9. WATER DRAINING FROM A BOTTLE When water flows out of a bottle, the potential energy of the water (Ep = mgh = pAh, where p is the density of water, A is the cross-sectional area of the bottle and h is the height of the water in the bottle) is turned into kinetic energy (Ek = mv² = pa(d)2, where a is the cross-sectional area of the hole out of which the water drains). Energy is conserved, so these quantities must be equal: we can simplify this equation to 1 Ek = Ep dh dt pa( dh dt ² =pAh = 2-h a Let's define the ratio A/a as the constant c, which we could measure. Then, taking the square root of both sides, we get the equation dh /2ch dt where we negative sign arises because we know the height of the water in the bottle decreases with time. a.) Solve the differential equation for h(t) assuming the initial height is h(0) = 1. Don't worry about units, and your answer will involve the known constant c. b.) Of course, the height of water in the bucket should decrease and, eventually, the bucket should become empty. If you've correctly solved part a, you'll find that the height of the water, h(t), does not behave this way. According to your solution from part a, what happens after the bucket becomes empty? This occurs because the differential equation actually has two solutions - the solution you found in part a is not unique. In fact, on problem 4 of homework 1 you looked at a similar problem. c.) Find all solutions to the differential equation d = -√2ch. One set of solutions will be like the one you found in part a but containing a constant, which you would determine from the initial condition; the other solution will be a single, simple solution like the one in problem 4 of homework 1. d.) The "true" solution follows the one you found in part a and then, when it crosses the other solution you found in part c, follows that solution. Sketch this function. Is that behavior consistent with how you'd expect the height of water in a leaky bucket to behave?