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Question-193391

Question Number 193391 by DAVONG last updated on 12/Jun/23 Answered by aba last updated on 12/Jun/23 $$\left(\mathrm{1}\right)\:\mathrm{log}_{\mathrm{a}} \left(\mathrm{6}\right)−\mathrm{log}_{\mathrm{a}} \left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:\mathrm{log}_{\mathrm{a}} \left(\frac{\mathrm{6}}{\mathrm{x}}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{ln}\left(\frac{\mathrm{6}}{\mathrm{x}}\right)=\mathrm{ln}\left(\sqrt{\mathrm{a}}\right)\:\Rightarrow\:\mathrm{x}=\frac{\mathrm{6}}{\:\sqrt{\mathrm{a}}}\:\checkmark \\ $$ Commented…

Question-193345

Question Number 193345 by yaslm last updated on 11/Jun/23 Answered by aleks041103 last updated on 13/Jun/23 $${The}\:{trajectry}\:{of}\:{the}\:{water}\:{is}\:{a}\:{freefall}\:{parabolla} \\ $$$$\Rightarrow{H}=\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \Rightarrow{t}=\sqrt{\frac{\mathrm{2}{H}}{{g}}} \\ $$$${D}={vt}={v}\sqrt{\frac{\mathrm{2}{H}}{{g}}} \\ $$$$\Rightarrow{v}=\sqrt{\frac{{gD}^{\mathrm{2}} }{\mathrm{2}{H}}}…