📚 Case study – Driving a golf ball | 高尔夫球击球的案例研究
Golf is a sport rich in mathematical modelling opportunities. When a golfer strikes a ball with a driver, the resulting flight can be described using principles of projectile motion, trigonometry and calculus. This case study explores how IB Mathematics can be applied to analyse the trajectory of a golf ball, determine the optimal launch conditions for maximum distance, and examine the effect of real-world factors such as elevation differences and air resistance. By building a simplified model and gradually refining it, we will demonstrate the power of mathematical reasoning in a practical sporting context.
高尔夫是一项蕴含丰富数学建模机会的运动。当球手用一号木击球时，球的飞行可以用抛体运动、三角学和微积分的原理来描述。本案例研究探讨如何运用 IB 数学来分析高尔夫球的轨迹、确定最大距离的最佳发射条件，并考察诸如高度差和空气阻力等现实因素的影响。通过构建一个简化模型并逐步完善，我们将展示数学推理在真实运动情境中的威力。
1. Introducing the Case Study | 案例介绍
This investigation is inspired by a typical Internal Assessment (IA) scenario in IB Mathematics: a golfer wishes to understand how the choice of club and swing technique affects the driving distance. We will focus on a single drive struck with an initial speed of 45 m s⁻¹, a realistic value for an amateur player using a modern driver. The ball is launched from a tee that is elevated above the fairway, introducing an initial height difference that alters the classic projectile range formula.
本探究的灵感源自 IB 数学中常见的内部评估 (IA) 情境：一位高尔夫球手希望了解球杆选择和挥杆技术如何影响开球距离。我们将聚焦于一次使用现代一号木杆的击球，其初始速度为 45 m s⁻¹，这是业余球手的真实数值。球从高于球道的发球台击出，这种初始高度差改变了经典的抛体射程公式。
We will begin with an idealised mathematical model assuming no air resistance and a uniform gravitational field. The primary goals are to predict the maximum height reached by the ball, the time it spends in the air, the horizontal distance covered, and the optimal launch angle for maximising range. Subsequent sections will refine the model by including a downhill landing zone and a qualitative discussion of drag forces.
我们将从理想化的数学模型开始，假设没有空气阻力且重力场均匀。主要目标是预测球达到的最大高度、在空中飞行的时间、水平距离以及使射程最大化的最佳发射角度。随后的小节将通过加入向下坡度的落点和定性讨论阻力来完善该模型。
2. Problem Statement and Assumptions | 问题表述与假设
The core question can be phrased as: Given an initial speed of 45 m s⁻¹ and a tee position 10 m above the landing area, what launch angle θ maximises the horizontal distance travelled by the ball? To make the problem tractable, we adopt the following assumptions: the golf ball is treated as a point mass; air resistance and wind are neglected; the ball’s spin and its resulting lift (Magnus effect) are ignored; and the acceleration due to gravity is constant at g = 9.8 m s⁻² acting vertically downwards.
核心问题可以表述为：给定 45 m s⁻¹ 的初速度以及发球台比落点区域高出 10 m 的条件，哪个发射角度 θ 能使球飞行的水平距离最大化？为了使问题可解，我们采用以下假设：将高尔夫球视为质点；忽略空气阻力和风；忽略球的自旋及其产生的升力（马格努斯效应）；重力加速度恒定为 g = 9.8 m s⁻²，方向竖直向下。
We also assume that the launch and landing surfaces are horizontal, with the y-axis taken as positive upwards. The origin is placed at ground level directly below the launch point, so the initial position is (0, h) where h = 10 m. The initial velocity vector has magnitude v₀ = 45 m s⁻¹ at an angle θ above the horizontal. The motion is confined to the x-y plane.
