Question Number 13681 by Tinkutara last updated on 22/May/17
Commented by Tinkutara last updated on 22/May/17
$$\mathrm{Please}\:\mathrm{answer}\:\mathrm{all}\:\mathrm{the}\:\mathrm{questions}. \\ $$
Commented by ajfour last updated on 22/May/17
$${not}\:{clickable},\:{upload}\:{again}.. \\ $$
Commented by prakash jain last updated on 22/May/17
$$\mathrm{I}\:\mathrm{think}\:\mathrm{you}\:\mathrm{should}\:\mathrm{use}\:\mathrm{camscanner} \\ $$$$\mathrm{for}\:\mathrm{taking}\:\mathrm{pictures}\:\mathrm{of}\:\mathrm{printed} \\ $$$$\mathrm{material}. \\ $$
Answered by ajfour last updated on 22/May/17
$${take}\:{upward}\:{direction}\:{positive}. \\ $$$$\boldsymbol{{a}}_{{A}} =−\boldsymbol{{g}} \\ $$$$\boldsymbol{{a}}_{\boldsymbol{{B}}} =−\boldsymbol{{g}} \\ $$$$\boldsymbol{{a}}_{\boldsymbol{{A}}} −\boldsymbol{{a}}_{\boldsymbol{{B}}} =\mathrm{0}\: \\ $$$$ \\ $$$$\boldsymbol{{v}}_{\boldsymbol{{A}}} =\boldsymbol{{u}}−\boldsymbol{{gt}}\:\:\:;\:\:\boldsymbol{{v}}_{\boldsymbol{{B}}} =−\boldsymbol{{gt}} \\ $$$${v}_{{A}} −{v}_{{B}} =\left({u}−{gt}\right)−\left(−{gt}\right)\:=\boldsymbol{{u}} \\ $$$$ \\ $$$${separation}\:=\:\boldsymbol{{h}}−\boldsymbol{{ut}}\: \\ $$$$\:\:\:\:\:\:\:\:\:=\boldsymbol{{initial}}\:\boldsymbol{{separation}}\:−\:\left(\boldsymbol{{relative}}\:\boldsymbol{{velocity}}\right)\left(\boldsymbol{{time}}\right)\:. \\ $$
Commented by Tinkutara last updated on 23/May/17
$$\mathrm{Can}\:\mathrm{you}\:\mathrm{explain}\:\mathrm{your}\:\mathrm{last}\:\mathrm{line}?\:\mathrm{Why} \\ $$$$\mathrm{separation}\:=\:\mathrm{Initial}\:\mathrm{separation}\:− \\ $$$$\mathrm{relative}\:\mathrm{velocity}\:×\:\mathrm{time}? \\ $$