The height (h) of water flowing out of a tank decreases at a rate proportional to the square root of the height of water still in the tank. If h=9 at t=0 and h=4 at t=5, what is the water’s height at t=15? What is the physical interpretation of this?
Note: time, t, is measured in minutes, and height, h, is measured in metres.
Let k>0, a constant.
The differential equation to be solved is given by: dh/dt = - k(h)^0.5.
Using 'separation of variables' gives the solution: 2(h)^0.5 = - kt + c (where c is an arbitary constant)
Using the given conditions, you can solve to find that: c =6, k = 0.4
Substituing for t=15 gives the final solution: at t=15, h=0 which implies that the tank is completely drained
SN
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