## darkwhale Group Title 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? one year ago one year ago

1. salini Group Title

u know the acceleration on earth and moon for a given distance say x calculate teh time of motion

2. darkwhale Group Title

2. t=Vo/A

3. imron07 Group Title

Find h with this: $v^2=v_0^2+2ah$ Since v=0 at highest point, you can find then: $0=v_0^2+2ah \\h=\sqrt{\frac{-v_0^2}{2a}}\\\frac{h_{m}}{h_{e}}=\frac{\sqrt{\frac{-v_0^2}{2a_m}}}{\sqrt{\frac{-v_0^2}{2a_e}}}$

4. darkwhale Group Title

$\frac{ 9.81 }{ 1.67 }12.0$ h=70.6, but how do u solve the second part?

5. imron07 Group Title

$v=v_0-gt$ again, $v=0$ then you can find time traveled by wrench for each g's

6. darkwhale Group Title

i'm not sure how to get Vo

7. darkwhale Group Title

oh is it 12m/s?

8. imron07 Group Title

You can find v0, but don't really need to. $0=v_0-gt\\t=\frac{v_0}{g}\\\frac{t_e}{t_m}=\frac{v_0/g_e}{v_0/g_m}$

9. darkwhale Group Title

but dont i need both Vo and t?

10. darkwhale Group Title

$\frac{ Te }{ Tm } =\frac{ 1.67 }{ 9.8 }$

11. imron07 Group Title

Correct :)

12. darkwhale Group Title

so its 0.170?

13. imron07 Group Title

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 $t_m=0.170t_e$.

14. darkwhale Group Title

yeah Δt=tm−te