Question Number 226514 by Tawa11 last updated on 01/Dec/25
Answered by fantastic2 last updated on 02/Dec/25
$${E}_{{p}_{{box}} } ={mgh} \\ $$$${when}\:{box}\:{hits}\:{ground}\: \\ $$$${E}_{{k}} ={E}_{{p}} ={mgh} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} ={mgh} \\ $$$$\Rightarrow{v}=\sqrt{\mathrm{2}{gh}}=\sqrt{\mathrm{2}×\mathrm{9}.\mathrm{8}×\mathrm{30}}=\sqrt{\mathrm{2}×\mathrm{2}×\mathrm{49}×\mathrm{3}}=\mathrm{14}\sqrt{\mathrm{3}}{ms}^{−\mathrm{1}} \\ $$
Commented by Tawa11 last updated on 02/Dec/25
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$