A large keg of height H and cross-sectional area A1 is filled with root beer. The top is open to the atmosphere. There is a spigot opening of area A2, which is much smaller than A1, at the bottom of the keg. (a) Show that when the height of the root beer is h, the speed of the root beer leaving the spigot is approximately (2 gh)^(1/2) **square root of 2gh** . (b) Show that if A2 << A1, the rate of change of the height h of the root beer is given by dh /dt = −( A2 /A1 )(2gh )^(1/2) . (c) Find h as a function of time if h = H at t = 0. (d) Find the total time needed to drain the keg if H = 2.00 m, A1 = 0.800 m2, and A2 = 1.00 × 10–4 A1. Assume laminar nonviscous flow. I have problems with the answers of (C) and (D). My attempt at (C): I just had to integrate (B) and solve for "h", but my answer is different than the one from the book. Please, help. My answer is: h(t)= ( -(A2/A1)t(2g)^(1/2) + H^1/2 ) ^2, where A2,A1 are the areas, t is time, and (2g)^1/2 is the square root of 2g. My solution for (D) follows from (C), but since my solution for C is presumably wrong, then my solution for D is wrong. Please, help. Thank you.