## iwanttogotostanford one year ago hhhlll

1. iwanttogotostanford

@YoungStudier @Nnesha @PrincessHush @pooja195

2. iwanttogotostanford

3. anonymous

@surjithayer

4. anonymous

$h=ut+\frac{ 1 }{ 2 }g t^2$ for each initial velocity u=0 m/s

5. anonymous

$g ~t^2=2h,g=\frac{ 2h }{ t^2 }$

6. anonymous

h is height t=time taken g=acceleration due to gravity

7. iwanttogotostanford

@surjithayer What is the acceleration due to gravity for the Earth, Moon, and Mars?

8. iwanttogotostanford

would t= 6.6 for the first empty slot?

9. anonymous

distance m=height=h Time T=t i solve first $g=\frac{ 2*2.25 }{ \left( 0.680 \right)^2 }=?$

10. iwanttogotostanford

9.7

11. anonymous

9.732

12. iwanttogotostanford

d = 1/2at2 is the equation for an object that starts at rest.

13. iwanttogotostanford

ok so- that is the one for the first box?

14. iwanttogotostanford

what would I do for the second box going horizontal?

15. anonymous

when the body falls from a certain height,initial velocity=0 m/s

16. iwanttogotostanford

So what how would I figure out acceleration due to gravity for the second box going across?

17. anonymous

h=3.08,t=0.793

18. iwanttogotostanford

so I would use that to figure out the second box?? @surjithayer

19. anonymous

first two values are for earth next two for Moon and last two for Mars.

20. iwanttogotostanford

so the very first box on the graph is 9.732 right???

21. iwanttogotostanford

@surjithayer

22. anonymous

correct

23. iwanttogotostanford

@surjithayer

24. iwanttogotostanford