Question Number 30738 by NECx last updated on 25/Feb/18
$${A}\:{boy}\:{can}\:{swim}\:{with}\:{a}\:{speed}\:{of} \\ $$$$\mathrm{26}{m}/{s}\:{in}\:{still}\:{water}.{He}\:{wants}\:{to} \\ $$$${swim}\:{across}\:{a}\:\mathrm{150}{m}\:{river}\:{from} \\ $$$${a}\:{point}\:{A}\:{to}\:{point}\:{B}\:{which}\:{is}\: \\ $$$${directly}\:{opposite}\:{the}\:{other}\:{side} \\ $$$${of}\:{the}\:{river}.{The}\:{river}\:{flows}\:{with} \\ $$$${a}\:{speed}\:{of}\:\mathrm{10}{m}/{s}. \\ $$$$\left.{i}\right){if}\:{he}\:{always}\:{swim}\:{in}\:{the}\: \\ $$$${direction}\:{parallel}\:{to}\:{AB},{find}\:{how} \\ $$$${far}\:{he}\:{lands}\:{downstream}\:{of}\:{B}. \\ $$$$\left.{ii}\right){In}\:{what}\:{direction}\:{relative}\:{to} \\ $$$${the}\:{bank}\:{must}\:{he}\:{swim}\:{so}\:{as}\:{to} \\ $$$${cross}\:{directly}\:{from}\:{A}\:{to}\:{B}. \\ $$
Commented by NECx last updated on 25/Feb/18
$${please}\:{help} \\ $$
Answered by mrW2 last updated on 26/Feb/18
$${assume}\:{the}\:{boy}\:{is}\:{a}\:{superman}\:{who} \\ $$$${can}\:{really}\:{swim}\:{with}\:{a}\:{speed}\:{of} \\ $$$$\mathrm{26}\:{m}/{s}\:\left(=\mathrm{93}.\mathrm{6}\:{km}/{h}\right). \\ $$$$\left({i}\right) \\ $$$${a}=\frac{\mathrm{150}}{\mathrm{26}}×\mathrm{10}=\mathrm{57}.\mathrm{7}\:{m} \\ $$$$\left({ii}\right) \\ $$$$\mathrm{cos}\:\theta=\frac{\mathrm{10}}{\mathrm{26}} \\ $$$$\Rightarrow\theta=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{10}}{\mathrm{26}}=\mathrm{67}.\mathrm{4}° \\ $$