A water tank is in the shape of a cone with vertical axis and vertex downward. The tank has a radius of 3m and is high 5m. At first the tank is full of water, but at time t=0 (in seconds), a small hole at the vertex is opened and the water begins to drain. When the height of the water in the tank has dropped to , the water is flowing out at a rate of 2m (cube)/s.
At what rate, in meters per seconds, is the water level dropping then?
I got this: dv/dt = 2m(cube)/s and we're looking for dh/dt.
We also know V=Pi(r^5)(h) /3 and r=3h/5.
However I don't know what to do afterwards

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