我们还假设发射面和着陆面是水平的，y 轴以向上为正。原点设在发射点正下方的地面处，因此初始位置为 (0, h)，其中 h = 10 m。初速度矢量的大小为 v₀ = 45 m s⁻¹，方向与水平面成 θ 角。运动被限制在 x-y 平面内。
3. Modelling the Projectile Motion | 抛物运动建模
Using Newtonian mechanics, the horizontal component of velocity remains constant because there is no horizontal force. Thus, the horizontal displacement as a function of time t is given by x(t) = v₀ cosθ · t. The vertical motion is governed by a constant downward acceleration g. Integrating yields the vertical velocity v_y(t) = v₀ sinθ – g t, and the vertical displacement y(t) = h + v₀ sinθ · t – ½ g t².
根据牛顿力学，水平方向的速度分量保持不变，因为没有水平力作用。因此，水平位移随时间 t 的函数为 x(t) = v₀ cosθ · t。垂直运动由恒定的向下加速度 g 控制。积分可得垂直速度 v_y(t) = v₀ sinθ – g t，以及垂直位移 y(t) = h + v₀ sinθ · t – ½ g t²。
These parametric equations fully describe the trajectory. The ball lands when y(t) = 0, which gives a quadratic equation in t. Solving this quadratic yields the flight time T. The range R is then found by substituting T into x(t). Explicitly, we solve: ½ g t² – v₀ sinθ · t – h = 0, where the positive root corresponds to the landing time.
这些参数方程完整描述了运动轨迹。当 y(t) = 0 时球落地，由此得到关于 t 的二次方程。解这个二次方程可得飞行时间 T。然后将 T 代入 x(t) 即可求得射程 R。具体地，我们求解：½ g t² – v₀ sinθ · t – h = 0，其中正根对应着陆时刻。
0 = h + (v₀ sinθ) t – ½ g t²
This is a standard projectile model with an initial height offset. When h = 0, the formula simplifies to the well-known range equation R = (v₀² sin 2θ) / g. Our model is slightly more complex, requiring us to use the quadratic formula to obtain T.
这是一个带有初始高度偏移的标准抛体模型。当 h = 0 时，公式简化为众所周知的射程方程 R = (v₀² sin 2θ) / g。我们的模型稍复杂一些，需要用求根公式来获取 T。
4. Setting Parameters | 参数设定
For our numerical investigation, we fix the following values: initial speed v₀ = 45 m s⁻¹, gravitational acceleration g = 9.8 m s⁻², and tee elevation h = 10 m. The launch angle θ will be varied from 10° to 50° in steps of 5° to observe its effect on range and flight time. All angles are measured in degrees but converted to radians for calculus operations.
在我们的数值研究中，固定以下数值：初速度 v₀ = 45 m s⁻¹，重力加速度 g = 9.8 m s⁻²，发球台高度 h = 10 m。发射角度 θ 将在 10° 到 50° 之间以 5° 为步长变化，以观察其对射程和飞行时间的影响。所有角度以度为单位，但在进行微积分运算时转换为弧度。
These parameters reflect typical conditions on a golf course. A launch speed of 45 m s⁻¹ corresponds to about 162 km/h, which is attainable by skilled amateurs. The 10 m elevation difference is plausible for a tee positioned on a hill overlooking the fairway. We will initially ignore the fact that a real golf ball’s launch angle with a driver is often lower (10°–15°) because of the club’s loft and the ball’s aerodynamic properties.
这些参数反映了高尔夫球场的典型状况。45 m s⁻¹ 的发射速度约合 162 km/h，熟练的业余球手可以达到。10 m 的高度差对于位于小丘上、俯瞰球道的发球台而言是合理的。我们起初会忽略这样一个事实：由于球杆的倾角和球的空气动力学特性，实际一号木开球的角度通常较低（10°–15°）。
5. Calculating Maximum Height | 计算最大高度
The maximum height H_max above the tee ground level occurs when the vertical velocity becomes zero. Setting v_y(t) = 0 gives t_peak = (v₀ sinθ) / g. Substituting this time into the y(t) equation yields H_max = h + (v₀² sin²θ) / (2g). Note that this height is measured from the fairway ground, because the initial height h is included.
