Question Number 94602 by peter frank last updated on 19/May/20
Answered by mr W last updated on 21/May/20
Commented by mr W last updated on 21/May/20
$${at}\:{time}\:{t}: \\ $$$${x}_{{M}} ={d} \\ $$$${y}_{{M}} ={h}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \\ $$$${x}_{{B}} ={u}\:\mathrm{cos}\:\theta\:{t} \\ $$$${y}_{{B}} ={u}\:\mathrm{sin}\:\theta\:{t}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \\ $$$${such}\:{that}\:{the}\:{bullet}\:{meets}\:{the}\:{monkey}, \\ $$$${u}\:\mathrm{cos}\:\theta\:{t}={d} \\ $$$$\Rightarrow{t}=\frac{{d}}{{u}\:\mathrm{cos}\:\theta} \\ $$$${u}\:\mathrm{sin}\:\theta\:{t}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} ={h}−\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \\ $$$${u}\:\mathrm{sin}\:\theta\:×\frac{{d}}{{u}\:\mathrm{cos}\:\theta}={h} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=\frac{{h}}{{d}} \\ $$$${this}\:{is}\:{true}\:{if}\:{the}\:{gun}\:{is}\:{directed}\:{to} \\ $$$${the}\:{monkey}\:{as}\:{it}\:{fires}\:{the}\:{bullet}. \\ $$$${that}\:{means}\:{the}\:\boldsymbol{{bullet}}\:\boldsymbol{{can}}\:\boldsymbol{{always}}\: \\ $$$$\boldsymbol{{meet}}\:\boldsymbol{{the}}\:\boldsymbol{{monkey}}. \\ $$$$ \\ $$$${such}\:{that}\:{the}\:{bullet}\:{meets}\:{the}\:{monkey} \\ $$$${in}\:{the}\:{air}: \\ $$$${h}−\frac{\mathrm{1}}{\mathrm{2}}{g}\left(\frac{{d}}{{u}\:\mathrm{cos}\:\theta}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$${u}^{\mathrm{2}} \geqslant\frac{{gd}^{\mathrm{2}} }{\mathrm{2}{h}}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta\right)=\frac{{gd}^{\mathrm{2}} }{\mathrm{2}{h}}\left(\mathrm{1}+\frac{{h}^{\mathrm{2}} }{{d}^{\mathrm{2}} }\right)=\frac{{g}\left({d}^{\mathrm{2}} +{h}^{\mathrm{2}} \right)}{\mathrm{2}{h}} \\ $$$$\Rightarrow{u}\geqslant\sqrt{\frac{{g}\left({d}^{\mathrm{2}} +{h}^{\mathrm{2}} \right)}{\mathrm{2}{h}}} \\ $$
Commented by peter frank last updated on 22/May/20
$$\mathrm{thank}\:\mathrm{you} \\ $$