Apparent Height in a Swimming Pool

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A swimmer 0.95 m below the water surface sees an image of a diving board 6.00 m above them due to refraction. The relevant equation for this problem involves the refractive indices and distances from the surface. The swimmer calculated the image distance incorrectly, leading to an incorrect final height of the diving board. The correct approach involves recognizing that the 6.00 m includes both the image distance and the swimmer's depth. Clarification on the variables and their meanings is essential for solving the problem accurately.
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Homework Statement


A person swimming 0.95 m below the surface of the water in a swimming pool looks at the diving board that is directly overhead and sees the image of the board that is formed by refraction at the surface of the water. This image is a height of 6.00 m above the swimmer. What is the actual height of the diving board above the surface of the water?


Homework Equations


(na/s)+(nb/s')=0


The Attempt at a Solution


Rearranging the equation, I found s'=(nb*s)/-na. Plugging in nb=1.00, s=6.00m, and na=1.33, I found s'=-4.51, adding the depth of the swimmer, my final solution was -3.56m. This however is not correct. I also tried solving for s (which was actually my first attempt, although masteringphysics says the variable to solve for is s') to no avail. Any suggestions on where I may have gone wrong?
 
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s is not the image distance from the surface of the water. 6.0 m is the sum of the image distance and the depth of the swimmer.
 
ah, thank you for your help
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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