Question Number 15740 by tawa tawa last updated on 13/Jun/17
Answered by ajfour last updated on 13/Jun/17
$${Let}\:{cliff}\:{height}\:{be}\:\boldsymbol{{h}}, \\ $$$$\:{Total}\:{time}\:{of}\:{fall}\:{from}\:{cliff} \\ $$$${to}\:{ground}\:={T}\:\:;\: \\ $$$${also}\:{let}\:{me}\:{call}\:\mathrm{1}.\mathrm{30}{s}\:=\:{t}_{\mathrm{0}} \\ $$$$\:{h}=\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \\ $$$$\frac{\mathrm{2}{h}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{2}}{g}\left({t}−{t}_{\mathrm{0}} \right)^{\mathrm{2}} \\ $$$${upon}\:{division}, \\ $$$$\:\:\frac{\mathrm{3}}{\mathrm{2}}=\frac{{t}^{\mathrm{2}} }{\left({t}−{t}_{\mathrm{0}} \right)^{\mathrm{2}} }\:\:\:\Rightarrow\:\:{t}−{t}_{\mathrm{0}} =\left(\sqrt{\frac{\mathrm{2}}{\mathrm{3}}}\right){t} \\ $$$$\:\:\:{or}\:\:\:{t}=\frac{{t}_{\mathrm{0}} }{\mathrm{1}−\sqrt{\mathrm{2}/\mathrm{3}}}\:=\frac{{t}_{\mathrm{0}} \left(\mathrm{1}+\sqrt{\mathrm{2}/\mathrm{3}}\right)}{\left(\mathrm{1}/\mathrm{3}\right)} \\ $$$${h}=\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \:=\:\frac{{g}}{\mathrm{2}}\left(\mathrm{9}{t}_{\mathrm{0}} ^{\mathrm{2}} \right)\left(\frac{\mathrm{5}}{\mathrm{3}}+\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$$\:\:\:=\frac{\mathrm{9}.\mathrm{8}}{\mathrm{2}}\left(\mathrm{9}×\mathrm{1}.\mathrm{3}×\mathrm{1}.\mathrm{3}\right)\left(\frac{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{3}}\right) \\ $$$$\:\:=\left(\mathrm{4}.\mathrm{9}×\mathrm{3}×\mathrm{1}.\mathrm{69}\right)\left(\mathrm{5}+\mathrm{2}×\mathrm{2}.\mathrm{45}\right) \\ $$$$\:\:=\left(\mathrm{14}.\mathrm{7}×\mathrm{1}.\mathrm{69}\right)\left(\mathrm{9}.\mathrm{9}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{h}\:\approx\:\mathrm{246}{m}\:. \\ $$
Commented by tawa tawa last updated on 13/Jun/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$