Question Number 47344 by Necxx last updated on 08/Nov/18
$${On}\:{the}\:{moon}\:{the}\:{acceleration}\:{of} \\ $$$${free}\:{fall}\:{is}\:{only}\:{about}\:\mathrm{1}.\mathrm{6}{ms}^{−\mathrm{2}} . \\ $$$${About}\:{how}\:{long}\:{should}\:{a}\:{boy}\:{be} \\ $$$${able}\:{to}\:{throw}\:{a}\:{ball}\:{there}\:{if}\:{he}\:{can} \\ $$$${throw}\:{it}\:\mathrm{10}{m}\:{high}\:{on}\:{earth}? \\ $$$$\left({g}=\mathrm{10}{ms}^{−\mathrm{2}} \right) \\ $$
Answered by MrW3 last updated on 08/Nov/18
$${Energy}=\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} ={mg}_{{e}} {h}_{{e}} ={mg}_{{m}} {h}_{{m}} \\ $$$$\Rightarrow{h}_{{m}} =\frac{{g}_{{e}} }{{g}_{{m}} }{h}_{{e}} =\frac{\mathrm{10}}{\mathrm{1}.\mathrm{6}}×\mathrm{10}=\mathrm{62}.\mathrm{5}{m} \\ $$$${he}\:{can}\:{throw}\:{that}\:{ball}\:\mathrm{62}.\mathrm{5}{m}\:{high}\:{on} \\ $$$${the}\:{moon}. \\ $$
Commented by Necxx last updated on 09/Nov/18
$${thank}\:{you}\:{sir}. \\ $$$${incase}\:{of}\:{other}\:{time}\:{how}\:{am}\:{i}\:{to} \\ $$$${know}\:{the}\:{particle}\:{concept}\:{to}\:{have} \\ $$$${in}\:{mind}\:{like}\:{you}\:{used}\:{energy}\:{for} \\ $$$${this}. \\ $$