在发球台地面以上的最大高度 H_max 出现在垂直速度为零的时刻。令 v_y(t) = 0，得到 t_peak = (v₀ sinθ) / g。将该时间代入 y(t) 方程，可得 H_max = h + (v₀² sin²θ) / (2g)。注意这个高度是从球道地面起算的，因为包含了初始高度 h。
For example, when θ = 40°, v₀ sin 40° ≈ 45 × 0.6428 = 28.925 m s⁻¹. Then t_peak ≈ 28.925 / 9.8 ≈ 2.95 s. The additional height gained is (28.925²) / (2 × 9.8) ≈ 836.6 / 19.6 ≈ 42.7 m. Adding h = 10 m, the highest point is about 52.7 m above the landing area. This illustrates how a significant portion of the ball’s energy is converted into vertical ascent.
例如，当 θ = 40° 时，v₀ sin 40° ≈ 45 × 0.6428 = 28.925 m s⁻¹。那么 t_peak ≈ 28.925 / 9.8 ≈ 2.95 s。额外获得的高度为 (28.925²) / (2 × 9.8) ≈ 836.6 / 19.6 ≈ 42.7 m。加上 h = 10 m，最高点约在落点区域上方 52.7 m 处。这表明球的很大一部分能量转化成了垂直爬升。
If we compare with a shallower angle of 20°, sin 20° ≈ 0.342, so v₀ sin 20° = 15.39 m s⁻¹, giving H_max = 10 + (15.39²)/(19.6) ≈ 10 + 12.1 = 22.1 m. Clearly, higher launch angles produce much steeper, higher trajectories. In the absence of air resistance, this is purely a trade-off between vertical and horizontal velocity components.
如果我们与较平的角度 20° 进行比较，sin 20° ≈ 0.342，因此 v₀ sin 20° = 15.39 m s⁻¹，得到 H_max = 10 + (15.39²)/(19.6) ≈ 10 + 12.1 = 22.1 m。显然，较高的发射角会产生更加陡峭、更高的轨迹。在没有空气阻力的情况下，这纯粹是垂直和水平速度分量之间的权衡。
6. Determining Time of Flight | 确定飞行时间
The total time of flight T is obtained by solving the quadratic equation for t when y(t) = 0: ½ g t² – (v₀ sinθ) t – h = 0. The positive root is given by the quadratic formula:
总飞行时间 T 通过求解 y(t) = 0 时的二次方程得到：½ g t² – (v₀ sinθ) t – h = 0。正根由求根公式给出：
T = [v₀ sinθ + √(v₀² sin²θ + 2 g h)] / g
This expression reduces to T = 2 v₀ sinθ / g when h = 0, which is the symmetric case. The additional term 2 g h inside the square root reflects the fact that the ball must fall an extra vertical distance h, which prolongs the flight. For our parameters, the flight time increases with launch angle but in a non-linear fashion due to the quadratic relationship.
当 h = 0 时，该表达式简化为 T = 2 v₀ sinθ / g，即对称情形。平方根内额外的 2 g h 项反映了球需要多下落一段垂直距离 h，这延长了飞行时间。对于我们的参数，飞行时间随着发射角的增加而增加，但由于二次关系而呈非线性变化。
For θ = 40°, v₀ sinθ = 28.925 m s⁻¹, so v₀² sin²θ ≈ 836.6 and 2 g h = 2 × 9.8 × 10 = 196. The square root becomes √(836.6 + 196) = √1032.6 ≈ 32.13. Then T = (28.925 + 32.13) / 9.8 ≈ 61.055 / 9.8 ≈ 6.23 s. For θ = 20°, v₀ sin 20° = 15.39, v₀² sin²θ ≈ 236.8; the square root is √(236.8 + 196) = √432.8 ≈ 20.80. Thus T ≈ (15.39 + 20.80)/9.8 ≈ 3.69 s. The difference is substantial, showing the steep angle keeps the ball airborne far longer.
