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Question Number 8389 by tawakalitu last updated on 09/Oct/16

Answered by ridwan balatif last updated on 10/Oct/16

$$\mathrm{velocity}\mathrm{of}\mathrm{alarm}\mathrm{clock}\mathrm{before}\mathrm{strikes}\mathrm{the}\mathrm{ground}\mathrm{is}$$$$\mathrm{v}=\sqrt{2\mathrm{gh}}$$$$=\sqrt{2\times 10\times 15}$$$$=\sqrt{300}\mathrm{m}/\mathrm{s}$$$$=10\sqrt{3}\mathrm{m}/\mathrm{s}$$$$\mathrm{f}=\frac{340}{340+10\sqrt{3}}\times 800$$$$\mathrm{f}=\frac{27200}{34+\sqrt{3}}$$$$\mathrm{f}=761.22\mathrm{Hz}$$

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