How to calculate the actual distance flown by a golf ball?

Question

I am interested in finding out what the actual distance is flown by a golf ball on a drive. You can assume no wind, 10 degree launch angle, goes straight down the middle of the fairway (no draw, no slice...) and no erratic backspin on the ball (so it wont "balloon" but will have stable flight). You can also assume that the ball flies 200 yards of linear distance (relative to the ground). So basically I would like to know how much "extra" distance does the ball fly relative to the 200 yards of linear distance? I was thinking somewhere on the order of 15% more (30 yards more) but that is just a guess.

The answer need not be exact.

Also I am not sure what tag I should use so if someone has a better one go ahead and change it or just tell me and I will change it.

Answer 1

With the assumptions that you make, the golf ball is basically free falling without losing horizontal speed and is therefore described by a parabola: \begin{equation} y=-ax^2+bx+c, \end{equation} where $x$ is the horizontal distance and $y$ is the height of the ball. We know that when $x=\pm 100$, the ball is at the ground, so \begin{equation} 0=-a*10000+b*100+c \end{equation} and \begin{equation} 0=-a*10000-b*100+c \end{equation} If we take the sum and the difference of these formulas, we get \begin{equation} 0=-a*20000+2*c \qquad \text{and}\qquad 0=b*200. \end{equation} This gives us two relations, resulting in $b=0$ and $c=10000*a$. The question is now to determine $c$ or $a$ and we have one given fact left: the 10 degree launch angle. How would we translate this launch angle into mathematics? A 10 degree angle corresponds to a slope of \begin{equation} \frac{dy}{dx}=\tan(10)\approx 0.176. \end{equation} This must be the slope at $x=-100$. So we get \begin{equation} \frac{dy}{dx}|_{x=-100}=-2ax+b|_{x=-100}=200a=0.176 \end{equation} So, we find $a=0.176/200=0.00088$ and we now have the total trajectory of the ball \begin{equation} y=-0.00088 x^2 +8.8. \end{equation} We can use this formula to find the length of the path that the ball follows. We also see immediately that 8.8 is the maximal height that the ball reaches. The length of the trajectory can be calculated by \begin{equation} \int_{-100}^{100} \sqrt{1+(\frac{dy}{dx})^2} dx=\int_{-100}^{100} \sqrt{1+(-2*0.00088*x)^2} dx \approx 201.028 \end{equation} This last integral is easily done with this link! or some graphical calculator. Or you could it algebraically as \begin{equation} \int_{-100}^{100} \sqrt{1+(-2*0.00088*x)^2} dx = \left(-2*0.00088 x \sqrt{1+(-2*0.00088)^2 x^2}+\frac{\sinh^{(-1)}{(-2*0.00088 x)}}{-4*0.00088}\right)\Big|_{-100}^{100}. \end{equation}

In short, 201.028 is the total length of the ball's trajectory. So, it does only 1.028 extra yards. This is not so much as you expected, so maybe the 10 degree angle is a bit on the small side.

You can probably retrace the steps above to find the trajectory length for another launch angle.

Answer 2

This is a non mathematical answer but still mildly interesting and worth mentioning. For someone that really wanted to know this but didn't have the math skills to solve it (like me), I suppose someone could record the flight of the ball on their smartphone such that the entire ballflight could be recorded. Then they would take a piece of string and trace the ballflight on the smartphone screen. Then they would simply measure the length of the string relative to the length of the flat path (along the ground). Solutions like this is why I didn't major in math in college. I would always want to do things in a simpler (non mathematical) way.

Answer 3

Suppose the curve that a ball moves is given by a parabola. This would actually be the case, if we conducted such an experiment in a vacuum - in the real world, air resistance etc. would make the problem much more complicated (but this approximation will suffice for us).

Now we launch the ball in our coordinate system at $x = 0$ and at an angle of $\alpha$. and the ball lands after a horizontal distance of $d$. Our model function $f(x) = ax^2 + bx + c$ must satisfy three conditions:

1. $f(0) = 0$
2. $f'(0) = \tan \alpha$
3. $f(d) = 0$

From the first condition we can conclude $c = 0$. From the second condition, we get $f'(0) = b = \tan \alpha$ and our third condition yields $0 = f(d) = ad^2 + bd$, i.e. $a = -\frac{b}{d}$. This means our model function is $f(x) = -\frac{b}{d}x^2 + bx$ with $b = \tan \alpha$.

Now we want to calculate the arc length of $f$ over the interval $[0, d]$, the formula for this is given by $$L = \int_0^d \sqrt{1 + f'(x)^2}\, dx = \int_0^d \sqrt{1 + \left(b - 2\frac{b}{d}x\right)^2} \, dx = d \cdot \int_0^1 \sqrt{1 + b^2(1 - 2x)^2} \, dt = d \frac{b \sqrt{1 + b^2} + \sinh^{-1}(b)}{2b}$$

The last integral was calculated via Wolfram alpha (because I am lazy).

Now if we actually plug in $\alpha = 10^\circ$ we get $b = \tan 10^\circ \approx 0.176$ and the golf ball travelled a length of approximately $1.005d$, which is $0.5\%$ more than the horizontal distance