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Astronauts on our moon must function with an acceleration due to gravity of 0.170g . If an astronaut can throw a certain wrench 12.0m vertically upward on earth, how high could he throw it on our moon if he gives it the same starting speed in both places? How much longer would it be in motion (going up and coming down) on the moon than on earth?

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u know the acceleration on earth and moon for a given distance say x calculate teh time of motion
2. t=Vo/A
Find h with this: \[v^2=v_0^2+2ah\] Since v=0 at highest point, you can find then: \[0=v_0^2+2ah \\h=\sqrt{\frac{-v_0^2}{2a}}\\\frac{h_{m}}{h_{e}}=\frac{\sqrt{\frac{-v_0^2}{2a_m}}}{\sqrt{\frac{-v_0^2}{2a_e}}} \]

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Other answers:

\[\frac{ 9.81 }{ 1.67 }12.0\] h=70.6, but how do u solve the second part?
\[v=v_0-gt\] again, \[v=0\] then you can find time traveled by wrench for each g's
i'm not sure how to get Vo
oh is it 12m/s?
You can find v0, but don't really need to. \[0=v_0-gt\\t=\frac{v_0}{g}\\\frac{t_e}{t_m}=\frac{v_0/g_e}{v_0/g_m} \]
but dont i need both Vo and t?
\[\frac{ Te }{ Tm } =\frac{ 1.67 }{ 9.8 }\]
Correct :)
so its 0.170?
Wait, i'm not native eng speaker. What do they mean when they say how much longer? If they mean \[\Delta t=t_m-t_e\] then we're wrong. What we found is \[t_m=0.170t_e\].
yeah Δt=tm−te

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