 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/ib/paper-1/17d/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

h

, the first part gives

U (h) = U_{0} 1 - 2 g h / U_{0}^{2}

and

A (h) U (h) = A_{0} U_{0}

. Insert these into the assumed local impingement force from part (ii). Static support requires upward force equal to

M g

ρ A_{0} U_{0}^{2} 1 - \frac{2 g h}{U _{0}^{2}} (1 - cos α) = M g .

Equivalently,

h = \frac{U _{0}^{2}}{2 g} [1 - (\frac{M g}{ρ A _{0} U _{0}^{2} ( 1 - c o s α )})^{2}] .

For a nonnegative support height the ratio in parentheses must not exceed one, and the prescribed

h ≪ U_{0}^{2} / (2 g)

requires that ratio to be close to one. If the weight exceeds the lift available at the nozzle, this model has no supported height above the nozzle.

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