Question Number 167730 by Tawa11 last updated on 23/Mar/22
Commented by Tawa11 last updated on 24/Mar/22
$$\mathrm{Engineering}\:\mathrm{course}\:\mathrm{sir}.\:\mathrm{Computer}. \\ $$
Commented by ajfour last updated on 24/Mar/22
$${what}\:{course}\:{do}\:{u}\:{pursue}? \\ $$
Answered by malwan last updated on 24/Mar/22
$${The}\:{time}\:{is}\:\mathrm{6}.\mathrm{47}{s} \\ $$
Commented by Tawa11 last updated on 24/Mar/22
$$\mathrm{Please}\:\mathrm{workings}\:\mathrm{sir}. \\ $$
Commented by malwan last updated on 24/Mar/22
$${y}={u}\left({sin}\theta\right){t}−\left(\frac{\mathrm{1}}{\mathrm{2}}\right){gt}^{\mathrm{2}} \\ $$$$\Rightarrow−\mathrm{80}=\mathrm{40}\left({sin}\mathrm{30}\right){t}−\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{10}{t}^{\mathrm{2}} \\ $$$$\Rightarrow{t}^{\mathrm{2}} −\mathrm{4}{t}−\mathrm{16}=\mathrm{0} \\ $$$$\therefore\:{t}=\frac{+\:\mathrm{4}\:+\sqrt{\left(−\mathrm{4}\right)^{\mathrm{2}} −\mathrm{4}×\mathrm{1}×\left(−\mathrm{16}\right)}}{\mathrm{2}×\mathrm{1}} \\ $$$$=\left(\mathrm{2}+\mathrm{4}\sqrt{\mathrm{5}}\right){s}\:\approx\mathrm{6}.\mathrm{47}\:{s} \\ $$
Commented by Tawa11 last updated on 24/Mar/22
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$