under gravity.
特殊情况包括从悬崖水平抛出（θ = 0），此时初始垂直速度为零，飞行时间由高度决定，t = √2h/g。另一种情况是竖直上抛（θ = 90°），这简化为重力作用下的一维运动。
12. Worked Example | 典型例题
height, (b) the total time of flight, and (c) the horizontal range. Assume g = 9.81 m s⁻².
一个高尔夫球以 30 m s⁻¹ 的初速度、与水平面成 30° 角被击出。计算：(a) 最大高度，(b) 总飞行时间，(c) 水平射程。取 g = 9.81 m s⁻²。
lution: First, find components: uₓ = 30 cos 30° ≈ 25.98 m s⁻¹, uᵧ = 30 sin 30° = 15 m s⁻¹.
解答：首先求分量：uₓ = 30 cos 30° ≈ 25.98 m s⁻¹，uᵧ = 30 sin 30° = 15 m s⁻¹。
× 3.06 ≈ 79.5 m, or directly using R = (u² sin 2θ) / g = (900 × sin 60°) / 9.81 ≈ 79.5 m.
(a) 最大高度 H = (uᵧ²) / (2 g) = (15²) / (2 × 9.81) ≈ 11.47 m。(b) 总飞行时间 t = 2 uᵧ / g = 30 / 9.81 ≈ 3.06 s。(c) 射程 R = uₓ × t = 25.98 × 3.06 ≈ 79.5 m，或直接用 R = (u² sin 2θ) / g = (900 × sin 60°) / 9.81 ≈ 79.5 m。
Notice how the range equation gives the same result, confirming the consistency of the approach.
注意射程方程给出相同结果，验证了方法的一致性。
13. Common Misconceptions | 常见误区
One common mistake is to think that the horizontal velocity changes due to gravity. Remember, g only affects the vertical component; the horizontal velocity remains constant. Another error is confusing the angle that gives maximum range; 45° is correct only when launch and landing heights are the same and there is no air resistance.
一个常见错误是认为水平速度会因重力而改变。请记住，g 只影响垂直分量；水平速度保持不变。另一个错误是混淆最大射程的角度；仅在发射点与落点高度相同且无空气阻力时，45° 才是正确的。
assigned: uₓ = u cos θ, uᵧ = u sin θ. Swapping these can lead to completely wrong answers.
学生有时即使在发射和着陆高度不同时也把垂直运动当作对称的；这是不正确的。当高度不同时，上升和下降的时间不相等。最后，务必检查分量是否指定正确：uₓ = u cos θ，uᵧ = u sin θ。若搞反则会导致完全错误的答案。
14. Experiment and Graphing | 实验与图像分析
In practical investigations, projectile motion can be studied using video analysis or stroboscopic photography. Plotting y against x yields a parabola. Plotting y against x² gives a straight line with gradient –g/(2 uₓ²) and intercept related to the initial angle. These graphical methods allow experimental determination of g or the initial speed.
在实践探究中，抛体运动可通过视频分析或频闪摄影来研究。绘制 y 对 x 的图像会得到一条抛物线。绘制 y 对 x² 的图像会得到一条直线，其斜率为 –g/(2 uₓ²)，截距与初始角度有关。这些图像方法可以实验测定 g 或初速度。
The independence of horizontal and vertical motions can be demonstrated by dropping one ball vertically while simultaneously projecting another horizontally from the same height; both hit the ground at the same time.
水平与垂直运动的独立性可以通过同时从同一高度竖直放下一个球和水平抛出一个球来演示；两个球同时落地。
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