Question Number 69207 by Rio Michael last updated on 21/Sep/19
$${help}\:{please}. \\ $$$$ \\ $$$${A}\:{river}\:{is}\:\mathrm{5}{m}\:{wide}\:{and}\:{flows}\:{at}\:\mathrm{3}.\mathrm{0}{ms}^{−\mathrm{1}} .\:{A}\:{man}\:{can}\:{swim}\:{at}\:\mathrm{2}.\mathrm{0}{ms}^{−\mathrm{1}} \\ $$$${in}\:{still}\:{water}.\:{if}\:{he}\:{sets}\:{off}\:{at}\:{an}\:{angle}\:{of}\:\mathrm{90}°\:{to}\:{the}\:{bank} \\ $$$${calculate} \\ $$$$\left.{a}\right)\:{the}\:{mans}\:{time}\:{and}\:{velocity} \\ $$$$\left.{b}\right)\:{his}\:{distance}\:{downstream}\:{from}\:{the}\:{starting}\:{point}\:{till} \\ $$$${when}\:{he}\:{reaches}\:{the}\:{other}\:{side}\:{of}\:{the}\:{river}\:{bank} \\ $$$$\left.{c}\right)\:{the}\:{actual}\:{distance}\:{he}\:{swims}\:{through}\:{the}\:{water}. \\ $$
Commented by mr W last updated on 21/Sep/19
$${when}\:{the}\:{river}\:{is}\:{only}\:\mathrm{5}\:{m}\:{wide},\:{i}'{ll} \\ $$$${try}\:{to}\:{jump}\:{to}\:{the}\:{other}\:{side}\:{directly} \\ $$$${instead}\:{of}\:{swim}. \\ $$
Commented by Rio Michael last updated on 21/Sep/19
$${lol}\:{man}\:{its}\:{just}\:{a}\:{question} \\ $$
Commented by Prithwish sen last updated on 21/Sep/19
$$\mathrm{ha}\:\mathrm{ha}\:\mathrm{ha}\:!\:\mathrm{Yes}\:\mathrm{mr}.\mathrm{W}\:\mathrm{sir}\:\mathrm{I}\:\mathrm{agree}\:\mathrm{with}\:\mathrm{you}. \\ $$
Answered by mr W last updated on 21/Sep/19
$$\left.{a}\right) \\ $$$${t}=\frac{\mathrm{5}}{\mathrm{2}}=\mathrm{2}.\mathrm{5}\:{sec} \\ $$$${v}=\sqrt{\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }=\sqrt{\mathrm{13}}\:{m}/{s} \\ $$$$\left.{b}\right) \\ $$$$\mathrm{2}.\mathrm{5}×\mathrm{3}=\mathrm{7}.\mathrm{5}\:{m} \\ $$$$\left.{c}\right) \\ $$$$\sqrt{\mathrm{5}^{\mathrm{2}} +\mathrm{7}.\mathrm{5}^{\mathrm{2}} }=\frac{\mathrm{5}\sqrt{\mathrm{13}}}{\mathrm{2}}\approx\mathrm{9}\:{m} \\ $$
Commented by Rio Michael last updated on 21/Sep/19
$${thanks}\:{sir} \\ $$