How high above sea level do your eyes have to be to see a point that is 4.1 miles away "as the crow flies"? There's a fireworks show going on tonight at a little town that's 4.1 miles away from my house, and I want to watch it from a hill near my house. So I thought I'd set up a simple geometry problem to figure out how high up my eyes need to be, assuming no obstacles & negligible atmospheric reflection. The town in question is 7 ft above sea level, so we don't need to worry about its elevation.
Here are my formulae:
$$\theta = \arccos{ \dfrac{r}{r+h} }$$
$$\text{Distance} = \theta \times 2\pi \times r$$
where h = height above sea level and
r = radius of Earth
After I plugged in an r of $3967.5467$ miles, I deduced that h should be approximately 3.409 inches. However, I think I made a mistake, because even without obstacles, the h I got seems too small.
 A: Let $F$ be the location of the fireworks, $O$ the centre of the Earth, and $H$ the top of the hill. For all practical purposes $HT=4.1$. If $h$ is the required height, then by the Pythagorean Theorem we have $(HT)^2 +(OF)^2=(OT)^2$, and therefore $(r+h)^2=(4.1)^2+r^2=(r+h)^2$.
Expand. We get $r^2+2rh+h^2=r^2+(4.1)^2$. The number $h^2$ is negligibly small compared to $2rh$, so we want $2rh\approx (4.1)^2$. 
Remark: In principle, you don't need much of a hill! However, unless the town is across the bay from you, there is the practical problem of obstructions, such as trees. Avoiding visual obstructions is here  the main practical reason for seeking to be reasonably high up. 
Let us change the problem a little. There are two ships at sea, at distance $d$ miles from each other. The top $T_1$ of the mast of Ship 1 is $h_1$ miles above sea level. What should be the height $h_2$ of the top $T_2$ of the mast of Ship 2  so that $T_1$ is (barely) visible from $T_2$?
The line $T_1T_2$ is now tangent to the Earth. A calculation much like the one above shows that $2r(h_1+h_2)\approx d^2$. 
