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
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.