对于 θ = 40°，v₀ sinθ = 28.925 m s⁻¹，因此 v₀² sin²θ ≈ 836.6，而 2 g h = 2 × 9.8 × 10 = 196。平方根变为 √(836.6 + 196) = √1032.6 ≈ 32.13。那么 T = (28.925 + 32.13) / 9.8 ≈ 61.055 / 9.8 ≈ 6.23 s。对于 θ = 20°，v₀ sin 20° = 15.39，v₀² sin²θ ≈ 236.8；平方根为 √(236.8 + 196) = √432.8 ≈ 20.80。因此 T ≈ (15.39 + 20.80)/9.8 ≈ 3.69 s。差异巨大，表明陡峭的角度使球的滞空时间远更长。
7. Computing Horizontal Range | 计算水平射程
The horizontal range R is simply the constant horizontal velocity multiplied by the flight time: R = v₀ cosθ × T. Substituting our expression for T leads to a rather unwieldy formula, but numerical evaluation is straightforward. For the symmetric case h = 0, we have R = (v₀² sin 2θ) / g. With h > 0, the range is generally increased because the ball has further to fall and spends more time moving forward.
水平射程 R 就是恒定的水平速度乘以飞行时间：R = v₀ cosθ × T。代入 T 的表达式会得到一个颇为冗长的公式，但数值计算是直接明了的。对于对称情况 h = 0，我们有 R = (v₀² sin 2θ) / g。当 h > 0 时，射程通常会增大，因为球需要下落更远，向前运动的时间更长。
R = (v₀ cosθ / g) × [v₀ sinθ + √(v₀² sin²θ + 2 g h)]
This formula can be programmed into a spreadsheet or calculator to generate values for different θ. Let’s compute a few key angles. For θ = 40°, v₀ cos 40° ≈ 45 × 0.7660 = 34.47 m s⁻¹. With T = 6.23 s, R ≈ 34.47 × 6.23 ≈ 214.7 m. For θ = 20°, v₀ cos 20° ≈ 45 × 0.9397 = 42.29 m s⁻¹, T ≈ 3.69 s, so R ≈ 42.29 × 3.69 ≈ 156.1 m. Thus the steeper angle gives a substantially longer drive.
该公式可编入电子表格或计算器，生成不同 θ 的值。我们来计算几个关键角度。对于 θ = 40°，v₀ cos 40° ≈ 45 × 0.7660 = 34.47 m s⁻¹，T = 6.23 s，则 R ≈ 34.47 × 6.23 ≈ 214.7 m。对于 θ = 20°，v₀ cos 20° ≈ 45 × 0.9397 = 42.29 m s⁻¹，T ≈ 3.69 s，因此 R ≈ 42.29 × 3.69 ≈ 156.1 m。可见较陡的角度带来了显著更远的开球距离。
If we test a very low angle of 10°, the high horizontal speed is offset by a very short flight time, producing R ≈ 120 m. At 45° with h=10 m, we get R ≈ 221.4 m, slightly higher than 40°. This hints that the presence of an initial height shifts the optimal angle downwards from 45°. We will investigate this systematically next.
如果我们测试一个很低的 10° 角，高水平的水平速度被极短的飞行时间所抵消，产生的 R ≈ 120 m。在 h = 10 m、45° 角时，R ≈ 221.4 m，略高于 40°。这暗示初始高度的存在使最佳角度从 45° 向下偏移。我们接下来将系统地探究这一点。
8. Optimising the Launch Angle | 优化发射角度
To find the angle that maximises R in the presence of height h, we can differentiate the range expression with respect to θ and set the derivative to zero, or use numerical methods. The calculus approach is lengthy, so we instead compute R for a fine grid of θ from 0° to 50° and locate the maximum. Using our spreadsheet, we observe that the maximum range occurs near θ = 42° when h = 10 m. For h = 0, the classic optimum is exactly 45°.
