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Question Number 47344 by Necxx last updated on 08/Nov/18

On the moon the acceleration of  free fall is only about 1.6ms^(−2) .  About how long should a boy be  able to throw a ball there if he can  throw it 10m high on earth?  (g=10ms^(−2) )

$${On}\:{the}\:{moon}\:{the}\:{acceleration}\:{of} \\ $$$${free}\:{fall}\:{is}\:{only}\:{about}\:\mathrm{1}.\mathrm{6}{ms}^{−\mathrm{2}} . \\ $$$${About}\:{how}\:{long}\:{should}\:{a}\:{boy}\:{be} \\ $$$${able}\:{to}\:{throw}\:{a}\:{ball}\:{there}\:{if}\:{he}\:{can} \\ $$$${throw}\:{it}\:\mathrm{10}{m}\:{high}\:{on}\:{earth}? \\ $$$$\left({g}=\mathrm{10}{ms}^{−\mathrm{2}} \right) \\ $$

Answered by MrW3 last updated on 08/Nov/18

Energy=(1/2)mv^2 =mg_e h_e =mg_m h_m   ⇒h_m =(g_e /g_m )h_e =((10)/(1.6))×10=62.5m  he can throw that ball 62.5m high on  the moon.

$${Energy}=\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} ={mg}_{{e}} {h}_{{e}} ={mg}_{{m}} {h}_{{m}} \\ $$$$\Rightarrow{h}_{{m}} =\frac{{g}_{{e}} }{{g}_{{m}} }{h}_{{e}} =\frac{\mathrm{10}}{\mathrm{1}.\mathrm{6}}×\mathrm{10}=\mathrm{62}.\mathrm{5}{m} \\ $$$${he}\:{can}\:{throw}\:{that}\:{ball}\:\mathrm{62}.\mathrm{5}{m}\:{high}\:{on} \\ $$$${the}\:{moon}. \\ $$

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.

$${thank}\:{you}\:{sir}. \\ $$$${incase}\:{of}\:{other}\:{time}\:{how}\:{am}\:{i}\:{to} \\ $$$${know}\:{the}\:{particle}\:{concept}\:{to}\:{have} \\ $$$${in}\:{mind}\:{like}\:{you}\:{used}\:{energy}\:{for} \\ $$$${this}. \\ $$

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