📚 An Exploration on the Effect of the Launch Angle of a Projectile on the Horizontal Range: Formula Derivation | 抛射体发射角对水平射程影响的探究：公式推导
Projectile motion is a fundamental topic in IB Physics, blending kinematics with mathematical modelling. Understanding how the launch angle affects the horizontal range not only reinforces concepts of vectors, independence of motion, and quadratic equations, but also paves the way for experimental design and critical analysis. This article walks you through the step-by-step derivation of the range equation, examines the role of the launch angle, and explores conditions for maximum range, complementary angles, and real-world implications.
抛射体运动是IB物理中的一个基础课题，它融合了运动学与数学建模。理解发射角如何影响水平射程，不仅能巩固矢量、运动独立性以及二次方程等概念，还能为实验设计与批判性分析打下基础。本文将一步步引导你推导射程公式，探讨发射角的作用，并分析取得最大射程的条件、互补角现象以及实际应用意义。
1. Introduction to Projectile Motion | 抛射体运动简介
Projectile motion refers to the two-dimensional motion of an object launched into the air, subject only to the acceleration due to gravity (neglecting air resistance). The path followed is a parabola. In IB Physics, we study projectiles launched from ground level onto the same horizontal plane, with an initial speed v₀ at an angle θ to the horizontal.
抛射体运动是指物体被抛入空中后，仅在重力加速度作用下（忽略空气阻力）所做的二维运动，其轨迹是一条抛物线。在IB物理课程中，我们通常研究从地面以初速度 v₀、与水平方向夹角 θ 发射、并落回同一水平面上的抛射体。
2. Resolving the Initial Velocity | 分解初速度
The first crucial step is to resolve the launch velocity into horizontal (x) and vertical (y) components. The horizontal component is v₀x = v₀ cos θ, and the vertical component is v₀y = v₀ sin θ. The horizontal velocity remains constant (assuming no air resistance), while the vertical motion is governed by constant acceleration g = 9.81 m s⁻² downwards.
关键的第一步是将发射速度分解为水平分量（x方向）和竖直分量（y方向）。水平分量为 v₀x = v₀ cos θ，竖直分量为 v₀y = v₀ sin θ。水平速度保持不变（假设无空气阻力），而竖直运动则受恒定加速度 g = 9.81 m s⁻² 向下的制约。
3. Vertical Motion and Time of Flight | 垂直运动与飞行时间
We analyse the vertical motion using the kinematic equation s = u t + ½ a t². Since the projectile returns to the same vertical level, the net vertical displacement is zero. Taking upward as positive, we set s_y = 0, initial vertical velocity u_y = v₀ sin θ, and acceleration a_y = -g.
我们利用运动学公式 s = u t + ½ a t² 分析垂直运动。由于抛射体落回同一水平面，净竖直位移为零。取向上为正，则有 s_y = 0，初速度竖直分量 u_y = v₀ sin θ，加速度 a_y = -g。
0 = (v₀ sin θ) t – ½ g t². Solving for t (excluding the trivial t=0 solution), we get the time of flight t = (2 v₀ sin θ) / g.
0 = (v₀ sin θ) t – ½ g t²。解出飞行时间 t（舍去 t = 0 的平凡解），得 t = (2 v₀ sin θ) / g。
This expression shows that the time of flight is directly proportional to the initial vertical component of velocity, and hence to sin θ.
此式表明，飞行时间与初速度的竖直分量成正比，因而也与 sin θ 成正比。
4. Horizontal Motion and Range | 水平运动与射程
Horizontally, there is no acceleration, so the velocity remains constant at v₀x = v₀ cos θ. The horizontal range R is simply the product of this constant horizontal velocity and the time of flight:
水平方向没有加速度，因此速度保持为 v₀x = v₀ cos θ 不变。水平射程 R 即为这一恒定水平速度与飞行时间的乘积：
R = v₀x · t = (v₀ cos θ) · (2 v₀ sin θ / g)
R = v₀x · t = (v₀ cos θ) · (2 v₀ sin θ / g)
5. Deriving the Range Equation | 推导射程公式
Simplify the expression using the double-angle identity 2 sin θ cos θ = sin 2θ. This gives the classic range equation for a projectile launched and landing on the same horizontal plane:
利用倍角公式 2 sin θ cos θ = sin 2θ 进行化简，便得到发射点和落点在同一水平面上的经典射程公式：
R = (v₀² sin 2θ) / g
R = (v₀² sin 2θ) / g
This equation encapsulates the dependence of range on launch speed, angle, and gravitational field strength. Notice that R is proportional to the square of the initial speed, indicating that doubling the launch speed quadruples the range.
该方程概括了射程与发射速度、角度以及重力场强度之间的依赖关系。注意 R 与初速度的平方成正比，说明初速度加倍将使射程变为原来的四倍。
6. The Effect of Launch Angle on Range | 发射角对射程的影响
For a fixed initial speed v₀ and constant g, R depends solely on sin 2θ. The sine function reaches its maximum value of 1 when its argument is 90°. Hence, the maximum range occurs when 2θ = 90°, i.e., θ = 45°. This is a key result: in the absence of air resistance, the optimal launch angle for maximum horizontal distance is 45°.
对于固定的初速度 v₀ 和恒定的 g，R 仅取决于 sin 2θ。正弦函数在角度为 90° 时取得最大值 1。因此，当 2θ = 90°，即 θ = 45° 时，射程最大。这是一个重要结论：在无空气阻力的情况下，获得最大水平距离的最佳发射角为 45°。
As θ increases from 0° to 45°, sin 2θ increases from 0 to 1, so the range increases. Beyond 45°, sin 2θ decreases, and the range falls. A launch angle of 0° yields zero range (the projectile simply slides along the ground), while 90° gives zero horizontal range (the projectile goes straight up and down).
随着 θ 从 0° 增大到 45°，sin 2θ 从 0 增至 1，射程逐渐增大。超过 45° 后，sin 2θ 减小，射程也随之减小。发射角为 0° 时射程为零（物体沿地面滑动），发射角为 90° 时水平射程也为零（物体垂直上抛后落回原处）。
7. Complementary Angles Produce Equal Ranges | 互补角产生相等的射程
Because sin 2θ = sin (180° – 2θ), any two launch angles that are complementary (i.e., add up to 90°) will produce the same horizontal range. For example, a projectile launched at 30° and another at 60° will land at the same distance, provided the initial speed is identical. The 60° trajectory has a higher apogee but the same total range.
由于 sin 2θ = sin (180° – 2θ)，任何两个互余的发射角（即两者之和为 90°）会产生相同的水平射程。例如，以 30° 和 60° 发射的抛射体，若初速度相同，将落在同一距离处。60° 的轨迹最高点更高，但总射程一致。
This symmetry is often verified in IB laboratory investigations using projectile launchers and carbon paper to record landing positions.
在IB实验探究中，常利用抛射体发射器和复写纸记录落点位置，以验证这一对称性。
8. Extending to Different Launch and Landing Heights | 拓展到发射点与落点高度不同的情况
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