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# Can anyone help with this seismic reflection problem? watch

1. I know that a reflection from a horizontal reflector arrives at a coincident source and reciever at a two-way time of 0.133 seconds. At a reciever 200m away from the source the same reflection has a two-way time of 0.149 seconds.

I have to work out the velocity to the reflector.

So far I've drawn a diagram that looks like the image attached.

Everything I try and work out I end up not having all the information to do it. If anyone could explain how to do it I would be very grateful.
Attached Files
2. Seismic reflection.docx (27.4 KB, 71 views)
3. You have a large right angled triangle in the diagram and can apply Pythagoras' Theorem for the length of the 3 sides.
Top side is 200m
Hypotenuse (red) is 0.149 x speed of wave
Other side (green) is 0.133 x speed of wave.
4. (Original post by Stonebridge)
You have a large right angled triangle in the diagram and can apply Pythagoras' Theorem for the length of the 3 sides.
Top side is 200m
Hypotenuse (red) is 0.149 x speed of wave
Other side (green) is 0.133 x speed of wave.
So:

(0.149xV)2 = 2002 + (0.133xV)2

and I have to use that to work out what V is?

Thank you for trying to help. Sorry, I'm really not good at maths and physics and haven't done any of either for nearly 4 years so I may sound really stupid.
5. That's correct.
Multiply out the terms (you'll get v squared) and collect the v squareds on one side of the equation.
(0.149 x 0.149) x V^2 = 200^2 + (0.133 x 0.133) x V^2 to start you off.
Then
(0.0222)V^2 + (0.17689)V^2 = 200^2
then
( )V^2=200^2
Then
V^2 =

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