A cylindrical can of height h and base area A is immersed in water to a depth ho and left to sink down. A small hole of area 'a' exists at the bottom of the base of the can. Determine how quickly the can sinks.
Thus, vo = sqrt(2gy)
AV = avo
V: velocity of fluid w.r.t. the container
The Attempt at a Solution
dy/dt = (a/A)sqrt(2gy)
or, dy/sqrt(y) = (a/A) sqrt(2g) dt
or, on integrating from ho to h, i get
t = sqrt(2/g)[sqrt(h) - sqrt(ho)]A/a
I am not getting the answer, where did i go wrong.