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In Exercise, use the differential equation for a leaking container, Eq. dydt=B2gyA(y)\frac{d y}{d t}=-\frac{B \sqrt{2 g y}}{A(y)}

Water leaks through a hole of area B=0.002 m2B=0.002 \mathrm{~m}^2 at the bottom of a cylindrical tank that is filled with water and has height 3 m and a base of area 10 m210 \mathrm{~m}^2. How long does it take (a) for half of the water to leak out and (b) for the tank to empty ?

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The model for a leak of a water from the container that has a hole of area BB in the bottom, and A(y)A(y) is the area of a horizontal cross section at the height yy is

dydt=B2gyA(y)\frac {dy } {dt }=-\frac {B\sqrt { 2gy} } {A(y) }

where gg is acceleration due to gravity.

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