Question Number 55191 by Tawa1 last updated on 19/Feb/19
Answered by tanmay.chaudhury50@gmail.com last updated on 19/Feb/19
$$\left.\mathrm{18}\right){at}\:{greatest}\:{heught}\:{vertical}\:{component}\:{of} \\ $$$${velocity}\:{is}\:{zero} \\ $$$$\mathrm{0}^{\mathrm{2}} =\left({usin}\theta\right)^{\mathrm{2}} −\mathrm{2}{gh}_{\mathrm{1}} \\ $$$$\mathrm{0}^{\mathrm{2}} =\left\{{usin}\left(\mathrm{90}−\theta\right)\right\}^{\mathrm{2}} −\mathrm{2}{gh}_{\mathrm{2}} \\ $$$$\frac{\mathrm{2}{gh}_{\mathrm{1}} }{\mathrm{2}{gh}_{\mathrm{2}} }=\frac{{u}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta}{{u}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta} \\ $$$${tan}\theta=\sqrt{\frac{{h}_{\mathrm{1}} }{{h}_{\mathrm{2}} }}\: \\ $$
Commented by Tawa1 last updated on 19/Feb/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 19/Feb/19
$$\left.\mathrm{19}\right){Range}=\left({ucos}\theta\right)×\left({time}\:{of}\:{flight}\right) \\ $$$${time}\:{of}\:{flight}={T}=\mathrm{2}{t}_{\mathrm{0}} \\ $$$$\mathrm{0}={usin}\theta−{gt}_{\mathrm{0}} \\ $$$${t}_{\mathrm{0}} =\frac{{usin}\theta}{{g}} \\ $$$${R}=\left({ucos}\theta\right)\left(\frac{\mathrm{2}{usin}\theta}{{g}}\right)=\frac{{u}^{\mathrm{2}} {sin}\mathrm{2}\theta}{{g}} \\ $$$${R}_{{max}} \:{when}\:{sin}\mathrm{2}\theta=\mathrm{1}={sin}\frac{\pi}{\mathrm{2}} \\ $$$$\theta=\frac{\pi}{\mathrm{4}} \\ $$$${sin}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\rightarrow{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\rightarrow\frac{\pi}{\mathrm{4}} \\ $$
Commented by Tawa1 last updated on 19/Feb/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$