为了在存在高度 h 的情况下找到使 R 最大化的角度，我们可以对射程表达式关于 θ 求导并令导数为零，或者使用数值方法。微积分方法较为冗长，因此我们改为对 0° 到 50° 之间的精细角度网格计算 R，并找出最大值。通过电子表格，我们观察到当 h = 10 m 时最大射程出现在 θ = 42° 附近。对于 h = 0，经典最优角恰好是 45°。
For h = 0: dR/dθ = (2 v₀² / g) cos 2θ = 0 ⇒ θ = 45°
When the launch point is elevated, the optimum angle is smaller because a flatter trajectory allows the ball to take better advantage of the additional time afforded by the height difference. The rule of thumb is that the larger the ratio h / (v₀²/(2g)), the further the optimum drops below 45°. In our case, h = 10 m and the “height capability” v₀²/(2g) ≈ 103.3 m, so the ratio is about 0.097, giving a reduction of a few degrees.
当发射点较高时，最优角会更小，因为更平缓的轨迹能让球更好地利用高度差所带来的额外时间。一条经验法则是，比值 h / (v₀²/(2g)) 越大，最优角就越低于 45°。在本案例中，h = 10 m，“高度能力” v₀²/(2g) ≈ 103.3 m，比值约为 0.097，从而使最优角降低了几度。
For a full exploration, students could derive an exact implicit equation for θ_opt by differentiating R and setting the derivative to zero. Alternatively, they could model a sloping fairway, where the landing surface is not horizontal, further complicating the optimisation.
为进行全面探究，学生可以对 R 求导并令导数为零，从而推导出 θ_opt 的精确隐式方程。或者，他们可以模拟倾斜的球道，即着陆面并非水平，这会进一步使优化问题复杂化。
9. The Effect of Elevation Difference | 高度差的影响
The elevation difference between tee and landing area has a profound impact on both flight time and range. To isolate this effect, we can keep θ = 40° fixed and vary h from 0 to 20 m. At h = 0 m, T = 2 × 45 × sin 40° / 9.8 ≈ 5.90 s, and R = (45² sin 80°) / 9.8 ≈ (2025 × 0.9848) / 9.8 ≈ 203.5 m. At h = 10 m, we already found R ≈ 214.7 m, an increase of about 11.2 m. At h = 20 m, with the same angle, R increases further to approximately 225 m.
发球台与落点区域之间的高度差对飞行时间和射程都有深远影响。为了单独考察这一效应，我们可以将 θ = 40° 固定，让 h 从 0 m 变化至 20 m。在 h = 0 m 时，T = 2 × 45 × sin 40° / 9.8 ≈ 5.90 s，R = (45² sin 80°) / 9.8 ≈ (2025 × 0.9848) / 9.8 ≈ 203.5 m。在 h = 10 m 时，我们已求得 R ≈ 214.7 m，增加了约 11.2 m。在 h = 20 m 且角度相同时，R 进一步增加至约 225 m。
This demonstrates that downhill drives can significantly boost distance, which is why elevated tees on golf courses often lead to longer shots. However, a higher h also changes the optimal angle: for h = 20 m, the best angle drops to around 39°. The physical reason is clear: the ball already has gravitational potential energy converted into extra kinetic energy downwards, so it benefits from a more forward-directed initial velocity.
这表明向下坡度的开球可以显著增加距离，这正是高尔夫球场上高处的发球台常常会带来更远击球的原因。然而，较大的 h 也会改变最佳角度：对于 h = 20 m，最佳角下降至约 39°。物理原因很清楚：球已经具有重力势能转化为额外的向下动能，因此它从更向前的初速度中获益。
If the landing area were higher than the tee (uphill drive), the range would be reduced and the optimal angle would increase above 45°. Such symmetry considerations help golfers make informed decisions on club selection and swing plane when faced with varying terrain.
