If I hear thunder 5 seconds after I see the lighting, can I calculate the distance to where lighting occurred? My kid asked me this the other day, and it got me thinking that it is really impossible to calculate.  We know the speed of sound (340.29 m/s) and speed of light (299,792,458 m/s) and I can calculate the offset of how far it travelled in 5 seconds... But since we don't know the origination point of the lighting and how long it took to get to me...can we really know the distance where the lighting occurred?
Am I missing something?
 A: You don't need to know the origination point(you just need to know its a fixed point). Now let the distance between you and where lightning occurred be $d$ meters. Since, $$time=\frac{distance}{speed}$$ you can assume the the time lightning takes to reach you be equal to $t_1$ and the time that sound takes be $t_2$, and you are given $t_2-t_1=5$ seconds. Now you have,$$t_1=\frac{d}{299792458}$$ and $$t_2=\frac{d}{340.29}$$ Subtract the equations, $$5=\frac{d}{340.29}-\frac{d}{299792458}$$ You can now solve for $d$. It is not important to take into account the second fraction.(Light reaches almost instantaneously).This  answer is from WolframAlpha $d=1701.45$ meters.
A: Light is so fast it doesn't matter. Lets suppose you start counting when you see the lightning, but you will not count to more than 2 minutes. Sound can travel 340.29 m/s, hence you will only notice lightning / thunder which is less than $$340.29 \frac{\text{m}}{\text{s}}\cdot 2 \cdot 60 \text{s} = 40834.8\text{m}$$ away.
Lightning needs
$$t = \frac{s}{v} = \frac{40834.8\text{m}}{299,792,458 \text{m}/\text{s}} = 0.000136210231145975 \text{s}$$
Even if you take a stopwatch you have to take typical reaction times into consideration. If you are VERY fast, you will need about 10ms = 0.01s (it is rather something like 300ms - test it via http://www.humanbenchmark.com/tests/reactiontime). So even if you take the worst circumstances (it's far away, you react very fast) the error introduced by your reaction time is much higher than the error you get by supposing you react instantly to the light.
A: Even if you're taking into account the fact that the light travel at non infinite distance, you can calculate the origination point.
If you have a first signal that travel at $v_1$, and a second signal that travel at $v_2$. You have this system :
$$\left\lbrace \begin{array}. d = t_1\times v_1 \\ d = t_2 \times v_2 \end{array} \right.$$
And you know $t = t_1-t_2$. This imply that 
$$\left\lbrace \begin{array}. d = t_1\times v_1 \\ 0 = t_1\times v_1-t_2 \times v_2 \end{array} \right.$$
$$\left\lbrace \begin{array}. d = t_1\times v_1 \\ 0 = t_1\times v_1-(t_1-t) \times v_2 \end{array} \right.$$
Hence
$$t_1 = \frac{tv_2}{v_2-v_1}$$
So 
$$ d = \frac{tv_1v_2}{v_2-v_1}$$
A: Quote from Wiki:
A flash of lightning, followed after some time by a rumble of thunder illustrates the fact that sound travels significantly slower than light. Using this difference, one can estimate how far away the bolt of lightning is by timing the interval between seeing the flash and hearing thunder. The speed of sound in dry air is approximately 343 m/s or 1,127 ft/s or 768 mph (1,236 km/h) at 20 °C (68 °F). This translates to 0.213 miles per second or about 5 seconds per mile; however, this figure is only an approximation.
The speed of light is high enough that it can be taken as infinite in this calculation because of the relatively small distance involved. Therefore, the lightning is approximately one kilometer distant for every 3 seconds that elapse between the visible flash and the first sound of thunder (or one mile for every 5 seconds). In the same five seconds, the light could have traveled the Lunar distance four times. (In this calculation, the initial shock wave, which travels at a rate faster than the speed of sound, but only extends outward for the first 30 feet, is ignored.) Thunder is seldom heard at distances over 20 kilometers (12 mi). A very bright flash of lightning and an almost simultaneous sharp "crack" of thunder, a thundercrack, therefore indicates that the lightning strike was very near.

'But since we don't know the origination point of the lighting and how long it took to get to me...can we really know the distance where the lighting occurred?' 
I guess the distance is irrelevant because on considering light speed to be infinite, the distance will be traversed in 0 seconds (That is to say,- Since only approximations are required, you can simplify the problem down to the quote from wiki. You're right only in case of desire of extremely high precision calculations, which I guess is not necessary here...)
