Question Number 30331 by Tinkutara last updated on 20/Feb/18
Answered by rahul 19 last updated on 20/Feb/18
$$\left(\mathrm{3}\right)\:\mathrm{10}\:\mathrm{s}. \\ $$
Answered by ajfour last updated on 20/Feb/18
$${max}\:{range}\:=\:\frac{{u}^{\mathrm{2}} }{{g}}=\mathrm{0}.\mathrm{8}{m}\:\:\:{at}\:\:\theta=\mathrm{45}° \\ $$$$\Rightarrow\:\:\:\:\:{u}=\sqrt{\mathrm{8}}\:=\mathrm{2}\sqrt{\mathrm{2}}\:{m}/{s} \\ $$$${Time}\:{of}\:{flight}\:\:\:\:\:{T}=\frac{\mathrm{2}{u}\mathrm{sin}\:\mathrm{45}°}{{g}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{T}\:=\:\mathrm{0}.\mathrm{4}\:{s} \\ $$$${Total}\:{time}\:=\:\left(\mathrm{0}.\mathrm{4}{s}\right)×{no}.\:{of}\:{jumps} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{0}.\mathrm{4}×\frac{\mathrm{20}}{\mathrm{0}.\mathrm{8}}\:=\:\mathrm{10}{s}\:. \\ $$
Commented by Tinkutara last updated on 23/Feb/18
Thank you very much Sir! I got the answer.