如果落点区域高于发球台（上坡开球），射程将会减小，最佳角度将上升至 45° 以上。这类对称性考虑有助于球手在面临多变地形时，就球杆选择和挥杆平面做出明智的决策。
10. Considering Air Resistance Qualitatively | 定性考虑空气阻力
In reality, a golf ball travelling at high speeds experiences significant drag force, which is approximately proportional to the square of its speed. This force continuously reduces both horizontal and vertical velocity components, resulting in a noticeably shorter range and a less symmetric trajectory. The optimal launch angle for maximum distance in the real world is substantially lower than the vacuum prediction—typically around 12° to 20° for a driver, depending on spin and dimple design.
在现实中，高速飞行的高尔夫球会受到显著的阻力，阻力大小大约与其速度的平方成正比。该力持续减小水平和垂直的速度分量，导致射程显著缩短并使得轨迹不对称。在真实世界中使距离最大化的最佳发射角远低于真空中的预测——对于一号木通常约为 12° 到 20°，具体取决于旋转和凹坑设计。
Incorporating drag makes the equations of motion non-linear and generally requires numerical solution, which is an excellent extension for an IB Mathematical Exploration. Students can model the drag force as F_drag = ½ C_d ρ A v², where C_d is the drag coefficient (~0.25 for a golf ball), ρ is air density (~1.2 kg/m³), and A is the cross-sectional area. They can then use Euler’s method or a spreadsheet to simulate the trajectory and find the optimal angle.
纳入阻力使得运动方程变为非线性，通常需要数值求解，这是 IB 数学探究的一个绝佳拓展。学生可以将阻力建模为 F_drag = ½ C_d ρ A v²，其中 C_d 为阻力系数（高尔夫球约为 0.25），ρ 为空气密度（约 1.2 kg/m³），A 为横截面积。然后他们可以使用欧拉方法或电子表格来模拟轨迹并找出最佳角度。
Additionally, the Magnus effect caused by backspin generates lift, allowing the ball to stay aloft longer, which partially counteracts the range loss due to drag. Professional golfers exploit this by optimising spin rates. Quantifying these effects provides rich opportunities for data collection and mathematical analysis.
此外，由后旋引起的马格努斯效应会产生升力，使球在空中停留更久，部分抵消了由阻力造成的射程损失。职业球手通过优化旋转速率来利用这一点。对这些效应进行量化，为数据收集和数学分析提供了丰富的机会。
11. Conclusion and Further Exploration | 结论与进一步探究
This case study has used algebraic, trigonometric and calculus methods to model the flight of a driven golf ball. We determined that for a tee elevated 10 m above the fairway and an initial speed of 45 m s⁻¹, the optimal vacuum launch angle is about 42°, yielding a range of approximately 221 m. The classic 45° optimum holds only for level terrain. Introducing height differences, air resistance and spin significantly alters these results and reveals the complexity of sports ballistics.
本案例研究运用代数、三角和微积分方法对高尔夫球的开球飞行进行了建模。我们确定，对于高出球道 10 m 的发球台和 45 m s⁻¹ 的初始速度，真空中的最佳发射角约为 42°，产生的射程约为 221 m。经典的 45° 最优角仅适用于平坦地形。引入高度差、空气阻力和旋转会显著改变这些结果，并揭示出运动弹道学的复杂性。
Students can extend this work by recording real launch data using a launch monitor, comparing the results with their theoretical model, and discussing the discrepancies. They could also investigate the impact of wind, altitude (changing g and air density), or the elasticity of the club–ball collision. The progression from a simple idealised equation to a more realistic numerical simulation is a perfect illustration of the mathematical modelling cycle, a key component of the IB Mathematics curriculum.
学生可以通过使用发射监测仪记录真实发射数据来拓展这项工作，将结果与其理论模型进行比较，并讨论差异。他们还可以探究风、海拔（改变 g 和空气密度）或球杆与球碰撞的弹性所带来的影响。从简单的理想化方程推进到更逼真的数值模拟，完美地展示了数学建模循环，这是 IB 数学课程的关键组成部分。
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