# Speed of light and thunder

1. Nov 26, 2008

### jimen113

1. The problem statement, all variables and given/known data

suppose you hear a clap of thunder 16.2s after seeing the associated lightning stroke. The speed of sound waves in air is 343 m/s and the speed of light in air is 3.00E8 m/s. How far are you from the lightning stroke?

2. Relevant equations

5556.6m is the correct answer, however, why is the speed of light in air ignored?

3. The attempt at a solution
v=$$\frac{d}{t}$$
d=V*T
343 m/s * 16.2s=5556.6m

2. Nov 26, 2008

### chaoseverlasting

Its ignored because the difference is too small. You can assume that the instant you saw the light is the instant the thunder clap took place.

3. Nov 26, 2008

### jimen113

Thanks! it now makes sense

4. Nov 26, 2008

### unscientific

299,792,458 m/s

5. Nov 27, 2008

### Mentallic

Yes unscientific, that IS how fast I can run :tongue2:

Not only is the difference between the actual thunder clap and seeing the lightning so insignificantly small (18.5 microseconds), the variations in the speed of sound in air can alter the true results quite more significantly. Rain/wind etc. can all factor into the speeds.

6. Nov 27, 2008

### HallsofIvy

If d is the distance,in meters, from the lighting to you, it would take the light d/c= (d/3)*10^-8 seconds to reach you. It would take the sound d/s= d/343 seconds to reach you. If you hear the sound 16.2 seconds after seeing the light, you know that d/343- (d/3)*10^-8= 16.2 Multiplying both sides by 343 and 3*10^8, 3(10^8)d- 343d= 1666980000000 or (300000000- 343)d= 299999657d= 1666980000000. d= 1666980000000/299999657= 5556.60635305 meters. Since the original time was only measured to three significant figures, that is no more accurate than 5560 meters.

Essentially, saying the time was "16.2 seconds" implies you are measuring to the nearest tenth of a second. Since it takes light much less than .1 second to reach you, to that accuracy, the time it takes the light to reach you is neglible.