Question Number 172604 by Mikenice last updated on 29/Jun/22
Answered by mr W last updated on 29/Jun/22
$${v}_{\mathrm{1}} =\sqrt{\mathrm{2}{gh}_{\mathrm{1}} }=\sqrt{\mathrm{2}×\mathrm{9}.\mathrm{8}×\mathrm{4}.\mathrm{9}}=\mathrm{9}.\mathrm{8}\:{m}/{s} \\ $$$${t}_{\mathrm{1}} =\frac{{v}_{\mathrm{1}} }{{g}}=\frac{\mathrm{9}.\mathrm{8}}{\mathrm{9}.\mathrm{8}}=\mathrm{1}\:{s} \\ $$$${t}_{\mathrm{2}} =\mathrm{4}−\mathrm{1}=\mathrm{3}\:{s} \\ $$$${h}={v}_{\mathrm{1}} {t}_{\mathrm{2}} =\mathrm{9}.\mathrm{8}×\mathrm{3}=\mathrm{29}.\mathrm{4}\:{m} \\ $$
Commented by Tawa11 last updated on 30/Jun/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$