 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-2/5a/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Since

z (t) = h_{I} - α t

A (t) = \frac{π r _{0}^{2} 2 g}{α} h_{I} - α t

until the emptying time

t_{e} = h_{I} / α

. Equivalently,

\frac{A ( t )}{A ( 0 )} = 1 - \frac{t}{t _{e}} .

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