Lightning Strikes Twice

Using a good sighting compass and a stopwatch it was observered that a lightning bolt struck the ground 16 ° 16° West of North from my position. The resulting thuder crack was heard 2.8 s 2.8 s later. Another lightning bolt was spotted hitting the ground at 44 ° 44° West of North. The thunder rumble associated with that bolt was heard 9.2 s 9.2 s after seeing the lightning. How far apart were the two lightning strikes expressed in k m km ?

Assume the speed of sound is 340 m s e c 340 \frac{m}{sec}

Image credit: Wikipedia أسامة الطيب


The answer is 2.331.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Darryl Dennis
Jan 12, 2015

The speed of light is so fast that the time of travel of the light from the lightening bolts can be considerd to be instantaneous for this problem.

Distance to bolt = speed of sound in m/sec times the length time sound took to get to my location.

First bolt distance 2.8 times 340 = 952 m

Second bolt distance 9.2 times 340 = 3128 m

Angular separation of sightings = 44 - 16 = 28 deg.

Second bolt vector in direction of first bolt = cos (28) times 3128 = 2761.86

Difference in direction of first bolt 2761.86 - 952 = 1809.68

Second bolt vector at right angle to first bolt = sin(28) times 3128 = 1468.5

Disance between strikes sqrt( 1809.7^2 + 1468.5^2) = 2330.6m

2.302 km (note I originaly posted an incorrect answer to my own question I am hoping to have it corrected)

Second bolt vector in direction of first bolt = cos (28) times 3128 = 2830.0

I get 2761.86 for this operation. What am I doing wrong?

Guiseppi Butel - 6 years, 5 months ago

Log in to reply

You're correct, Guiseppi. The distance should be

( 3128 cos ( 2 8 ) 952 ) 2 + ( 3128 sin ( 2 8 ) ) 2 = 2330.688 \sqrt{(3128\cos(28^{\circ}) - 952)^{2} + (3128\sin(28^{\circ}))^{2}} = 2330.688 m,

which is 2.33 2.33 km to 3 3 significant figures. I used the Cosine Law to solve the problem, such that the desired distance D D is given by

D = 312 8 2 + 95 2 2 2 ( 3128 ) ( 952 ) cos ( 2 8 ) D = \sqrt{3128^{2} + 952^{2} - 2(3128)(952)\cos(28^{\circ})} ,

in essence just a rearrangement of the previous expression.

P.S.. For the knife problem, I found that the younger brother gets 6 6 sheep on his last turn.

Brian Charlesworth - 6 years, 5 months ago

Log in to reply

That's what I found also. My mistake (6 rupees)

Then if the knife belonged solely to the older brother its worth was 4 rupees. If it was owned in concert then its true value was 8 rupees.

How did you get 2?

Guiseppi Butel - 6 years, 5 months ago

Log in to reply

@Guiseppi Butel Assuming the knife belongs to the older brother, since he ends up with 4 4 more rupees than his younger brother he needs to give something of his worth 2 2 rupees to his brother to even things out, since he'll then lose 2 2 rupees of "wealth" and his brother will gain 2 2 . It's the same as if he had just given his brother 2 2 of his actual rupees, but instead he gives him his knife in place of the rupees. Another way of looking at it is that the end of this process, they have both gained the same amount of wealth.

It was a weird question; that knife just came out of nowhere. :)

Brian Charlesworth - 6 years, 5 months ago

Log in to reply

@Brian Charlesworth "I see!" said the blind man. Thanks, I fool so feelish now. :(

Footnote: 'the knife' implies that it was discussed before but it wasn't.

Guiseppi Butel - 6 years, 5 months ago

Log in to reply

@Guiseppi Butel Haha. No problem. I guess 'the knife' was mentioned in the title of the question, but why we should be concerned with it and it's shiny side is beyond me. Perhaps it's a reference to a story in Indian culture that I'm unaware of; a quick google search came up empty, however.

Brian Charlesworth - 6 years, 5 months ago

Thanks, I have updated the answer to 2.331.

Darryl was using radians instead of degrees in his calculation, hence the error.

Calvin Lin Staff - 6 years, 5 months ago

You are absolutely correct. I am having a very bad day with my calculator. I have not used it in years. I neglected to set it to degrees, it was giving me the cos and sin of 28 rad. Very sorry for the errors. I have corrected my solution to show what I believe is the corect answer.

Thanks for pointing this out.

Darryl Dennis - 6 years, 5 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...