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Find the time interval between the above two events. 
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#1
Jan813, 10:27 PM

P: 6

1. The problem statement, all variables and given/known data
A foreign fishing submarine moves straight north through Canadian waters at a constant velocity of 12 m/s and a constant depth of 150m below the surface. A Canadian helicopter is in pursuit, flying in exactly the same direction at a constant velocity of 52 m/s and a constant altitude of 550m above the surface. The helicopter, lightly armed, shoots a narrow laser beam which enters the water and strikes the submarine. At one instant, the laser beam leaving the helicopter is directed at an angle of 10° below the horizontal to hit the target. At a later time, the beam has to be tilted down to 30° below the horizontal to maintain contact. Given that the index of refraction is 1.00 for air and 1.33 for water, find the time interval between the above events. Possible Answers; a) 33.3 b) 55.1 c) 68.9 d) 46.7 e) 78 2. Relevant equations n1xsin∅1 = n2xsin∅2 3. The attempt at a solution I found the two refractive angles; 1xsin10° = 1.33xsin∅1 ∅1 = 7.5° 1xsin30° = 1.33xsin∅2 ∅2 = 22.08° After that, I'm stuck. I have no idea how to use that to find the time interval between each degree. 


#2
Jan813, 10:38 PM

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P: 4,761

Hello 7even. Welcome to PF!
In Snell's law are the angles measured from the surface or from the normal to the surface? 


#3
Jan813, 10:40 PM

P: 6

From the horizontal, so yes, the surface!



#4
Jan813, 10:45 PM

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P: 4,761

Find the time interval between the above two events.
See here



#5
Jan813, 10:51 PM

P: 6

I see, so those angles should be changed from 10 to 80 and 30 to 60.
Therefore ∅1 = 47.77° and ∅2=40.63 


#6
Jan813, 10:52 PM

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P: 4,761

Yes. good. Consider the first firing of the laser. Can you figure out the horizontal distance between the helicopter and submarine at this instant?



#7
Jan813, 10:56 PM

P: 6

Create a right angle triangle; use the vertical length between the submarine and the helicopter and solve for the horizontal distance.
Therefore, tan10°=700m/x ... x = 3969.9m 


#8
Jan813, 11:04 PM

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P: 4,761

No. You're going to need two right triangles. One for the helicopter and one for the submarine.



#9
Jan813, 11:09 PM

P: 6

I see. For the first instant;
tan10°=550m/x .... x = 3,119.2m tan47.77°=x/150m ... x = 165.25m 


#10
Jan813, 11:16 PM

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So you can figure out the total horizontal distance between the two at the first firing. Repeat for the second firing and see if you can use these distances to determine the time.



#11
Jan813, 11:28 PM

P: 6

Oh I get it, thanks so much for the help!



#12
Jan813, 11:36 PM

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If you were on the submarine how far away horizontally from you was the helicopter at the first firing? How far away horizontally was the helicopter away from you at the second firing? Thus, conclude how much closer the helicopter got (horizontally) to the submarine between the two firings. 


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