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: 
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](http://www.wolframalpha.com/input/?i=integrate%20sqrt(1%20%2B%20b%5E2*(1-2x)%5E2)%20from%200%20to%201) (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
To expand the above calculations to describe a real golf ball trajectory, we need to include the effects of *drag* and *lift* (the *Magnus* force), when the equations become:

$\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}= - k\_D|\mathbf{v}|\mathbf{v} +
 k\_L\, |\mathbf{v}|(\boldsymbol{\hat \omega}\times \mathbf{v})+ \mathbf{g}, \quad (\text{m/s}^2)$

where $\mathbf{v}(x,y,z)$ is the velocity vector, $\hat \omega$ the unit angular velocity vector, $\mathbf{g}$ gravitational acceleration vector, $t \ge t\_0$ time, and $k\_D$ and $k\_L$ are coefficients for drag and lift forces respectively. These coefficients in turn are dependent upon the balls's spin-rate $S(t)$ and are usually defined empirically by suitable correlations.
It is the lift resulting from golf ball back-spin and surface dimples that largely contributes to the extra distance travelled, over and above that which would be achieved without this effect.

Finally, ambient conditions have to be taken into account, e.g. wind speed, air density, etc.

A fully expanded mathematical and physical description can be found in:

Griffiths (2016), *Numerical Methods Using R*, Cambridge University Press., Chap 9.

The full computer code (in open source language R) to perform various simulations can be downloaded from [www.pdecomp.net](http://www.pdecomp.net)