Question Number 177055 by ajfour last updated on 30/Sep/22
Commented by ajfour last updated on 30/Sep/22
$${Our}\:{boy}\:{throws}\:{a}\:{ball}\:{with}\:{speed}\:{u} \\ $$$${and}\:{starts}\:{climbing}\:{down}\:{at}\:{same} \\ $$$${speed}\:{u}\:{and}\:{runs}\:{forward}\:{to}\: \\ $$$${catch}\:{the}\:{ball},\:{and}\:{does}\:{so}\:{success} \\ $$$$-{fully}.\:{Find}\:\theta\:{in}\:{terms}\:{of}\:{u},{g},{h}. \\ $$
Commented by peter frank last updated on 30/Sep/22
$$\mathrm{how}\:\mathrm{do}\:\mathrm{you}\:\mathrm{draw}\:\mathrm{this}? \\ $$
Commented by mr W last updated on 01/Oct/22
$${every}\:{body}\:{can}\:{draw}\:{with}\:{his}\:{smart} \\ $$$$\left.{phone}.\:{also}\:{i}\:{can}\:{do}\::\right).\:\:{but}\:{whether} \\ $$$${you}\:{can}\:{draw}\:{as}\:{nicely}\:{as}\:{ajfour}\:{sir},\: \\ $$$${is}\:{an}\:{other}\:{question}. \\ $$
Commented by mr W last updated on 01/Oct/22
Commented by peter frank last updated on 01/Oct/22
$$\mathrm{what}\:\mathrm{the}\:\mathrm{name}\:\mathrm{of}\:\mathrm{the}\:\mathrm{app} \\ $$
Commented by mr W last updated on 02/Oct/22
$${every}\:{smart}\:{phone}\:{has}\:{an}\:{app}\:{for} \\ $$$${drawing}/{editing}\:{pictures}!\:{the}\:{name} \\ $$$${of}\:{the}\:{app}\:{doesn}'{t}\:{matter}. \\ $$
Answered by mr W last updated on 30/Sep/22
$${ut}={h}+{u}\:\mathrm{cos}\:\theta\:{t} \\ $$$$\Rightarrow{t}=\frac{{h}}{{u}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)} \\ $$$${u}\:\mathrm{sin}\:\theta\:{t}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}=−{h} \\ $$$${u}\:\mathrm{sin}\:\theta×\frac{{h}}{{u}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}−\frac{{g}}{\mathrm{2}}×\left(\frac{{h}}{{u}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}\right)^{\mathrm{2}} =−{h} \\ $$$$\mathrm{sin}\:\theta+\mathrm{1}−\mathrm{cos}\:\theta−\frac{{gh}}{\mathrm{2}{u}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\:\theta\right)}=\mathrm{0} \\ $$$$\mathrm{sin}\:\frac{\theta}{\mathrm{2}}\:\mathrm{cos}\:\frac{\theta}{\mathrm{2}}+\mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}−\frac{{gh}}{\mathrm{8}{u}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}}=\mathrm{0} \\ $$$${with}\:\lambda=\frac{{gh}}{\mathrm{8}{u}^{\mathrm{2}} } \\ $$$$\mathrm{sin}^{\mathrm{3}} \:\frac{\theta}{\mathrm{2}}\:\mathrm{cos}\:\frac{\theta}{\mathrm{2}}+\mathrm{sin}^{\mathrm{4}} \:\frac{\theta}{\mathrm{2}}−\lambda=\mathrm{0} \\ $$$$\mathrm{sin}^{\mathrm{3}} \:\frac{\theta}{\mathrm{2}}\:\mathrm{cos}\:\frac{\theta}{\mathrm{2}}=\lambda−\mathrm{sin}^{\mathrm{4}} \:\frac{\theta}{\mathrm{2}} \\ $$$$\mathrm{sin}^{\mathrm{6}} \:\frac{\theta}{\mathrm{2}}\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}\right)=\lambda^{\mathrm{2}} −\mathrm{2}\lambda\:\mathrm{sin}^{\mathrm{4}} \:\frac{\theta}{\mathrm{2}}+\mathrm{sin}^{\mathrm{8}} \:\frac{\theta}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}\:\mathrm{sin}^{\mathrm{8}} \:\frac{\theta}{\mathrm{2}}−\mathrm{sin}^{\mathrm{6}} \:\frac{\theta}{\mathrm{2}}−\mathrm{2}\lambda\:\mathrm{sin}^{\mathrm{4}} \:\frac{\theta}{\mathrm{2}}−\lambda^{\mathrm{2}} =\mathrm{0} \\ $$$${example}: \\ $$$${u}=\mathrm{5}\:{m}/{s},\:{h}=\mathrm{5}{m}\:\Rightarrow\lambda=\frac{\mathrm{10}×\mathrm{5}}{\mathrm{8}×\mathrm{5}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\theta\approx\mathrm{68}.\mathrm{7325}° \\ $$
Commented by ajfour last updated on 30/Sep/22
$${Thanks}\:{sir},\:{dont}\:{seem}\:{a}\:{simpler} \\ $$$${way}\:{out}.. \\ $$