Question Number 71053 by jagannath19 last updated on 11/Oct/19
Commented by jagannath19 last updated on 11/Oct/19
$${explain} \\ $$
Answered by mr W last updated on 11/Oct/19
$${let}\:{V}={speed}\:{on}\:{the}\:{ground} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{mV}^{\mathrm{2}} ={mgH}+\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} \\ $$$$\Rightarrow{V}=\sqrt{\mathrm{2}{gH}+{v}^{\mathrm{2}} } \\ $$$${V}\:{is}\:{independent}\:{from}\:{the}\:{direction} \\ $$$$\Rightarrow\left(\mathrm{4}\right)\:{is}\:{correct} \\ $$