Question Number 52818 by Tawa1 last updated on 13/Jan/19 | ||
Answered by tanmay.chaudhury50@gmail.com last updated on 13/Jan/19 | ||
$${V}_{{cart}} =\mathrm{4}{i} \\ $$$$\left({V}^{} \right)_{{cart}} ^{{stone}} ={velocity}\:{of}\:{stone}\:{w}.{r}.{t}\:{man}\left({cart}\right) \\ $$$${V}_{{cart}} ^{{stone}} ={V}_{{earth}} ^{{stone}} −{V}_{{earth}} ^{{cart}} \\ $$$$\mathrm{6}{cos}\mathrm{30}^{{o}} {k}+\mathrm{6}{sin}\mathrm{30}^{\mathrm{0}} {j}+\mathrm{4}{i}={V}_{{earth}} ^{{stone}} \\ $$$${m}_{{stone}} ×{V}_{{earth}} ^{{stone}} =\left({m}_{{stone}} +{m}_{{object}} \right){V}_{{earth}} ^{{combined}} \\ $$$${at}\:{highest}\:{point}\:{V}_{{z}} \:{component}=\mathrm{0}\:{for}\:{stone} \\ $$$${V}_{{earth}} ^{{combined}} \:{at}\:{highest}\:{point}=\frac{\mathrm{0}×{k}+\mathrm{6}{sin}\mathrm{30}^{{o}} {j}+\mathrm{4}{i}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{0}×{k}+\mathrm{3}{j}+\mathrm{4}{i}}{\mathrm{2}}=\mathrm{2}{i}+\frac{\mathrm{3}}{\mathrm{2}}{j}+\mathrm{0}×{k} \\ $$$${speed}\:{combined}=\mid{V}_{{earth}} ^{{combined}} \mid=\sqrt{\mathrm{2}^{\mathrm{2}} +\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} }\:=\sqrt{\mathrm{4}+\mathrm{2}.\mathrm{25}}\:=\mathrm{2}.\mathrm{5}{m}/{sec} \\ $$$$ \\ $$$$ \\ $$$$\left.{b}\right){v}=\sqrt{{gl}}\:\:\:{l}=\frac{{v}^{\mathrm{2}} }{{g}}=\frac{\mathrm{6}.\mathrm{25}}{\mathrm{9}.\mathrm{8}}=\mathrm{0}.\mathrm{64}{meter} \\ $$ | ||
Commented by Tawa1 last updated on 13/Jan/19 | ||
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$ | ||
Commented by tanmay.chaudhury50@gmail.com last updated on 13/Jan/19 | ||
$${is}\:{my}\:{answer}\:{correct}... \\ $$ | ||