Weekend Assignment sheet - question

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In summary, the question involves an observer located at an unknown distance from a cannon and a cannonball fired at 425 m/s. The sound is heard 0.20s after the cannonball passes the observer. Using the equations speed = distance * time and the speed of sound being 345 m/s, the solution can be found by setting two equations equal to each other and solving for time, then plugging that value back into the original equations to find the distance between the cannon and the observer.
  • #1
mike_302
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Weekend Assignment sheet -- question

Homework Statement



An observer is located at a unknown distance away from a cannon. A cannonball is fired from the cannon at 425 m/s. If the cannonball goes past the observer 0.20s before the sound is heard, how far away is the cannon from the observer.


Homework Equations



speed= distance * time ?


The Attempt at a Solution




I don't even know where to begin on this one. I can SEE that SOMEHOW you are comparing the two speeds to tell how far the SOUND went while the cannonball was travelling, but I don't have a very clear idea of how to answer this one in my head. I don't necessarily want the answer to this as much as I do the way to go about finding it. Thanks!
 
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  • #2
were you given the speed of sound?
 
  • #3
OH! Jeeze... Yes, I forgot that part, 345 m/s.. SORRY! :S lol
that was the ONLY sentence I was missing... "The speed of sound is 345 m/s"
 
  • #4
mike_302 said:
OH! Jeeze... Yes, I forgot that part, 345 m/s.. SORRY! :S lol
that was the ONLY sentence I was missing... "The speed of sound is 345 m/s"
Ok, so look again at your relevant equation. It's mixed up a bit.
Once you correct it, set up 2 equations, one using the speed of sound, the other using the speed of the cannonball, and note that the sound is heard 0.2 seconds after the cannonball whizzes by the observer in time 't'.
 
  • #5
345*(tsound)=d
425*(tcannonball)=d
tcannonball=tsound+.2s or tsound=tcannonball-.2s

This is what I'm understanding but... Struggling to piece together. Am I on the right track? Can you point me in the right direction with these equations if I am? Thanks... I'll get this!, just with a little help required.
 
  • #6
mike_302 said:
345*(tsound)=d
425*(tcannonball)=d
tcannonball=tsound+.2s or tsound=tcannonball-.2s

This is what I'm understanding but... Struggling to piece together. Am I on the right track? Can you point me in the right direction with these equations if I am? Thanks... I'll get this!, just with a little help required.
You are on track...but which takes longer to reach the observer, the sound or the cannonball? Mind your plus and minus signs, they'll wear you down if you don't. Then can you solve the 2 equations with 2 unknowns?
 
  • #7
wow.. this is proving to be further out of my reach than I can handle... can someone possibly go one step further and provide the equations I need in order to complete this, and how I would use them together (I can do the actual math at that point).

I know this is an assignment and all but... At this point, I am much more intrigued at the theory behind answering this question than anything so... Thanks for your help so far! Hope to hear back soon.
 
  • #8
I have these two equations but I don't know how correct they are:

345*(t+0.20)=d which is equal to 345t+69=d
425*t=d
 
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  • #9
I believe they are correct. Now, since the distance 'd' for both equations is the same, i.e. d = d, set the two equations equal to each other and solve for 't'.

345*t + 69 = 425*t

To be sure you have it correct, after you solve for 't', put that answer back in the original equations, and the solution 'd' should be equal.

Good luck.
 
  • #10
HA! wow! that was silly of me. How didn't I notice that and put them together. Thanks for your help! appreciate it.
 
  • #11
ya.. ha, i just noticd that... thanks. annnd it worked out.. thanks very much for your help! I guess I had the right idea mathematically, but i was too sily to see that last little piece.. d=d ... This site is DEFIANTELY friendly and helpful
thanks all
 

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