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Velocity of image of submerged object when liquid surface moves

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Speed of the image (as seen by an observer in air) of the coin if the liquid surface is raised with speed $8\ m/s$

I know how to deal with when the object is moving but how to do when surface moves?

asked Apr 8 in Physics Problems by n3 (508 points)
edited Apr 8 by sammy gerbil
If the distance between observer and object is fixed, you can write an equation for the distance $y$ of the image from the observer in terms of the depth $h$ of the liquid, using the apparent depth concept. The velocity of the image is $dy/dt$ whereas the velocity of the surface of the liquid is $dh/dt=8m/s$.
Using apparent depth, $y=\frac{h}{4/3}\implies dy/dt=8\cdot\frac{3}{4}=6$, but they say it to be $2$.

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Suppose the observer is a fixed distance $H$ above the coin. Then the distance from the observer to the water surface is $H-h$ and the apparent distance of the coin below the surface is $\frac34 h$. So the apparent distance of the coin from the observer (measuring downwards) is $$y=(H-h)+\frac34h=H-\frac14h$$

The velocity of the water surface is $\frac{dh}{dt}=+8m/s$ (plus because $h$ is increasing) so the rate at which the image of the coin is moving downwards (in the direction of increasing y) is $$\frac{dy}{dt}=-\frac14\frac{dh}{dt}=-2m/s$$ The minus sign indicates that the image is moving upwards towards the observer, whereas y increases downwards.

answered Apr 8 by sammy gerbil (28,746 points)
selected Apr 8 by n3
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